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https://www.hsslive.co.in/2022/07/ap-board-class-8th-maths-chapter-10-direct-and-inverse-proportions-ex-10-4-textbook-solutions-pdf.html | 1,713,555,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00866.warc.gz | 739,433,756 | 45,946 | # HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## AP Board Class 8 Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Book Answers
AP Board Class 8 Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Book Answers
## Andhra Pradesh State Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Books Solutions
Board AP Board Materials Textbook Solutions/Guide Format DOC/PDF Class 8th Subject Maths Chapters Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Provider Hsslive
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## AP Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:
Question 1.
Rice costing ₹480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?
Solution:
Method – 1: Number of men and rice required to them are in inverse proportion.
Number of men ∝ 1 No. of days
⇒ Compound ratio of 8:12 and 20: 15
= 812=2015 = 89 …………….. (2)
From (1), (2)
480 : x = 8 : 9
⇒ 480𝑥=89
⇒ x = 480×98 = ₹540
∴ The cost of required rice is ₹ 540
Method – II :
𝑀1𝐷1𝑊1=𝑀2𝐷2𝑊2
M= No. of men
D1 = No .of days
W1 = Cost of rice
∴ M1 = 8
D1 = 20
W1 = ₹ 480
M2 = 12
D2 = 15
W2 = ? (x)
⇒ x = 45 x 12 = ₹ 540
The cost of required rice = ₹ 540/-
Question 2.
10 men can lay a road 75 km. long in 5 days. In how many days can 15 men lay a road 45 km. long?
Solution:
𝑀1𝐷1𝑊1=𝑀2𝐷2𝑊2
∴ M1 = 10
D1 = 5
W1 = 75
M2 = 15
D2 = ?
W2 = 45
∴ x = 2
∴ No. of days are required = 2
Question 3.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days can 20 men working at 9 hours per day do the same work?
Solution:
M1D1H1 = M2D2H2
∴ M1 = 24
D1 = 15 days
H1 = 8 hrs
M2 = 20
D2 = ?
H2 = 9 hrs
⇒ 24 × 15 × 8 = 20 × x × 9
∴ No. of days are required = 16
[ ∵ No. of men and working hours are in inverse]
Question 4.
175 men can dig a canal 3150 m long in 36 days. How many men are required to dig a canal 3900 m. long in 24 days?
Solution:
𝑀1𝐷1𝑊1=𝑀2𝐷2𝑊2
M1 = 175
D1 = 36
W1 = 3150
M2 = ?
D2 = 24
W2 = 3900
∴ No. of workers are required = 325
Question 5.
If 14 typists typing 6 hours a day can take 12 days to complete the manuscript of a book, then how many days will 4 typists, working 7 hours a day, can take to do the same job?
Solution:
M1D1H1 = M2D2H2
M1 = 14
D1 = 12 days
H1 = 6
M2 = 4
D2 = ?
H2 = 7
⇒ 14 × 12 × 6 = 4 × x × 7
⇒ x = 36
∴ No. of days are required = 36
[ ∵ No of men and working hours are in inverse proportion]
## Andhra Pradesh Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbooks for Exam Preparations
Andhra Pradesh Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 exam preparation. The AP Board STD 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Books State Board syllabus with maximum efficiency.
## FAQs Regarding Andhra Pradesh Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Solutions
#### Can we get a Andhra Pradesh State Board Book PDF for all Classes?
Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.
## Important Terms
Andhra Pradesh Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4, AP Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbooks, Andhra Pradesh State Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4, Andhra Pradesh State Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook solutions, AP Board Class 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbooks Solutions, Andhra Pradesh Board STD 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4, AP Board STD 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbooks, Andhra Pradesh State Board STD 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4, Andhra Pradesh State Board STD 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook solutions, AP Board STD 8th Maths Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbooks Solutions,
Share: | 1,753 | 5,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-18 | latest | en | 0.821513 |
http://en.m.wikibooks.org/wiki/Linear_Algebra/Topic:_Geometry_of_Linear_Maps | 1,409,727,919,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535925433.20/warc/CC-MAIN-20140901014525-00268-ip-10-180-136-8.ec2.internal.warc.gz | 181,449,356 | 15,145 | Linear Algebra/Topic: Geometry of Linear Maps
Linear Algebra
← Topic: Line of Best Fit Topic: Geometry of Linear Maps Topic: Markov Chains →
The pictures below contrast $f_1(x)=e^x$ and $f_2(x)=x^2$, which are nonlinear, with $h_1(x)=2x$ and $h_2(x)=-x$, which are linear. Each of the four pictures shows the domain $\mathbb{R}^1$ on the left mapped to the codomain $\mathbb{R}^1$ on the right. Arrows trace out where each map sends $x=0$, $x=1$, $x=2$, $x=-1$, and $x=-2$. Note how the nonlinear maps distort the domain in transforming it into the range. For instance, $f_1(1)$ is further from $f_1(2)$ than it is from $f_1(0)$ — the map is spreading the domain out unevenly so that an interval near $x=2$ is spread apart more than is an interval near $x=0$ when they are carried over to the range.
The linear maps are nicer, more regular, in that for each map all of the domain is spread by the same factor.
The only linear maps from $\mathbb{R}^1$ to $\mathbb{R}^1$ are multiplications by a scalar. In higher dimensions more can happen. For instance, this linear transformation of $\mathbb{R}^2$, rotates vectors counterclockwise, and is not just a scalar multiplication.
The transformation of $\mathbb{R}^3$ which projects vectors into the $xz$-plane is also not just a rescaling.
Nonetheless, even in higher dimensions the situation isn't too complicated.
Below, we use the standard bases to represent each linear map $h:\mathbb{R}^n\to \mathbb{R}^m$ by a matrix $H$. Recall that any $H$ can be factored $H=PBQ$, where $P$ and $Q$ are nonsingular and $B$ is a partial-identity matrix. Further, recall that nonsingular matrices factor into elementary matrices $PBQ=T_nT_{n-1}\cdots T_jBT_{j-1}\cdots T_1$, which are matrices that are obtained from the identity $I$ with one Gaussian step
$I\xrightarrow[]{k\rho_i}M_i(k) \qquad I\xrightarrow[]{\rho_i\leftrightarrow\rho_j}P_{i,j} \qquad I\xrightarrow[]{k\rho_i+\rho_j}C_{i,j}(k)$
($i\neq j$, $k\neq 0$). So if we understand the effect of a linear map described by a partial-identity matrix, and the effect of linear mapss described by the elementary matrices, then we will in some sense understand the effect of any linear map. (The pictures below stick to transformations of $\mathbb{R}^2$ for ease of drawing, but the statements hold for maps from any $\mathbb{R}^n$ to any $\mathbb{R}^m$.)
The geometric effect of the linear transformation represented by a partial-identity matrix is projection.
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \quad\xrightarrow{\begin{pmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &0 \end{pmatrix}_{\mathcal{E}_3,\mathcal{E}_3}}\quad \begin{pmatrix} x \\ y \\ 0 \end{pmatrix}$
For the $M_i(k)$ matrices, the geometric action of a transformation represented by such a matrix (with respect to the standard basis) is to stretch vectors by a factor of $k$ along the $i$-th axis. This map stretches by a factor of $3$ along the $x$-axis.
Note that if $0\leq k<1$ or if $k<0$ then the $i$-th component goes the other way; here, toward the left.
Either of these is a dilation.
The action of a transformation represented by a $P_{i,j}$ permutation matrix is to interchange the $i$-th and $j$-th axes; this is a particular kind of reflection.
In higher dimensions, permutations involving many axes can be decomposed into a combination of swaps of pairs of axes— see Problem 5.
The remaining case is that of matrices of the form $C_{i,j}(k)$. Recall that, for instance, that $C_{1,2}(2)$ performs $2\rho_1+\rho_2$.
$\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{\begin{pmatrix} 1 &0 \\ 2 &1 \end{pmatrix}_{\mathcal{E}_2,\mathcal{E}_2}} \qquad \begin{pmatrix} x \\ 2x+y \end{pmatrix}$
In the picture below, the vector $\vec{u}$ with the first component of $1$ is affected less than the vector $\vec{v}$ with the first component of $2$$h(\vec{u})$ is only $2$ higher than $\vec{u}$ while $h(\vec{v})$ is $4$ higher than $\vec{v}$.
Any vector with a first component of $1$ would be affected as is $\vec{u}$; it would be slid up by $2$. And any vector with a first component of $2$ would be slid up $4$, as was $\vec{v}$. That is, the transformation represented by $C_{i,j}(k)$ affects vectors depending on their $i$-th component.
Another way to see this same point is to consider the action of this map on the unit square. In the next picture, vectors with a first component of $0$, like the origin, are not pushed vertically at all but vectors with a positive first component are slid up. Here, all vectors with a first component of $1$— the entire right side of the square— is affected to the same extent. More generally, vectors on the same vertical line are slid up the same amount, namely, they are slid up by twice their first component. The resulting shape, a rhombus, has the same base and height as the square (and thus the same area) but the right angles are gone.
For contrast the next picture shows the effect of the map represented by $C_{2,1}(1)$. In this case, vectors are affected according to their second component. The vector $\binom{x}{y}$ is slid horozontally by twice $y$.
Because of this action, this kind of map is called a skew.
With that, we have covered the geometric effect of the four types of components in the expansion $H=T_nT_{n-1}\cdots T_jBT_{j-1}\cdots T_1$, the partial-identity projection $B$ and the elementary $T_i$'s. Since we understand its components, we in some sense understand the action of any $H$. As an illustration of this assertion, recall that under a linear map, the image of a subspace is a subspace and thus the linear transformation $h$ represented by $H$ maps lines through the origin to lines through the origin. (The dimension of the image space cannot be greater than the dimension of the domain space, so a line can't map onto, say, a plane.) We will extend that to show that any line, not just those through the origin, is mapped by $h$ to a line. The proof is simply that the partial-identity projection $B$ and the elementary $T_i$'s each turn a line input into a line output (verifying the four cases is Problem 6), and therefore their composition also preserves lines. Thus, by understanding its components we can understand arbitrary square matrices $H$, in the sense that we can prove things about them.
An understanding of the geometric effect of linear transformations on $\mathbb{R}^n$ is very important in mathematics. Here is a familiar application from calculus. On the left is a picture of the action of the nonlinear function $y(x)=x^2+x$. As at that start of this Topic, overall the geometric effect of this map is irregular in that at different domain points it has different effects (e.g., as the domain point $x$ goes from $2$ to $-2$, the associated range point $f(x)$ at first decreases, then pauses instantaneously, and then increases).
But in calculus we don't focus on the map overall, we focus instead on the local effect of the map.
At $x=1$ the derivative is $y^\prime(1)=3$, so that near $x=1$ we have $\Delta y\approx 3\cdot\Delta x$.
That is, in a neighborhood of $x=1$, in carrying the domain to the codomain this map causes it to grow by a factor of $3$ — it is, locally, approximately, a dilation.
The picture below shows a small interval in the domain $(x-\Delta x\,..\,x+\Delta x)$ carried over to an interval in the codomain $(y-\Delta y\,..\,y+\Delta y)$ that is three times as wide: $\Delta y \approx 3\cdot \Delta x$.
(When the above picture is drawn in the traditional cartesian way then the prior sentence about the rate of growth of $y(x)$ is usually stated: the derivative $y^\prime(1)=3$ gives the slope of the line tangent to the graph at the point $(1,2)$.)
In higher dimensions, the idea is the same but the approximation is not just the $\mathbb{R}^1$-to-$\mathbb{R}^1$ scalar multiplication case. Instead, for a function $y:\mathbb{R}^n\to \mathbb{R}^m$ and a point $\vec{x}\in\mathbb{R}^n$, the derivative is defined to be the linear map $h:\mathbb{R}^n\to \mathbb{R}^m$ best approximating how $y$ changes near $y(\vec{x})$. So the geometry studied above applies.
We will close this Topic by remarking how this point of view makes clear an often-misunderstood, but very important, result about derivatives: the derivative of the composition of two functions is computed by using the Chain Rule for combining their derivatives. Recall that (with suitable conditions on the two functions)
$\frac{d\,(g\circ f)}{dx}(x) = \frac{dg}{dx}(f(x))\cdot\frac{df}{dx}(x)$
so that, for instance, the derivative of $\sin(x^2+3x)$ is $\cos(x^2+3x)\cdot(2x+3)$. How does this combination arise? From this picture of the action of the composition.
The first map $f$ dilates the neighborhood of $x$ by a factor of
$\frac{df}{dx}(x)$
and the second map $g$ dilates some more, this time dilating a neighborhood of $f(x)$ by a factor of
$\frac{dg}{dx}(\,f(x)\,)$
and as a result, the composition dilates by the product of these two.
In higher dimensions the map expressing how a function changes near a point is a linear map, and is expressed as a matrix. (So we understand the basic geometry of higher-dimensional derivatives; they are compositions of dilations, interchanges of axes, shears, and a projection). And, the Chain Rule just multiplies the matrices.
Thus, the geometry of linear maps $h:\mathbb{R}^n\to \mathbb{R}^m$ is appealing both for its simplicity and for its usefulness.
ExercisesEdit
Problem 1
Let $h:\mathbb{R}^2\to \mathbb{R}^2$ be the transformation that rotates vectors clockwise by $\pi/4$ radians.
1. Find the matrix $H$ representing $h$ with respect to the standard bases. Use Gauss' method to reduce $H$ to the identity.
2. Translate the row reduction to to a matrix equation $T_jT_{j-1}\cdots T_1H=I$ (the prior item shows both that $H$ is similar to $I$, and that no column operations are needed to derive $I$ from $H$).
3. Solve this matrix equation for $H$.
4. Sketch the geometric effect matrix, that is, sketch how $H$ is expressed as a combination of dilations, flips, skews, and projections (the identity is a trivial projection).
Problem 2
What combination of dilations, flips, skews, and projections produces a rotation counterclockwise by $2\pi/3$ radians?
Problem 3
What combination of dilations, flips, skews, and projections produces the map $h:\mathbb{R}^3\to \mathbb{R}^3$ represented with respect to the standard bases by this matrix?
$\begin{pmatrix} 1 &2 &1 \\ 3 &6 &0 \\ 1 &2 &2 \end{pmatrix}$
Problem 4
Show that any linear transformation of $\mathbb{R}^1$ is the map that multiplies by a scalar $x\mapsto kx$.
Problem 5
Show that for any permutation (that is, reordering) $p$ of the numbers $1$, ..., $n$, the map
$\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} \mapsto \begin{pmatrix} x_{p(1)} \\ x_{p(2)} \\ \vdots \\ x_{p(n)} \end{pmatrix}$
can be accomplished with a composition of maps, each of which only swaps a single pair of coordinates. Hint: it can be done by induction on $n$. (Remark: in the fourth chapter we will show this and we will also show that the parity of the number of swaps used is determined by $p$. That is, although a particular permutation could be accomplished in two different ways with two different numbers of swaps, either both ways use an even number of swaps, or both use an odd number.)
Problem 6
Show that linear maps preserve the linear structures of a space.
1. Show that for any linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$, the image of any line is a line. The image may be a degenerate line, that is, a single point.
2. Show that the image of any linear surface is a linear surface. This generalizes the result that under a linear map the image of a subspace is a subspace.
3. Linear maps preserve other linear ideas. Show that linear maps preserve "betweeness": if the point $B$ is between $A$ and $C$ then the image of $B$ is between the image of $A$ and the image of $C$.
Problem 7
Use a picture like the one that appears in the discussion of the Chain Rule to answer: if a function $f:\mathbb{R}\to \mathbb{R}$ has an inverse, what's the relationship between how the function — locally, approximately — dilates space, and how its inverse dilates space (assuming, of course, that it has an inverse)?
Solutions
Linear Algebra
← Topic: Line of Best Fit Topic: Geometry of Linear Maps Topic: Markov Chains → | 3,392 | 12,334 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 152, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2014-35 | longest | en | 0.821348 |
https://www.worksheetscorner.com/multiplying-with-interactive-activities/ | 1,726,837,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00002.warc.gz | 985,737,813 | 14,611 | Date: May 14, 2023
# Multiplying with Interactive Activities Worksheet Test
## Part 1: Multiple Choice
Choose the correct answer from the options given.
1. What is the product of 7 and 8?
• A) 15
• B) 56
• C) 64
• D) 72
2. What is the product of 9 and 0?
• A) 0
• B) 9
• C) 1
• D) 90
3. What is the product of 3 and 5?
• A) 8
• B) 12
• C) 15
• D) 35
4. What is the product of 10 and 10?
• A) 100
• B) 20
• C) 50
• D) 1000
5. What is the product of 2 and 6?
• A) 12
• B) 8
• C) 14
• D) 26
## Part 2: Fill in the Blanks
Fill in the blank with the correct number.
1. 4 x 3 = ____
2. 9 x 2 = ____
3. 5 x 5 = ____
4. 7 x 1 = ____
5. 2 x 8 = ____
## Part 3: True or False
Write “True” if the statement is correct. Write “False” if the statement is incorrect.
1. When you multiply any number by 1, the answer is always that number.
2. The product of 0 and any number is always 0.
3. The order of the numbers being multiplied doesn’t matter.
4. The product of two even numbers is always even.
5. The product of two odd numbers is always odd.
## Part 4: Word Problems
Solve the following word problems.
1. A store has 6 packs of cookies with 9 cookies in each pack. How many cookies are there in total?
2. A farmer has 12 baskets of apples with 8 apples in each basket. How many apples does the farmer have in total?
3. A group of friends want to buy 4 pizzas. Each pizza has 10 slices. How many slices of pizza do they have in total?
4. If a pencil costs 7 cents, how much would 9 pencils cost?
5. A class has 25 students. Each student has 5 pencils. How many pencils does the class have in total?
## Part 5: Interactive Activities
1. Go to https://www.mathsisfun.com/numbers/multiplication-table.html and complete the multiplication table up to 12×12.
2. Go to https://www.mathplayground.com/ASB_Index.html and complete the “Multiplication Grand Prix” game. Take a screenshot of your final score and submit it with your completed worksheet.
Note: This worksheet is subject to review and modification by the teacher. | 630 | 2,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-38 | latest | en | 0.886684 |
http://discrete.prof.ninja/513/ | 1,571,145,867,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986659097.10/warc/CC-MAIN-20191015131723-20191015155223-00222.warc.gz | 52,427,712 | 2,407 | # May 13th
## Complexity Theory Basics
When analyzing algorithms we want to reason (quickly) about the running time of algorithms. We use the RAM model in which the following are considered to cost 1 CPU operation:
• Addition or subtraction of two numbers
• Multiplication of two numbers
• Comparison of two numbers
• A function call
• An if statement
• Any memory access
Other attributes:
• Memory is unlimited
• Recursive calls, for loops, and while loops all require analysis
The goal is to allow you to count operations and CPU cost without needing to know the specific architecture you are running on.
Some notes on algorithm analysis.
## Big-Oh Analysis
Big-oh, $$\mathcal{O}( \cdot )$$, has nothing to do with CPU cost, persay, it is just a way to fuzzily compare functions. We say that $$f(n) = \mathcal{O}(g(n))$$ if and only if there is some cutoff $$N$$ and positive real $$c$$ such that for all $$n \geq N$$ we have $$|f(n)| \leq c \cdot |g(n)|$$.
The official definition is there to get you out of trouble when you feel lost, but it seems way more complicated than it really is. It's really used to compare functions.
### Intro Problems
1. Prove, using the definition, that $$3n^2 = \mathcal{O}(n^2)$$.
2. Do the same to show that $$\frac{1}{5} n = \mathcal{O}(n^3)$$.
3. Show that $$2^n \neq \mathcal{O}(n^4)$$
### Properties of $$\mathcal{O}$$
For $$f, g, f_1, g_1$$ functions from $$\mathbb{N} \to \mathbb{R}$$ we know that:
1. If $$f = \mathcal{O}(g)$$ then $$f + g = \mathcal{O}(g)$$
2. If $$f = \mathcal{O}(f_1)$$ and $$g = \mathcal{O}(g_1)$$ then $$f \cdot g = \mathcal{O}(f_1 \cdot g_1)$$
3. $$c \cdot f = \mathcal{O}(f)$$ for any constant real $$c$$
Upshot is that you pretty much ignore small terms.
### Next batch
1. Prove that $$a_n x^n + \cdots + a_0 = \mathcal{O}(x^n)$$
2. For $$b \geq a$$ prove that $$n^a = \mathcal{O}(n^b)$$.
3. More problems to practice with here
### Putting it all together
So to use big-oh in algorithms we ballpark the cost of loops and multiplications ignoring specific details.
Sample algorithms to practice on: sorting!
This is insert sort done via Folk Dance!
or a gif:
1. What is the best input for this algorithm?
2. What is the worst input for this algorithm?
3. Analyze the number of comparisons in insertion sort for the worst-case on $$n$$ items. | 698 | 2,331 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-43 | latest | en | 0.830069 |
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Example Keywords: intel -slippers \$76
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# Hexadecimal ( Binary Arithmetic )
Rank: 100%
In and , hexadecimal (also base 16, or hex) is a positional with a , or base, of 16. It uses sixteen distinct symbols, most often the symbols "0"–"9" to represent values to nine, and "A"–"F" (or alternatively "a"–"f") to represent values ten to fifteen.
Hexadecimal numerals are widely used by computer system designers and programmers, as they provide a more human-friendly representation of values. Each hexadecimal digit represents four , also known as a , which is half a . For example, a single byte can have values ranging from 0000 0000 to 1111 1111 in binary form, which can be more conveniently represented as 00 to FF in hexadecimal.
In mathematics, a subscript is typically used to specify the . For example, the decimal value would be expressed in hexadecimal as . In programming, a number of notations are used to support hexadecimal representation, usually involving a prefix or suffix. The prefix 0x is used in C and related languages, which would denote this value by 0x{{hexadecimal|10995|no}}.
Hexadecimal is used in the transfer encoding Base16, in which each byte of the plaintext is broken into two 4-bit values and represented by two hexadecimal digits.
Representation
Written representation
Using 0–9 and A–F
In contexts where the is not clear, hexadecimal numbers can be ambiguous and confused with numbers expressed in other bases. There are several conventions for expressing values unambiguously. A numerical subscript (itself written in decimal) can give the base explicitly: 15910 is decimal 159; 15916 is hexadecimal 159, which is equal to 34510. Some authors prefer a text subscript, such as 159decimal and 159hex, or 159d and 159h.
In linear text systems, such as those used in most computer programming environments, a variety of methods have arisen:
• In (including ), character codes are written as hexadecimal pairs prefixed with %: <nowiki></nowiki> where %20 is the space (blank) character, code point 20 in hex, 32 in decimal.
• In and , characters can be expressed as hexadecimal numeric character references using the notation &#x''code'';, where the x denotes that code is a hex code point (of 1- to 6-digits) assigned to the character in the standard. Thus &#x2019; represents the right single quotation mark (’), Unicode code point number 2019 in hex, 8217 (thus &#8217; in decimal).
• In the standard, a character value is represented with U+ followed by the hex value, e.g. U+20AC is the (€).
• in HTML, CSS and X Window can be expressed with six hexadecimal digits (two each for the red, green and blue components, in that order) prefixed with #: white, for example, is represented as #FFFFFF. CSS also allows 3-hexdigit abbreviations with one hexdigit per component: #FA3 abbreviates #FFAA33 (a golden orange: ).
• Unix (and related) shells, AT&T assembly language and likewise the C programming language (and its syntactic descendants such as C++, C#, D, Java, , Python and Windows PowerShell) use the prefix 0x for numeric constants represented in hex: 0x5A3. Character and string constants may express character codes in hexadecimal with the prefix \x followed by two hex digits: '\x1B' represents the control character; "\x1B[0m\x1B[25;1H" is a string containing 11 characters (plus a trailing NUL to mark the end of the string) with two embedded Esc characters.The string "\x1B[0m\x1B[25;1H" specifies the character sequence Esc [ 0 m Esc [ 2 5 ; 1 H Nul. These are the escape sequences used on an ANSI terminal that reset the character set and color, and then move the cursor to line 25. To output an integer as hexadecimal with the function family, the format conversion code %X or %x is used.
• In (e-mail extensions) encoding, characters that cannot be represented as literal characters are represented by their codes as two hexadecimal digits (in ASCII) prefixed by an equal to sign =, as in Espa=F1a to send "España" (Spain). (Hexadecimal F1, equal to decimal 241, is the code number for the lower case n with tilde in the ISO/IEC 8859-1 character set.)
• In Intel-derived assembly languages and Modula-2, hexadecimal is denoted with a suffixed H or h: FFh or 05A3H. Some implementations require a leading zero when the first hexadecimal digit character is not a decimal digit, so one would write 0FFh instead of FFh
• Other assembly languages (6502, ), Pascal, , some versions of (), , Godot and Forth use \$ as a prefix: \$5A3.
• Some assembly languages (Microchip) use the notation H'ABCD' (for ABCD16). Similarly, Fortran 95 uses Z'ABCD'.
• Ada and enclose hexadecimal numerals in based "numeric quotes": 16#5A3#. For bit vector constants uses the notation x"5A3".The VHDL MINI-REFERENCE: VHDL IDENTIFIERS, NUMBERS, STRINGS, AND EXPRESSIONS
• represents hexadecimal constants in the form 8'hFF, where 8 is the number of bits in the value and FF is the hexadecimal constant.
• The language uses the prefix 16r: 16r5A3
• and the and its derivatives denote hex with prefix 16#: 16#5A3. For PostScript, binary data (such as image ) can be expressed as unprefixed consecutive hexadecimal pairs: AA213FD51B3801043FBC...
• uses the prefixes #x and #16r. Setting the variables *read-base* and *print-base* to 16 can also be used to switch the reader and printer of a Common Lisp system to Hexadecimal number representation for reading and printing numbers. Thus Hexadecimal numbers can be represented without the #x or #16r prefix code, when the input or output base has been changed to 16.
• , MSX is Coming — Part 2: Inside MSX Compute!, issue 56, January 1985, p. 52 , and prefix hexadecimal numbers with &H: &H5A3
• and use & for hex.BBC BASIC programs are not fully portable to (without modification) since the latter takes & to prefix values. (Microsoft BASIC primarily uses &O to prefix octal, and it uses &H to prefix hexadecimal, but the ampersand alone yields a default interpretation as an octal prefix.
• TI-89 and 92 series uses a 0h prefix: 0h5A3
• ALGOL 68 uses the prefix 16r to denote hexadecimal numbers: 16r5a3. Binary, quaternary (base-4) and octal numbers can be specified similarly.
• The most common format for hexadecimal on IBM mainframes () and midrange computers (IBM System i) running the traditional OS's (zOS, zVSE, zVM, TPF, ) is X'5A3', and is used in Assembler, PL/I, , JCL, scripts, commands and other places. This format was common on other (and now obsolete) IBM systems as well. Occasionally quotation marks were used instead of apostrophes.
• introduced the use of a particular typeface to represent a particular radix in his book The TeXbook.Donald E. Knuth. The TeXbook (Computers and Typesetting, Volume A). Reading, Massachusetts: Addison–Wesley, 1984. . The source code of the book in TeX (and a required set of macros ftp://tug.ctan.org/pub/tex-archive/systems/knuth/lib/manmac.tex) is available online on . Hexadecimal representations are written there in a typewriter typeface: 5A3
• Any IPv6 address can be written as eight groups of four hexadecimal digits (sometimes called hextets), where each group is separated by a colon (:). This, for example, is a valid IPv6 address: 2001:0db8:85a3:0000:0000:8a2e:0370:7334; this can be abbreviated as 2001:db8:85a3::8a2e:370:7334. By contrast, IPv4 addresses are usually written in decimal.
• Globally unique identifiers are written as thirty-two hexadecimal digits, often in unequal hyphen-separated groupings, for example {3F2504E0-4F89-41D3-9A0C-0305E82C3301}.
There is no universal convention to use lowercase or uppercase for the letter digits, and each is prevalent or preferred in particular environments by community standards or convention.
History of written representations
The use of the letters A through F to represent the digits above 9 was not universal in the early history of computers.
• During the 1950s, some installations favored using the digits 0 through 5 with an to denote the values 10–15 as , , , , and .
• The SWAC (1950) and Bendix G-15 (1956) computers used the lowercase letters u, v, w, x, y and z for the values 10 to 15.
• The (1952) computer used the uppercase letters K, S, N, J, F and L for the values 10 to 15.
• The Librascope LGP-30 (1956) used the letters F, G, J, K, Q and W for the values 10 to 15.
• The Datamatic D-1000 (1957) used the lowercase letters b, c, d, e, f, and g whereas the 100 (1967) used the uppercase letters B, C, D, E, F and G for the values 10 to 15.
• The (1960) used the letters S, T, U, V, W and X for the values 10 to 15.
• The computer (1960) used the letters D, G, H, J, K (and possibly V) for values 10–15.
• The Pacific Data Systems 1020 (1964) used the letters L, C, A, S, M and D for the values 10 to 15.
• New numeric symbols and names were introduced in the notation by in 1968. This notation did not become very popular.
• Bruce Alan Martin of Brookhaven National Laboratory considered the choice of A–F "ridiculous". In a 1968 letter to the editor of the CACM, he proposed an entirely new set of symbols based on the bit locations, which did not gain much acceptance.
• Soviet programmable calculators Б3-34 (1980) and similar used the symbols "−", "L", "C", "Г", "E", " " (space) for the values 10 to 15 on their displays.
• Seven-segment display decoder chips used various schemes for outputting values above nine. The Texas Instruments 7446/7447/7448/7449 and 74246/74247/74248/74249 use truncated versions of "2", "3", "4", "5" and "6" for the values 10 to 14. Value 15 (1111 binary) was blank.
Verbal and digital representations
There are no traditional numerals to represent the quantities from ten to fifteen – letters are used as a substitute – and most languages lack non-decimal names for the numerals above ten. Even though English has names for several non-decimal powers ( pair for the first binary power, score for the first power, , gross and for the first three powers), no English name describes the hexadecimal powers (decimal 16, 256, 4096, 65536, ... ). Some people read hexadecimal numbers digit by digit like a phone number, or using the NATO phonetic alphabet, the Joint Army/Navy Phonetic Alphabet, or a similar ad hoc system.
Systems of counting on digits have been devised for both binary and hexadecimal. Arthur C. Clarke suggested using each finger as an on/off bit, allowing finger counting from zero to 102310 on ten fingers.
(2019). 9780007289981, Ballantine.
Another system for counting up to FF16 (25510) is illustrated on the right.
Signs
The hexadecimal system can express negative numbers the same way as in decimal: −2A to represent −4210 and so on.
Hexadecimal can also be used to express the exact bit patterns used in the processor, so a sequence of hexadecimal digits may represent a or even a value. This way, the negative number −4210 can be written as FFFF FFD6 in a 32-bit CPU register (in two's-complement), as C228 0000 in a 32-bit FPU register or C045 0000 0000 0000 in a 64-bit FPU register (in the IEEE floating-point standard).
Just as decimal numbers can be represented in exponential notation, so too can hexadecimal numbers. By convention, the letter P (or p, for "power") represents times two raised to the power of, whereas E (or e) serves a similar purpose in decimal as part of the . The number after the P is decimal and represents the binary exponent.
Usually the number is normalised so that the leading hexadecimal digit is 1 (unless the value is exactly 0).
Example: 1.3DEp42 represents .
Hexadecimal exponential notation is required by the IEEE 754-2008 binary floating-point standard. This notation can be used for floating-point literals in the C99 edition of the C programming language. Using the %a or %A conversion specifiers, this notation can be produced by implementations of the family of functions following the C99 specification and Single Unix Specification (IEEE Std 1003.1) standard.
Conversion
Binary conversion
Most computers manipulate binary data, but it is difficult for humans to work with the large number of digits for even a relatively small binary number. Although most humans are familiar with the base 10 system, it is much easier to map binary to hexadecimal than to decimal because each hexadecimal digit maps to a whole number of bits (410). This example converts 11112 to base ten. Since each position in a binary numeral can contain either a 1 or a 0, its value may be easily determined by its position from the right:
• 00012 = 110
• 00102 = 210
• 01002 = 410
• 10002 = 810
Therefore:
= 810 + 410 + 210 + 110
= 1510
With little practice, mapping 11112 to F16 in one step becomes easy: see table in written representation. The advantage of using hexadecimal rather than decimal increases rapidly with the size of the number. When the number becomes large, conversion to decimal is very tedious. However, when mapping to hexadecimal, it is trivial to regard the binary string as 4-digit groups and map each to a single hexadecimal digit.
This example shows the conversion of a binary number to decimal, mapping each digit to the decimal value, and adding the results.
= 26214410 + 6553610 + 3276810 + 1638410 + 819210 + 204810 + 51210 + 25610 + 6410 + 1610 + 210 = 38792210
Compare this to the conversion to hexadecimal, where each group of four digits can be considered independently, and converted directly:
00102
216
5EB5216
The conversion from hexadecimal to binary is equally direct.
Other simple conversions
Although quaternary (base 4) is little used, it can easily be converted to and from hexadecimal or binary. Each hexadecimal digit corresponds to a pair of quaternary digits and each quaternary digit corresponds to a pair of binary digits. In the above example 5 E B 5 216 = 11 32 23 11 024.
The (base 8) system can also be converted with relative ease, although not quite as trivially as with bases 2 and 4. Each octal digit corresponds to three binary digits, rather than four. Therefore we can convert between octal and hexadecimal via an intermediate conversion to binary followed by regrouping the binary digits in groups of either three or four.
Division-remainder in source base
As with all bases there is a simple for converting a representation of a number to hexadecimal by doing integer division and remainder operations in the source base. In theory, this is possible from any base, but for most humans only decimal and for most computers only binary (which can be converted by far more efficient methods) can be easily handled with this method.
Let d be the number to represent in hexadecimal, and the series hihi−1...h2h1 be the hexadecimal digits representing the number.
1. i ← 1
2. hi ← d mod 16
3. d ← (d − hi) / 16
4. If d = 0 (return series hi) else increment i and go to step 2
"16" may be replaced with any other base that may be desired.
The following is a implementation of the above algorithm for converting any number to a hexadecimal in String representation. Its purpose is to illustrate the above algorithm. To work with data seriously, however, it is much more advisable to work with bitwise operators.
function toHex(d) {
``` var r = d % 16;
if (d - r == 0) {
}
```
}
function toChar(n) {
``` const alpha = "0123456789ABCDEF";
return alpha.charAt(n);
```
}
It is also possible to make the conversion by assigning each place in the source base the hexadecimal representation of its place value — before carrying out multiplication and addition to get the final representation. For example, to convert the number B3AD to decimal, one can split the hexadecimal number into its digits: B (1110), 3 (310), A (1010) and D (1310), and then get the final result by multiplying each decimal representation by 16 p ( p being the corresponding hex digit position, counting from right to left, beginning with 0). In this case, we have that:
which is 45997 in base 10.
Tools for conversion
Most modern computer systems with graphical user interfaces provide a built-in calculator utility capable of performing conversions between the various radices, and in most cases would include the hexadecimal as well.
In Microsoft Windows, the Calculator utility can be set to Scientific mode (called Programmer mode in some versions), which allows conversions between radix 16 (hexadecimal), 10 (decimal), 8 () and 2 (binary), the bases most commonly used by programmers. In Scientific Mode, the on-screen includes the hexadecimal digits A through F, which are active when "Hex" is selected. In hex mode, however, the Windows Calculator supports only integers.
Elementary arithmetic
Elementary operations such additions, subtractions, multiplications and divisions can be carried out indirectly through conversion to an alternate , such as the decimal system, since it's the most commonly adopted system, or the binary system, since each hex digit corresponds to four binary digits,
Alternatively, one can also perform elementary operations directly within the hex system itself — by relying on its addition/multiplication tables and its corresponding standard algorithms such as and the traditional subtraction algorithm.
Real numbers
Rational numbers
As with other numeral systems, the hexadecimal system can be used to represent , although repeating expansions are common since sixteen (1016) has only a single prime factor; two.
For any base, 0.1 (or "1/10") is always equivalent to one divided by the representation of that base value in its own number system. Thus, whether dividing one by two for binary or dividing one by sixteen for hexadecimal, both of these fractions are written as 0.1. Because the radix 16 is a (42), fractions expressed in hexadecimal have an odd period much more often than decimal ones, and there are no (other than trivial single digits). Recurring digits are exhibited when the denominator in lowest terms has a not found in the radix; thus, when using hexadecimal notation, all fractions with denominators that are not a power of two result in an infinite string of recurring digits (such as thirds and fifths). This makes hexadecimal (and binary) less convenient than for representing rational numbers since a larger proportion lie outside its range of finite representation.
All rational numbers finitely representable in hexadecimal are also finitely representable in decimal, and : that is, any hexadecimal number with a finite number of digits also has a finite number of digits when expressed in those other bases. Conversely, only a fraction of those finitely representable in the latter bases are finitely representable in hexadecimal. For example, decimal 0.1 corresponds to the infinite recurring representation 0.1 in hexadecimal. However, hexadecimal is more efficient than duodecimal and sexagesimal for representing fractions with powers of two in the denominator. For example, 0.062510 (one sixteenth) is equivalent to 0.116, 0.0912, and 0;3,4560.
21/2 0.50.8 1/2
31/3 0.3333... = 0.0.5555... = 0. 1/3
41/4 0.250.4 1/4
51/5 0.20. 1/5
61/6,0.10.2,1/6
71/770.0.71/7
81/8 0.1250.2 1/8
91/9 0.0. 1/9
101/10,0.10.1,1/A
111/11 0.0.B1/B
121/12,0.080.1,1/C
131/13130.0.D1/D
141/14, 70.00.1, 71/E
151/15,0.00.,1/F
161/16 0.06250.1 1/10
171/17170.0. 1/11
181/18,0.00.0,1/12
191/19190.0.131/13
201/20,0.050.0,1/14
211/21, 70.0., 71/15
221/22,0.00.0, B1/16
231/23230.0.171/17
241/24,0.0410.0,1/18
251/25 0.040. 1/19
261/26, 130.00.0, D1/1A
271/27 0.0. 1/1B
281/28, 70.030.0, 71/1C
291/29290.0.1D1/1D
301/30, ,0.00.0, ,1/1E
311/31310.0.1F1/1F
321/32 0.031250.08 1/20
331/33,0.0., B1/21
341/34, 170.00.0,1/22
351/35, 70.00., 71/23
361/36,0.020.0,1/24
Irrational numbers
The table below gives the expansions of some common irrational numbers in decimal and hexadecimal.
...1.6A09E667F3BCD...
...1.BB67AE8584CAA...
...2.3C6EF372FE95...
...1.9E3779B97F4A...
...
3.243F6A8885A308D313198A2E0
3707344A4093822299F31D008...
...2.B7E151628AED2A6B...
...0.6996 9669 9669 6996...
...0.93C467E37DB0C7A4D1B...
Powers
Powers of two have very simple expansions in hexadecimal. The first sixteen powers of two are shown below.
1
2
4
8
16dec
32dec
64dec
128dec
256dec
512dec
1024dec
2048dec
4096dec
8192dec
16,384dec
32,768dec
65,536dec
Cultural
Etymology
Use in Chinese culture
The traditional Chinese units of measurement were base-16. For example, one jīn (斤) in the old system equals sixteen . The (Chinese ) can be used to perform hexadecimal calculations such as additions and subtractions.
Primary numeral system
As with the system, there have been occasional attempts to promote hexadecimal as the preferred numeral system. These attempts often propose specific pronunciation and symbols for the individual numerals. Some proposals unify standard measures so that they are multiples of 16.
An example of unified standard measures is , which subdivides a day by 16 so that there are 16 "hexhours" in a day.
Base16 (Transfer encoding)
Base16 (as a proper name without a space) can also refer to a binary to text encoding belonging to the same family as Base32, Base58, and Base64.
In this case, data is broken into 4-bit sequences, and each value (between 0 and 15 inclusively) is encoded using 16 symbols from the character set. Although any 16 symbols from the ASCII character set can be used, in practice the ASCII digits '0'-'9' and the letters 'A'-'F' (or the lowercase 'a'-'f') are always chosen in order to align with standard written notation for hexadecimal numbers.
There are several advantages of Base16 encoding:
• Being exactly half a byte, 4-bits is easier to process than the 5 or 6 bits of Base32 and Base64 respectively
• The symbols 0-9 and A-F are universal in hexadecimal notation, so it is easily understood at a glance without needing to rely on a symbol lookup table
• Many CPU architectures have dedicated instructions that allow access to a half-byte (otherwise known as a ""), making it more efficient in hardware than Base32 and Base64
The main disadvantages of Base16 encoding are:
• Space efficiency is only 50%, since each 4-bit value from the original data will be encoded as an 8-bit byte. In contrast, Base32 and Base64 encodings have a space efficiency of 63% and 75% respectively.
• Possible added complexity of having to accept both uppercase and lowercase letters
Support for Base16 encoding is ubiquitous in modern computing. It is the basis for the W3C standard for , where a character is replaced with a percent sign "%" and its Base16-encoded form. Most modern programming languages directly include support for formatting and parsing Base16-encoded numbers.
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1s Time | 5,941 | 22,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-30 | latest | en | 0.884939 |
http://hi.gher.space/forum/viewtopic.php?f=25&p=29304 | 1,722,694,517,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00457.warc.gz | 10,071,106 | 11,528 | ## Quickfur's renders
Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.
### Re: Quickfur's renders
Here's the next 4D Catalan, as promised:
This is what the Polytope Wiki calls the Triangular Antitegmatic Hexacontatetrachoron, the dual of the runcinated tesseract. It's obviously the direct analogue of the deltoidal icositetrahedron, the dual of the rhombicuboctahedron.
Interestingly enough, the projection envelope of this 4D baby is exactly the deltoidal icositetrahedron, having exactly the same proportions. This leads to a conjecture that all duals of x4o...o3x have this projective relationship with each other. I haven't worked out the proof yet, nor have I checked the 5D analogue yet, but I'm fairly confident this is the case. I find this remarkable, because most of the 4D Catalans (that I've checked so far) don't have such direct proportions with their 3D analogues. It seems that x4o...o3x is one of those cases where there's a direct analogy across dimensions -- if this conjecture is true. It's also remarkable that this polychoron has 3 different edge lengths, whereas the deltoidal icositetrahedron has only 2 different edge lengths, yet in the projection the edge length proportions line up nicely. So this projective relationship isn't as trivial as it may seem. In any case, it's an interesting observation.
quickfur
Pentonian
Posts: 2984
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
that's right. with the runcinate xP..Qx, the dual is an antigum along the line between the vertices of xP...Qo and its dual. so the antitegum is based on what is the ...
I use the Conway term strombiate.
The dream you dream alone is only a dream
the dream we dream together is reality.
\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
wendy
Pentonian
Posts: 2025
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Location: Brisbane, Australia
### Re: Quickfur's renders
Next Catalan?
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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### Re: Quickfur's renders
Oops, I actually had it ready 2 weeks ago, but got busy and totally forgot to post it on the website. I'll get around to it sometime today, hopefully.
quickfur
Pentonian
Posts: 2984
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
OK, got it set up faster than expected. Here it is:
This is the square antitegmatic hecatontetracontatetrachoron (what a mouthful!), the dual of the runcinated 24-cell. This time round I actually have two series of structural renders, one the traditional layer-by-layer cell breakdown, and a bonus one showing the Hopf fibration structure of a subset of the cells. Very interesting, check it out!
quickfur
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Posts: 2984
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
This month's POM (polytope of the month) is the dual of the uniform pentagonal prism, the (non-Johnson) pentagonal bipyramid:
Nothing special in itself, but it does feature as a cell in a 4D Catalan which will be posted next month. So stay tuned!
quickfur
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Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
That one then ought be the dual of the rectified 600-cell, cf. https://bendwavy.org/klitzing/incmats/o3m3o5o.htm, which also is known as "pentagonal-bipyramidal heptacosiicosachoron".
In fact yourself provided a first render in here already, cf. http://hi.gher.space/forum/viewtopic.php?p=16141#p16141, however the pic itself seems lost since (but then was saved and still is shown on my page, hehe).
--- rk
Klitzing
Pentonian
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### Re: Quickfur's renders
Very nice! So you already know what's coming.
quickfur
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Location: The Great White North
### Re: Quickfur's renders
And here it is:
This is the joined 120-cell, sporting 720 pentagonal bipyramids. It is the dual of the rectified 600-cell, one of the rare semiregular polychora.
quickfur
Pentonian
Posts: 2984
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
Hmm, apparently I neglected to post this month's POM:
This is, of course, the famous rhombic triacontahedron, which is one of the projection images of the castellated rhombicuboctahedral prism, the first CRF we discovered that sported bilunabirotunda cells in a non-trivial manner. The order-5 vertices correspond to the apices of the castellated prism's pentagonal pyramid cells, and the order-3 vertices correspond with the tetrahedral cells.
quickfur
Pentonian
Posts: 2984
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
Thank you for going to the effort to continue to post these after all these years!
I'm hoping to spend a little more time reading up on these topics again. What would you say were the most interesting discoveries of the last 5 years or so?
Keiji
Posts: 1985
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England
### Re: Quickfur's renders
Hey Keiji, good to have you active here again, it's been a while!
Sad to say, I have hardly been keeping up with discoveries lately. Mostly just doing what I can to keep my website updating in a semi-regular fashion (has been better since last year when I made a large enough buffer of Catalan solid entries that I could continue posting monthly in spite of often having no time at all to do anything 4D or 3D related). I'm dreading the day my buffer runs out, then I might have to go on hiatus again.
//
Anyway, since I'm at it, might as well include the latest entry on my website, that I haven't gotten around to posting here until now:
This is the so-called "bidecachoron", the dual of the bitruncated 5-cell. (Not perfectly happy with the name, I got the name from the Polytope Wiki, apparently coined by Bowers. But for lack of a better name, it will have to suffice for now.) Interestingly enough, the projection envelope is a cube, and the layout of cells in it embodies the two ways one could inscribe a tetrahedron in a cube. I'm sure there's a rational reason behind it all, of course, but it's just an interesting coincidence that struck me when I was doing these renders.
quickfur
Pentonian
Posts: 2984
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
Just to provide some context: That name was chosen more out of analogy with a different family than the Catalans - the family of convex vertex-transitive polychora formed from the convex hull of two inversely oriented 5-cell truncates, so you have names like the bitruncatodecachoon rfo the hull of two opposite truncated 5-cells, etc. As you can tell from the coordinates listed on your site, the dual of the bitruncated 5-cell is just the convex hull of the compound of two dual 5-cells.
A full investigation into these (along with the 24-cell family analogs of them) was done by those of us on the Discord/wiki a few years back. Interestingly, for some there are several related forms based on taking the hull of two similar polychora, but with the edge lengths varied - any Wythoffian uniform polychoron with more than one ringed node in the CD has infinite non-uniform (but still isogonal) variations formed by adjusting edge lengths, and in some cases (such as that of the cantellated 5-cell) taking the hull of two of these variants will give results with different types of cells, rather than just similar edge-length variations. There are in fact a total of 24 of these isogonal polychora with doubled 5-cell symmetry and up to 240 vertices, and each has a corresponding one with doubled 24-cell symmetry. The 5-cell ones actually have a page on the Bowers polychoron site now if you are at all interested - he also gets into even stranger chiral varations, because you can do this with variations of the (non-uniform but still isogonal) snub 5-cell as well.
Trionian
Posts: 128
Joined: Sat Mar 18, 2017 1:42 pm
### Re: Quickfur's renders
I've noticed that even among the 3D Catalans, the coordinates are essentially a union of the coordinates of 2 or more of xNo3o, oNx3o, and oNo3x, variously scaled. I.e. the Catalans are the convex hull of various compounds of the regular and quasi-regular polyhedra of their respective families.
Apparently in 4D this is also the case. I surmise it's true in all dimensions.
quickfur
Pentonian
Posts: 2984
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Location: The Great White North
### Re: Quickfur's renders
quickfur wrote:I've noticed that even among the 3D Catalans, the coordinates are essentially a union of the coordinates of 2 or more of xNo3o, oNx3o, and oNo3x, variously scaled. I.e. the Catalans are the convex hull of various compounds of the regular and quasi-regular polyhedra of their respective families.
Apparently in 4D this is also the case. I surmise it's true in all dimensions.
Correct, at least if the original polytope is Wythoffian. And it is easy to figure out which specific vertex sets are involved, based on the facets of the original. As you know the facets of a Wythoffian uniform polytope can be fonud from the CD by removing one of the nodes and all adjacent edges. However, sometimes this will yield a lower-dimensional element instead - take x4x3x3o, where removing the third node gives x4x o, representing only an octagonal face between two great rhombicuboctahedra. Its dual then does not use the o4o3o3o rectate with that node (that is o4o3x3o) but does use the others.
For non-Wythoffian cases of course things get more complicated, so this will not work for stuff like the snub 24-cell if you get around to its dual.
Trionian
Posts: 128
Joined: Sat Mar 18, 2017 1:42 pm
### Re: Quickfur's renders
Forgot to post this here, this is the latest 4D Catalan:
This is the joined 16-cell, the dual of the rectified tesseract. Interestingly enough, the projection envelope is a regular octahedron.
quickfur
Pentonian
Posts: 2984
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
Hmm, looks like I forgot to post last month's polytope of the month:
And yesterday I just posted this month's polytope:
(Both images are linked to the respective pages.)
quickfur
Pentonian
Posts: 2984
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Quickfur's renders
Looks like again I forgot to post last month's POM:
This is the dual of the rhombicosidodecahedron, as yall probably already know.
//
Meanwhile, this month's POM is already up:
This looks like just the vertex-first projection of the tesseract... but actually, it's the triangular antitegmatic icosachoron, the dual of the runcinated 5-cell. It's analogous to the rhombic dodecahedron (as the dual of the face-expanded tetrahedron aka cuboctahedron), which also projects to a hexagon with 3 rhombus faces that looks like a vertex-first projection of a cube. Both are n-cube lookalikes in this angle, although of course, from other POVs their difference becomes obvious.
quickfur
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Previous | 3,244 | 11,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.912871 |
http://bredow.me/?page_id=79 | 1,621,388,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00418.warc.gz | 10,689,710 | 6,728 | # Counting and Summing up
### COUNTING FORMULA EXAMPLES
• =COUNTIF(Region,”North”)
• Counts the number of rows in which the value in Region is “North
• =COUNTIF(AvgCost,3400
• Counts the number of rows in which AvgCost equals 400
• =COUNTIF(Sales,”>500”)
• Counts the number of rows in which Sales greater than 500
• =COUNTIF(Sales,”<>100”)
• Counts the number of rows in which Sales <> 100
• =COUNTIF(Area,”?????”)
• Counts the number of rows in which Area contains five letters
• =COUNTIF(Region,”*h*”)
• Counts the number of rows in which Region contains the letter H (not case sensitive)
• =COUNTIFS(Month,”Jan”,Sales,”>200”)
• Counts the number of rows in which Month = “Jan” and Sales > 200
• {=SUM((Month=”Jan”)*(Sales>200))}
• An array formula that counts the number of rows in which Month = “Jan” and Sales > 200
• =COUNTIFS(Month,”Jan”,Region,”North”)
• Counts the number of rows in which Month = “Jan” and Region = “North”
• {=SUM((Month=”Jan”)*(Region=”North”))}
• An array formula that counts the number of rows in which Month = “Jan” and Region = “North”
• =COUNTIFS(Month,”Jan”,Region,”North”)+COUNTIFS(Month,”Jan”,Region,”South”)
• Counts the number of rows in which Month = “Jan” and Region = “North” or “South”
• {=SUM((Month=”Jan”)*((Region=”North”)+(Region=”South”)))}
• An array formula that counts the number of rows in which Month = “Jan” andRegion = “North” or “South”
• =COUNTIFS(Sales,”>=300”,Sales,”<=400”)
• Counts the number of rows in which Sales is between 300 and 400
• {=SUM((Sales>=300)*(Sales<=400))}
• An array formula that counts the number of rows in which Sales is between 300 and 400
### SUMMING FORMULA EXAMPLES
• =SUMIF(Sales,”>200”)
• Sum of all Sales over 200
• =SUMIF(Month,”Jan”,Sales)
• Sum of Sales in which Month = “Jan”
• =SUMIF(Month,”Jan”,Sales)+SUMIF(Month,”Feb”,Sales) or “Feb”
• Sum of Sales in which Month =”Jan”
• =SUMIFS(Sales,Month,”Jan”,Region,”North”)
• Sum of Sales in which Month=”Jan” and Region=”North”
• =SUMIFS(Sales,Month,”Jan”,Region,”North”)
• Sum of Sales in which Month=”Jan” and Region=”North”
• {=SUM((Month=”Jan”)*(Region=”North”)*Sales)}
• An array formula that returns the sum of Sales in which Month=”Jan” andRegion=”North”
• =SUMIFS(Sales,Month,”Jan”,Region,”<>North”)
• Sum of Sales in which Month=”Jan” and Region <> “North”
• {=SUM((Month=”Jan”)*(Region<>”North”)*Sales)}
• An array formula that returns the sum of Sales in which Month=”Jan” and Region <> “North”
• =SUMIFS(Sales,Month,”Jan”,Sales,”>=200”)
• Sum of Sales in which Month=”Jan” and Sales>=200 (Excel 2007 only)
• {=SUM((Month=”Jan”)*(Sales>=200)*(Sales))}
• An array formula that returns the sum of Sales in which Month=”Jan” and Sales>=200
• =SUMIFS(Sales,Sales,”>=300”,Sales,”<=400”)
• Sum of Sales between 300 and 400 (Excel 2007 only)
• {=SUM((Sales>=300)*(Sales<=400)*(Sales))}
• An array formula that returns the sum of Sales between 300 and 400 | 876 | 2,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-21 | latest | en | 0.829665 |
https://www.exceldemy.com/page/281/ | 1,656,577,475,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00645.warc.gz | 827,413,595 | 42,447 | ### One and Two Ways (Variables) Sensitivity Analysis in Excel!
Sensitivity analysis studies how different sources of uncertainty can affect the final output in a mathematical model. Recently one of my friends ...
### Practice Excel Indirect Function with 3 Case Studies
Today I want to talk about one of the most important Excel functions - INDIRECT function. Before doing that, I need to give you a brief ...
### How to Make a Control Chart with Excel VBA!
In this article, I will show you How to Make a Control Chart with Excel VBA. Control charts are widely used to monitor process stability and control. ...
### Excel Charts with Dynamic Title and Legend Labels
Dynamism is everywhere! Then why not with Excel? Creating dynamic charts in Excel is possible in many ways. What I am going to cover in this ...
### How different is VBA from other programming languages?
THE EXCEL CONTRAPTION A beginner's journey into the world of Excel is always filled with many “OMG it can do that too ?!? Amazing!”. For what all ...
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Excel pivot tables are beautiful! Then why do you need to reverse your pivot tables or pivoted data? Because data generated by businesses are ...
### How to Use INDEX Function in Excel (8 Examples)
The INDEX Function is one of the top used 10 Excel functions. In this tutorial, you will get a complete idea of how the INDEX function works in Excel ...
### Excel Advanced Filter [Multiple Columns & Criteria, Using Formula & with Wildcards]
I was performing keyword analysis for one of my new projects using SemRush.com, a competitive analysis tool. Actually, I was checking for how many ...
### Separating Numbers from a Text – [A Formula Challenge for You]!
At Udemy, I run several courses on Excel and Data Analysis. In one of my free courses (Excel Formulas and Functions with Excel Formulas Cheat Sheet), ...
### Stock return analysis using histograms & 4 skewness of histograms
A histogram is a widely used way to summarize a large data set. Creating a histogram using Data Analysis ToolPak is very easy. This blog is aimed to ...
### How to insert Equation in Excel using Equation Editor (Easy Guide)
If you do mathematics related report or assignment with Excel, you have to know how to use the Equation Editor. In this tutorial, we shall ... | 486 | 2,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-27 | latest | en | 0.861568 |
https://www.printablemultiplication.com/free-printable-2s-multiplication-worksheets-2/ | 1,621,098,515,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990551.51/warc/CC-MAIN-20210515161657-20210515191657-00185.warc.gz | 987,901,515 | 78,583 | # Free Printable 2’s Multiplication Worksheets
Free Printable 2’s Multiplication Worksheets – Multiplication worksheets are an effective method to aid young children in exercising their multiplication expertise. The multiplication tables that kids find out make up the basic base on which a number of other advanced and more modern ideas are taught in later steps. Multiplication plays a very important part in expanding research and mathematics marks in schools. It might be considered to be a fantastic tool to improve children’s expertise at this stage, when standard comprehending remains to be very limited. Multiplication worksheet for kids shows multiplication by means of a number of calculations.
There are 2 ways multiplication worksheets can be utilized. You may get them to oneself. On the other hand, purchase them from some professional items available in the market. In making your own personal multiplication worksheets, it really is vital that you comprehend every single idea engaged. It can help to first do plenty of calculations to make sure that you fully grasp every method. Multiplication worksheets not just help children master multiplication problems in a fast manner, but in addition assist them to comprehend the subtleties of numerous ideas. It not merely means they are comfortable but in addition helps them to imagine their estimations within their thoughts, thereby building additional capabilities.
In Free Printable 2’s Multiplication Worksheets, you will come across simple addition, multiplication, department and subtraction problems. These issues could have to be perfected just before they are often employed in typical mathematics course. These worksheets will assist youngsters resolve an addition difficulty, subtraction issue and department difficulty. By way of example, when your young child wants to know how significantly to give to his good friends together with his wallet dollars, he has to comprehensive the subsequent Multiplication Worksheets. After accomplishing this, your child will have comprehended and been able to fix the next multiplication difficulty: Addition: Let by = a b by.
Free Printable 2’s Multiplication Worksheets aid youngsters recognize all the arithmetic conditions and principles. In case your youngster desires to multiply buck monthly bills and feels he will be needing to know addition, he will very first execute a Multiplication Worksheet, for example. Soon after doing it, he will know that the value of a b has to be equivalent to c. For that reason, the solution to the multiplication problem could be is multiplied by value of a b. Some other multiplication worksheets help children comprehend section troubles such as Flourish by Two, By A number of, by Seven and through 12. The replies to these divisions is dependent upon exactly how the numbers are multiplied.
Math worksheets will even supply some practice exercise routines to your young child. These may help him increase his multiplication and subtraction expertise. They will also enable your kids to develop time management techniques, that can give her a good edge in competing assessments. As soon as they mastered the simple steps, these math worksheets may also be excellent to inspire kids to practice their multiplication specifics. Your son or daughter could have exciting performing these worksheets and will sense achieved at the conclusion of your day.
Multiplication worksheets are really useful for children and mothers and fathers equally. They are going to utilize them to discover and practice various math information and remedy issues daily. They will likely produce math expertise whilst having fun. You will be surprised by your kid’s improvements after using these worksheets. | 663 | 3,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-21 | longest | en | 0.953825 |
https://mail.python.org/pipermail/python-list/2009-May/535647.html | 1,566,321,058,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315551.61/warc/CC-MAIN-20190820154633-20190820180633-00145.warc.gz | 522,292,246 | 2,364 | # list comprehension question
Shane Geiger sgeiger at councilforeconed.org
Fri May 1 18:24:05 CEST 2009
```from goopy.functional import flatten #
http://sourceforge.net/projects/goog-goopy/
b = [(1,2), (3,4), (5,6)]
print flatten(b)
#from goopy.functional import flatten #
http://sourceforge.net/projects/goog-goopy/
def flatten(seq):
"""
Returns a list of the contents of seq with sublists and tuples "exploded".
The resulting list does not contain any sequences, and all inner sequences
are exploded. For example:
>>> flatten([7,(6,[5,4],3),2,1])
[7,6,5,4,3,2,1]
"""
lst = []
for el in seq:
if type(el) == list or type(el) is tuple:
lst.extend(flatten(el))
else:
lst.append(el)
return lst
Chris Rebert wrote:
> On Thu, Apr 30, 2009 at 5:56 PM, Ross <ross.jett at gmail.com> wrote:
>
>> If I have a list of tuples a = [(1,2), (3,4), (5,6)], and I want to
>> return a new list of each individual element in these tuples, I can do
>> it with a nested for loop but when I try to do it using the list
>> comprehension b = [j for j in i for i in a], my output is b =
>> [5,5,5,6,6,6] instead of the correct b = [1,2,3,4,5,6]. What am I
>> doing wrong?
>>
>
> Your comprehension is the identity comprehension (i.e. it effectively
> just copies the list as-is).
> What you're trying to do is difficult if not impossible to do as a
> comprehension.
>
> Here's another approach:
> b = list(itertools.chain.from_iterable(a))
>
> And without using a library function:
> b = []
> for pair in a:
> for item in pair:
> b.append(item)
>
> Cheers,
> Chris
>
--
Shane Geiger, IT Director
Council For Economic Education / www.councilforeconed.org
sgeiger at councilforeconed.org / 402-438-8958
Teaching Opportunity
``` | 518 | 1,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-35 | longest | en | 0.765217 |
https://openforecast.org/sba/whatIsProbability.html | 1,685,838,420,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00403.warc.gz | 481,962,145 | 9,153 | This book is in Open Review. I want your feedback to make the book better for you and other readers. To add your annotation, select some text and then click the on the pop-up menu. To see the annotations of others, click the button in the upper right hand corner of the page
## 2.1 What is probability?
We start with a classical example: tossing a coin. If you have one, take it in your hands, look at it, and answer a question: what outcome will you have if you toss it? Toss it once and, let’s say, it ended up showing heads. Can you predict the outcome of the next toss based on this observation? What if you toss it again and end up with tails? Would that change your prediction for the next toss?
What we could do in this situation to predict future outcomes is to write down the results of tosses as zeroes (for heads) and ones (for tails). We will then have a set of observations of a style:
1 0 0 1 0 1 1 1 0 1
If we then take the mean of this series, we will see that the expected outcome based on our sample is 0.6. We would call this value the empirical probability. It shows us that roughly in 60% of the cases we get tails. But this is based on just 10 experiments. If we continue tossing the coin for many more times, this probability (in the case of a fair coin) will be equal to 0.5, meaning that in the 50% of the cases the coin will show heads and in the other 50% it will be tails. In fact, we know that there are only two possible outcomes in this experiment, and that in case of a fair coin, there are no specific forces that could change the outcome and lead to more tails than heads. In this case, we can say that the theoretical probability of having tails is 0.5. Note that this does not tell us anything about each specific outcome, but only demonstrates what happens on average, when we repeat the experiment many times.
Definition 2.1 Probability is the measure of how likely an event is expected to occur if we observe it many times.
This definition implies that we cannot tell what the next outcome of the experiment will be (whether the coin toss will result in heads or tails). Instead, we can say what will happen on average if the experiment is repeated many times. By definition, the probability lies between 0 and 1, where 0 means that the event will not occur and 1 implies that it will always occur.
We could do other similar experiments, for example rolling a six-sided dice, and calculating the probability of a specific outcome. In the simple cases with coins, cards, dices etc, we can even tell the probability without running the experiments. All we need to do is calculate the number of outcomes of interests and divide them by all the possible outcomes. For example, the probability of getting 3 on a 6-sided dice is $$\frac{1}{6}$$, because there are overall six outcomes: 1, 2, 3, 4, 5 and 6, and the probability of getting any one of them is the same for all of them. The probability of getting 5 is $$\frac{1}{6}$$ as well for the same reason: all the six outcomes are considered equally possible and will happen on average every sixth roll.
Remark. In some tabletop games, the number of dices and their outcomes are encoded as $$a \mathrm{d} b$$, where $$a$$ is the number of dices, $$b$$ is the number of sides and d stands for the word “dice”. In our example, the 10-sided dice can be encoded as 1d10, while the classical 6-sided one is 1d6.
Mathematically, we will denote probability as $$\mathrm{P}(y)$$, where $$y$$ represents a specific outcome. We can write, for example, that the probability of having 3 in the dice roll experiment is: $$$\mathrm{P}(y=3) = \frac{1}{6} . \tag{2.1}$$$ We can calculate more complicated probabilities. For example, what is the probability of having an odd number when rolling a 1d6? We need to calculate the number of events of interest and divide that number by the number of possible outcomes. In our case, the former is 1, 3, and 5 (three numbers), while the latter is any integer number from 1 to 6 (six numbers). This means that: $$$\mathrm{P}(y \text{ is odd}) = \frac{3}{6} = \frac{1}{2}. \tag{2.2}$$$ | 1,010 | 4,107 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-23 | latest | en | 0.953491 |
https://link.springer.com/article/10.1007/s10817-017-9424-6 | 1,506,333,644,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690591.29/warc/CC-MAIN-20170925092813-20170925112813-00429.warc.gz | 687,054,274 | 41,252 | # Deciding Univariate Polynomial Problems Using Untrusted Certificates in Isabelle/HOL
• Wenda Li
• Grant Olney Passmore
• Lawrence C. Paulson
Open Access
Article
## Abstract
We present a proof procedure for univariate real polynomial problems in Isabelle/HOL. The core mathematics of our procedure is based on univariate cylindrical algebraic decomposition. We follow the approach of untrusted certificates, separating solving from verifying: efficient external tools perform expensive real algebraic computations, producing evidence that is formally checked within Isabelle’s logic. This allows us to exploit highly-tuned computer algebra systems like Mathematica to guide our procedure without impacting the correctness of its results. We present experiments demonstrating the efficacy of this approach, in many cases yielding orders of magnitude improvements over previous methods.
### Keywords
Interactive theorem proving Isabelle/HOL Decision procedure Cylindrical algebraic decomposition
## 1 Introduction
Nonlinear polynomial systems are ubiquitous in science and engineering. As real-world applications of formal verification continue to grow and diversify, there is an increasing need for proof assistants (e.g., ACL2, Coq, Isabelle [27], HOL Light and PVS) to provide automation for reasoning about nonlinear systems over the reals [17, 24, 25].
Cylindrical algebraic decomposition (CAD) [8] is one of the most powerful known techniques for analysing non-linear polynomial systems. CAD-based methods have been implemented in various systems such as Z3 [9], QEPCAD [3], Mathematica and Maple. However, implementing CAD-based decision procedures within proof assistants has been hindered by the difficulty in formalising the mathematics justifying CAD computations.
In this paper, we present a formally verified procedure1 based on CAD for univariate polynomial problems with rational coefficients. Goals such as
\begin{aligned}&\forall x.\, \left( x^2>2 \wedge x^{10}-2x^5+1 \ge 0\right) \vee x<2\\&\qquad \exists x.\, \left( x^2=2 \wedge (x>1 \vee x<0)\right) \end{aligned}
can be discharged by our tactic automatically. It should be noted that certifying a general multivariate CAD procedure is much harder, and the univariate version we describe in the paper is only a first step in that direction.
A key feature of our procedure is its certificate-based design in which an external untrusted (but ideally highly efficient) program is used to find certificates, and those certificates are then checked by verified internal procedures. Overall, the soundness of our procedure depends solely on the soundness of Isabelle’s logic (and code generation2) rather than trusted external oracles. This is much like Isabelle’s sledgehammer tactic, which sceptically incorporates various external tools.
Our main contributions are:
• An efficient formalised theory of Tarski queries,
• An efficient approach to univariate sign determination at real algebraic points,
• A practical formally verified procedure for real algebraic problems based on univariate CAD.
The paper continues at follows: A motivating example (Sect. 2) and a description of the overall design (Sect. 3) sketch the general idea of our procedure. The construction and manipulation of real algebraic numbers is developed in (Sect. 4), including a sign determination procedure for evaluating polynomials at real algebraic points (Sect. 5). The main proof is described in (Sect. 6), which is followed by a discussion of interaction with external solvers (Sect. 7). Next, experiments and related work (Sect. 8) are described along with further discussion of our tactic (Sect. 9). We then conclude with a look towards the future (Sect. 10).
## 2 A Motivating Example
Unlike the general case of $$\mathbb {R}^n$$, the restriction of CAD to univariate problems (i.e., to $$\mathbb {R}^1$$) is relatively straight-forward. Suppose we wish to prove
\begin{aligned} \forall x.\, P(x)>0 \vee Q(x) \ge 0 \end{aligned}
where
\begin{aligned}&P(x)=\frac{1}{2}x^2-1\\&Q(x)=x+3. \end{aligned}
To do so, we can decompose$$\mathbb {R}$$ into disjoint connected components induced by the roots of P and Q. This is illustrated in Fig. 1:
and it can be observed that both P and Q have invariant signs over each of these components. For example, as can be seen from Fig. 1, $$P(x)<0$$ and $$Q(x)>0$$ hold for all $$x \in (-\sqrt{2},\sqrt{2})$$. To decide the conjecture, we can pick sample points from each of these components and evaluate $$\lambda x.\, P(x) > 0 \vee Q(x) \ge 0$$ at these points. That is,
\begin{aligned}&\forall x.\, P(x)>0 \vee Q(x) \ge 0 \nonumber \\&\quad = \forall D \in \mathfrak {D}.\, \forall x \in D.\, P(x)>0 \vee Q(x) \ge 0 \nonumber \\&\quad = \forall x \in \{-4,-3,-2,-\sqrt{2}, 0, \sqrt{2}, 2 \}.\, P(x)>0 \vee Q(x) \ge 0 \nonumber \\&\quad = (P(-4)>0 \vee Q(-4) \ge 0) \wedge (P(-3)>0 \vee Q(-3) \ge 0) \wedge \dots \nonumber \\&\qquad \wedge (P(2)>0 \vee Q(2) \ge 0) \nonumber \\&\quad = \mathrm {True} \end{aligned}
(1)
since
\begin{aligned} -4\in & {} (-\infty ,-3) \\ -3\in & {} \{-3\} \\ -2\in & {} (-3,-\sqrt{2}) \\ -\sqrt{2}\in & {} \{-\sqrt{2}\} \\ 0\in & {} (-\sqrt{2},\sqrt{2})\\ \sqrt{2}\in & {} \{\sqrt{2}\}\\ 2\in & {} (\sqrt{2},\infty ). \end{aligned}
Analogously, to decide an existential formula
\begin{aligned} \exists x.\, P(x) =0 \wedge Q(x) >0, \end{aligned}
we have
\begin{aligned}&\exists x.\, P(x)=0 \wedge Q(x)> 0 \nonumber \\&\quad = \exists D \in \mathfrak {D}.\, \exists x \in D.\, P(x)=0 \wedge Q(x)> 0 \nonumber \\&\quad = \exists x \in \{-4,-3,-2,-\sqrt{2}, 0, \sqrt{2}, 2 \}.\, P(x)=0 \wedge Q(x)> 0 \nonumber \\&\quad = (P(-4)=0 \wedge Q(-4)> 0) \vee (P(-3)=0 \wedge Q(-3)> 0) \vee \dots \nonumber \\&\qquad \vee (P(2)=0 \wedge Q(2) > 0)\nonumber \\&\quad = \mathrm {True}. \end{aligned}
(2)
In performing these arguments, there were a few “obvious” subtleties:
• The decomposition of $$\mathbb {R}$$ into the seven regions given covered the entire real line. That is,
\begin{aligned} (-\infty ,-3) \cup \{-3\} \cup (-3,-\sqrt{2}) \cup \{-\sqrt{2}\} \cup (-\sqrt{2},\sqrt{2})\cup \{\sqrt{2}\} \cup (\sqrt{2},\infty ) = \mathbb {R}. \end{aligned}
• The “sign-invariance” of P and Q over each region was exploited to allow only a single sample point to be selected from each region. This property holds as by the Intermediate Value Theorem, P and Q can only change sign by passing through a root.
• The signs of univariate polynomials were evaluated at irrational real algebraic points like $$\sqrt{2}$$ to determine the truth values of atomic formulas.
In creating our automatic proof procedure, all of this routine reasoning must, of course, be formalised. Moreover, the isolation of polynomial roots (and thus sign-invariant regions) and the sign determination for polynomials at real algebraic points are computationally expensive operations. Computer algebra systems like Mathematica have decades of tuning in their implementations of these core algebraic algorithms. To have a practical proof procedure, we wish to take advantage of these highly tuned external tools as much as possible. Let us next describe how this can be done.
## 3 A Sketch of Our Certificate-Based Design
There is a rich history of certificate-based, sceptical integrations between proof assistants and external solvers. Examples include John Harrison’s sums-of-squares method [17] and the Sledgehammer [31] command in Isabelle.
Certificate-based approaches are motivated by many observations, including:
• External solvers are often highly tuned and run much faster than verified ones.
• Verification of certificates from external solvers is usually much easier than finding them. Such verification ensures the soundness of the overall tactic.
• Switching between different external solvers does not require changes in formal proofs.
Algorithm 1 sketches our idea for univariate universal formulas. In particular, in line 3, we use external programs to return real roots of polynomials (i.e., $$\mathfrak {P}$$) from the quantifier-free part of the formula (i.e., F(x)). Those roots (i.e., $$roots$$) correspond to a decomposition such that each polynomial from $$\mathfrak {P}$$ has a constant sign over each component of this decomposition. Since the roots are returned by untrusted programs, in line 5, we not only check $$\forall x \in samples .\, F(x)$$ as in Eq. (1) but also certify that these roots are indeed all real roots of $$\mathfrak {P}$$.
The step in line 3 in Algorithm 1 is more commonly referred as (real) root isolation, which is a classic and well-studied topic in symbolic computing. Although we can in principle formalise our own root isolation procedure (e.g., using the Sturm–Tarski theorem), it is utterly unlikely that our implementation will be competitive with state-of-the-art ones, especially for polynomials of high degree, large bit-width, or whose roots are very close together. Therefore, we delegate this computationally expensive step to external tools.
With existential formulas, the situation is even simpler as illustrated in Algorithm 2, since we do not need to deal with the decomposition internally. Rather, all we need is a real algebraic witness that satisfies $$\lambda x.\, F(x)$$ to certify $$\exists x.\, F(x)$$. What is more interesting is that the satisfaction problem for $$\lambda x.\, F(x)$$ can be not only solved by a CAD procedure, which is complete but not very fast due to its symbolic nature, but also be complemented by highly efficient incomplete numerical methods. Thus it is natural to externalize the step in line 2 in Algorithm 2.
## 4 Encoding Real Algebraic Numbers
External programs in either Algorithms 1 and 2 can return real algebraic numbers (e.g. $$\sqrt{2}$$). In this section, we see how to formalise such numbers in Isabelle/HOL.
The real algebraic numbers ($$\mathbb {R}_{\mathrm {alg}}$$) are real roots of non-zero polynomials with integer (equivalently, rational) coefficients. They form a countable, computable subfield of the real numbers. To encode them, we use a polynomial with integer coefficients and a root selection method to “pin down” the root in question. Common root selection methods include isolating intervals, root indices or Thom encodings. We use the root interval approach, that is, a real algebraic number $$r \in \mathbb {R}_{\mathrm {alg}}$$ will be given by
• A polynomial $$p \in \mathbb {Z}[x]$$ s.t. $$p(r) = 0$$, and
• Two rationals $$a,b \in \mathbb {Q}$$ s.t. r is the only root of p contained in [ab].
To reason over the reals, we define a function to embed those real algebraic numbers into the reals: where is a polynomial with integer coefficients and the two arguments represent an interval. Note, a in Isabelle/HOL is a dyadic rational number of the form
\begin{aligned} a 2^b \quad \text {where} \quad a,b \in \mathbb {Z}. \end{aligned}
Compared to our previous work [21], where a pair of rational numbers is used to represent an interval, the dyadic rational approach is more efficient due to the elimination of ubiquitous greatest common divisor (gcd) operations within rational arithmetic.
In Isabelle/HOL, a real number is represented as a Cauchy sequence of type , where a Cauchy sequence is defined as We then convert an encoding of a real algebraic number into a sequence of type . The idea is to bisect the isolating interval through each recursive call, and proceed with the half where the sign of the polynomial changes at its end points: where evaluates the polynomial at the point . Note, encodes a real algebraic number here (rather than ), as we can embed and into .
It can be then shown that the sequence constructed by is indeed a Cauchy sequence and the real number represented by this sequence resides within the interval $$[ lb , ub ]$$, provided $$lb < ub$$: Note, the function of type constructs a real number from its underlying representation (i.e. a Cauchy sequence).
Finally, we can finish the definition of : where ensures
With the help of , we can now encode the real algebraic number $$\sqrt{2}$$ as where corresponds to the polynomial $$-2 x^0+ 0 x^1 + 1 x^2 = x^2-2$$, and 1 and 2 are the lower bound and upper bound respectively, such that $$\sqrt{2}$$ is the only root of $$x^2-2$$ within the interval (1, 2).
Furthermore, we can formally derive that is indeed a root of within the interval : where embeds the integer polynomial into a real one.
## 5 Deciding the Sign of a Univariate Polynomial at Real Algebraic Points
In the previous section, we described how to encode a real algebraic number as an integer polynomial and two dyadic rational numbers. Now, suppose we have
\begin{aligned} \sqrt{2} = (x^2-2,1,2) \end{aligned}
where $$(x^2-2,1,2)$$ is abbreviated from for the sake of readability. How can we computationally prove that
\begin{aligned} P(\sqrt{2}) = 0 \quad \text {where} \quad P(x) = \frac{1}{2} x^2 -1 \ ? \end{aligned}
Considering that $$\mathbb {R}_{\mathrm {alg}}$$ is a computable subfield of $$\mathbb {R}$$ and has decidable arithmetic and comparison operations, it is natural to evaluate such formulas through algebraic arithmetic:
\begin{aligned} P(\sqrt{2})= & {} \frac{1}{2} \times _{\mathrm {alg}} (x^2-2,1,2) \times _{\mathrm {alg}} (x^2-2,1,2) -_{\mathrm {alg}} 1\\= & {} \frac{1}{2} \times _{\mathrm {alg}} (x-2,1,3) -_{\mathrm {alg}} 1 \\= & {} \left( x-1,\frac{1}{2},\frac{3}{2}\right) -_{\mathrm {alg}} 1 \\= & {} 0, \end{aligned}
where $$\times _{\mathrm {alg}}$$ and $$-_{\mathrm {alg}}$$ are exact algebraic arithmetic operations that usually involve calculation of bivariate resultants. Although such computations are currently possible in Isabelle/HOL [21, 36], they are far from efficient.
In this section, we describe a verified procedure to decide the sign of univariate polynomials with rational coefficients at real algebraic points which uses only rational (or dyadic rational) arithmetic rather than costly algebraic arithmetic.
### 5.1 The Sturm–Tarski Theorem
We abbreviate $$\mathbb {R} \cup \{-\infty ,\infty \}$$ as $$\overline{\mathbb {R}}$$, the extended real numbers.
### Definition 1
(Tarski Query) The Tarski query $$\mathrm {TaQ}(Q,P,a,b)$$ is
\begin{aligned} \mathrm {TaQ}(Q,P,a,b) = \sum _{x \in (a,b), P(x)=0} \mathrm {sgn}(Q(x)) \end{aligned}
where $$a,b \in \overline{\mathbb {R}}$$, $$P, Q \in \mathbb {R}[X]$$, $$P\ne 0$$ and $$\mathrm {sgn}: \mathbb {R} \rightarrow \{-1,0,1\}$$ is the sign function.
The Sturm–Tarski theorem [23, Chapter 8] (or Tarski’s theorem [2, Chapter 2]) is essentially an effective way to compute Tarski queries through some remainder sequences:
### Theorem 1
(Sturm–Tarski) The Sturm–Tarski theorem states
\begin{aligned} \mathrm {TaQ}(Q,P,a,b) = \mathrm {Var}(\mathrm {SRemS}(P,P'Q);a,b) \end{aligned}
where $$P \ne 0$$, $$P,Q\in \mathbb {R}[X]$$, $$P'$$ is the first derivative of P, $$a,b \in \overline{\mathbb {R}}$$, $$a<b$$ and are not roots of P, $$\mathrm {SRemS}(P,P'Q)$$ is the signed remainder sequence of P and $$P'Q$$, and
\begin{aligned}&\mathrm {Var}([p_0, p_1, \dots , p_n];a,b) \\&\quad = \mathrm {Var}([p_0(a), p_1(a), \dots , p_n(a)]) - \mathrm {Var}([p_0(b), p_1(b), \dots , p_n(b)]) \end{aligned}
is the difference in the number of sign variations (after removing zeroes) in the polynomial sequence $$[p_0 , p_1 , \dots , p_n ]$$ evaluated at a and b.
Note that the more famous Sturm’s theorem, which counts the number of distinct real roots (of a univariate polynomial) within an interval, is a special case of the Sturm–Tarski theorem when $$Q=1$$.
### 5.2 A Formal Proof of the Sturm–Tarski Theorem
Our proof of the Sturm–Tarski theorem in Isabelle is based on Basu et al. [2, Chapter 2] and Cohen’s formalisation in Coq [6].
The core idea of our formal proof is built around the Cauchy index. First defined by Cauchy in 1837, the Cauchy index of a real rational function encodes deep properties of its roots and poles, and can be used as the basis of an algebraic method for computing Tarski queries.3
### Definition 2
Given $$P,Q \in \mathbb {R}[x]$$ and $$x \in \mathbb {R}$$, $$\mathrm {jump}(P,Q,x)$$ is defined as
\begin{aligned} \mathrm {jump}(P,Q,x) = {\left\{ \begin{array}{ll} -1 &{} \text{ if } \lim _{u \rightarrow x^-} \frac{Q(u)}{P(u)}=\infty \text{ and } \lim _{u \rightarrow x^+} \frac{Q(u)}{P(u)}=-\infty \\ 1 &{} \text{ if } \lim _{u \rightarrow x^-} \frac{Q(u)}{P(u)}=-\infty \text{ and } \lim _{u \rightarrow x^+} \frac{Q(u)}{P(u)}=\infty \\ 0 &{} \text{ otherwise. } \end{array}\right. } \end{aligned}
For example, let $$Q=x-4$$ and $$P=(x-3)(x-1)^2(x+1)$$. The graph of Q / P is shown in Fig. 2. We have
\begin{aligned} \mathrm {jump}(P,Q,x) = {\left\{ \begin{array}{ll} 1 &{} \text { when } x=-1\\ -1 &{} \text { when } x=3\\ 0 &{} \text{ otherwise. } \\ \end{array}\right. } \end{aligned}
The Cauchy index is the sum of the jumps of q / p over the interval (ab):
By case analysis, we can prove a connection between the Tarski query and the Cauchy index: where is a formal definition of the Tarski query and is the first derivative of .
Moreover, the Cauchy index can be related to Euclidean division () on polynomials by a recurrence: where
\begin{aligned} \mathrm {cross}\ p\ a\ b = {\left\{ \begin{array}{ll} 0 &{} \text{ if } p(a)p(b) \ge 0\\ 1 &{} \text{ if } p(a)p(b)< 0 \text{ and } p(a)<p(b)\\ -1 &{} \text{ if } p(a)p(b) < 0 \text{ and } p(a) \ge p(b).\\ \end{array}\right. } \end{aligned}
A similar recurrence relation holds for the number of sign variations of the signed remainder sequences (): where is defined as and the signed remainder sequence () is defined as and returns the number of sign changes when evaluating a list of polynomials () at .
Finally, by combining , and , we derive the Sturm–Tarski theorem: Note, this is just the bounded case of the Sturm–Tarski theorem. Proofs for the unbounded and half-bounded cases are similar.
### 5.3 Sign Determination Through the Sturm–Tarski Theorem
Given a polynomial q with rational coefficients and our encoding of a real algebraic number $$\alpha$$
\begin{aligned} \alpha = (p, lb , ub ) \end{aligned}
where p is an integer polynomial, and $$lb$$ and $$ub$$ are dyadic rationals, we can effectively decide the sign of $$q(\alpha )$$ using the Sturm–Tarski theorem, provided holds. The rationale behind is that ensures $$\alpha$$ is the only root of p within the interval $$( lb , ub )$$, hence
\begin{aligned} \mathrm {sgn}(q(\alpha ))&= \sum _{x \in ( lb , ub ), p(x)=0} \mathrm {sgn}(q(x))\\&= \mathrm {TaQ}(q,p,lb,ub)\\&= \mathrm {Var}(\mathrm {SRemS}(p,p'q); lb , ub ). \end{aligned}
Importantly, it can be observed that evaluating $$\mathrm {Var}(\mathrm {SRemS}(p,p'q); lb , ub )$$ requires only rational arithmetic rather than costly algebraic arithmetic.
To be even more efficient, we refine the procedure further to make use of dyadic rational arithmetic. The main advantage of dyadic rational arithmetic over rational arithmetic are reduced normalization steps and possible bit-level operations. For example, consider two rational numbers $$\frac{a_1}{b_1}$$ and $$\frac{a_1}{b_2}$$ where $$a_1,b_1,a_2,b_2 \in \mathbb {Z}$$, their sum is
\begin{aligned}&\frac{a_1}{b_1} + \frac{a_2}{b_2} = \frac{a_1 b_2 + a_2 b_1}{b_1 b_2} = \frac{(a_1 b_2 + a_2 b_1)/c}{(b_1 b_2)/c} \\&\quad \text {where} \ c=\gcd (a_1 b_2 + a_2 b_1,b_1 b_2). \end{aligned}
To counter the growth in the size of representations, we usually need to normalize the result by factoring out the gcd. Such gcd operations can be the source of major computational expense. Thankfully, they are unnecessary in the context of dyadic rationals. The sum of two dyadic rationals $$(a_1,e_1)$$ and $$(a_2,e_2)$$ where $$a_1,e_1,a_2,b_2 \in \mathbb {Z}$$ is
\begin{aligned} a_1 2^{e_1} + a_2 2^{e_2} = {\left\{ \begin{array}{ll} (a_1 2^{e_1-e_2}+a_2) 2^{e_2} &{} \text{ if } e_1>e_2\\ (a_1+a_2 2^{e_2-e_1}) 2^{e_1} &{} \text{ otherwise. } \\ \end{array}\right. } \end{aligned}
Moreover, multiplications by powers of two, such as $$a_1 2^{e_1-e_2}$$, can be optimised by shift operations.
However, the problem with dyadic rational numbers is that they do not have the division operation (e.g. $$1 \times 2^0$$ divided by $$3 \times 2^0$$ is no longer a dyadic rational), hence they do not form a field, while Euclidean division only works for polynomials over a field. This problem can be solved if we switch from Euclidean division ( and ):
\begin{aligned} P = (P \mathbin {\mathrm {div}}Q)\, Q + (P \mathbin {\mathrm {mod}}Q) \ \text { and } \ (Q = 0 \vee \deg (P \mathbin {\mathrm {mod}}Q) < \deg (Q)) \end{aligned}
to pseudo-division ( and ) [10]:
\begin{aligned}&\mathrm {lc}(Q)^{1+\deg (P) -\deg (Q)} P = (P \mathbin {\mathrm {pdiv}}Q)\, Q + (P \mathbin {\mathrm {pmod}}Q) \\&\quad \text { and } \ (Q = 0 \vee \deg (P \mathbin {\mathrm {mod}}Q) < \deg (Q)) \\&\quad \text {where lc({ Q}) is the leading coefficient of { Q},} \end{aligned}
since pseudo-division can be carried out by polynomials over an integral domain (rather than a field).
Based on pseudo-division, the signed pseudo-remainder sequence ($$\mathrm {SPRemS}$$) can be defined: where is the scalar product on polynomials and is the leading coefficient of . Accordingly, the function to count the difference in sign variations can be refined: and linked to the previous one based on signed remainder sequences ($$\mathrm {SRemS}$$): where embeds a into and coverts a (i.e. polynomial with dyadic rational coefficients) to a by embedding each of the coefficients into .
Finally, we define a function that returns the sign of a univariate polynomial at some point: Note, for now, if either or any coefficient of is an irrational real number (e.g. an irrational real algebraic number), evaluating will raise an exception, as Isabelle/HOL, by default, only supports rational arithmetic. Although we can eliminate some such exceptions by loading any of the recent algebraic arithmetic libraries [21, 36], we consider exact algebraic arithmetic too slow for our purpose as stated at the beginning of Sect. 5. Alternatively, by proving some code equations, we can restore the executability of when is constructed by and coefficients of are rational reals: where
And note that evaluating requires only dyadic arithmetic, which is much more efficient than exact algebraic arithmetic.
Moreover, the executability of is restored similarly as well: where checks if the polynomial p has exactly one real root within the interval $$( lb , ub )$$ by exploiting Sturm’s theorem (a special case of our formalised Sturm–Tarski theorem).
After restoring executability of on real algebraic numbers, we can now check the sign of $$P(x)=\frac{1}{2}x^2-1$$ at $$\sqrt{2}$$ by typing the following command: which returns 0 (i.e. $$P(\sqrt{2})=0$$).
### 5.4 Remark
A formal proof of the Sturm–Tarski theorem is not new among proof assistants: it has been formalised in PVS [25] and Coq [6]. However, as far as we know, we are the first to exploit this theorem to build a verified sign determination procedure of real algebraic numbers, which uses only rational or dyadic rational arithmetic.
Real algebraic numbers are essential in symbolic computing, and well studied. In general, exact real algebrac arithmetic is rarely used in modern computer algebra systems due to its extreme inefficiency. For example, consider the problem of isolating the real roots of a polynomial with real algebraic coefficients. Modern approaches usually use sophisticated techniques to soundly approximate those coefficients to a certain precision rather than carrying out exact algebraic arithmetic [5, 33, 35], relying on exact symbolic procedures as a fall-back in degenerate cases.
Following these efficient modern approaches, our sign determination procedure can be improved in at least the following ways:
• Sophisticated interval arithmetic can be used to decide the sign before resorting to a remainder sequence, as has been done in Z3 [10]. This approach should help when the sign is non-zero.
• Pseudo-division, which we are currently using for building remainder sequences, is not good for controlling coefficients growth. More sophisticated approaches, such as subresultant sequences and modular methods, can be used to optimise the calculation of remainder sequences.
## 6 The Formal Development of the Decision Procedure
In this section, we describe the main proof underlying our tactic.
### 6.1 Parsing Formulas
The first step of our tactic is to parse the target formula into a structured form. This process is usually referred as reification [4] in Isabelle/HOL. More specifically, given an Isabelle/HOL term e of type $$\tau$$, we define a (more structured) datatype $$\delta$$ and an interpretation function $$interp$$ of type $$\delta \Rightarrow \tau \ list \Rightarrow \tau$$, such that for some e‘ of type $$\delta$$
\begin{aligned} e = interp \ e\ xs \end{aligned}
where $$xs$$ is a list of free variables in e. Subsequently, instead of directly dealing with e, we now convert it into a more pleasant form $$interp \ e\ xs$$ where e‘ is in fact a formal language that captures the structure of e.
The datatypes we defined to capture the structure of target univariate formulas are as follows: and the interpretation functions:
Given the definition of a (structured) datatype and the corresponding interpretation function , target formulas can now be parsed. For example, we can convert a univariate formula into an equivalent form In particular, note in which inequalities have been parsed into a polynomial sign determination problem.
On the contrary, a bivariate non-closed formula such as will be converted into where the constructor indicates that such formula is not supported by our current tactic.
### 6.2 Existential Case
To discharge a univariate existential formula is easy: we can computationally check if a certificate (i.e., a real algebraic number) returned by an external solver satisfies the quantifier-free part of the formula: where of type is a certificate that is supposed to be instantiated by an external solver. The function converts from to . In other words, to prove an existential formula: we can computationally check the truth value of the quantifier-free part of the formula at : which is possible due to the sign determination procedure described in Sect. 5.
### 6.3 Universal Case
For the universal case, the core lemma is as follows: where states that is a bijective function between the decomposition and the sample points . Essentially, what the lemma shows is that given a predicate , an unbounded universal formula is equivalent to a bounded one , if the truth value of is constant over each component of the decomposition: .
On top of the lemma , we similarly convert an unbounded univariate real formula into a bounded one: where
Most importantly, all assumptions of the lemma and its right-hand side can be computationally checked, through which we can prove an unbounded univariate universal formula: .
## 7 Linking to an External Solver
Certificates for both existential and universal cases can be produced by any program performing univariate CAD. For now, we implement the program on top of Mathematica. More specifically, the universal certificates are constructed by the Mathematica command SemialgebraicComponentInstances, which gives sample points in each connected component of a semialgebraic set. The existential certificates are constructed by the command FindInstance, which incorporates powerful numerical methods to accelerate the search for real algebraic sample points.
Also, it may be worth mentioning that after a certificate has been found, our tactic will record it (as a string) so that repeating the proof no longer requires the external solver. This is much like the sums-of-squares tactic [17].
In general, the certificate-based design grants us much flexibility: We can easily switch to a more efficient external solver without modifying existing formal proofs. In fact, we were first using an implementation of univariate CAD built within MetiTarski, which turned out to be not very efficient, and we simply switched to the current one based on Mathematica. In the future, we plan to experiment with other open-source CAD implementations such as Z3 and QEPCAD to provide more options with external solvers.
## 8 Experiments and Related Work
The most relevant work is the recent tarski strategy by Narkawicz et al. [25] in PVS. Both their work and ours rely on a formal proof of the Sturm–Tarski theorem (which they call Tarski’s theorem) and handle roughly the same class of problems4 (i.e., first-order univariate formulas over reals). There are two main differences between their work and ours:
• Their procedure resembles Tarski’s original quantifier elimination [2, Chapter 2] and Cyril Cohen’s quantifier elimination procedure in Coq [6, Chapter 12] by making use of both the Sturm–Tarski theorem and matrices. In contrast, our tactic is based on CAD and real algebraic numbers (instead of matrices).
• Their procedure is entirely built within PVS, while ours sceptically makes use of efficient external programs to generate certificates.
To compare both tactics empirically, we have conducted experiments on several typical examples from their paper5 and the MetiTarski project6 [29]. The experiments are run on a desktop with an Intel Core 2 Quad Q9400 (quad core, 2.66 GHz) CPU and 8 gigabytes RAM. Results of the experiments are illustrated in Fig. 3, where our tactic includes both certificate searching and checking process, while the does the checking part only (when repeating a proof with certificates already recorded as a string).
In general, the experiments indicate that our tactic outperforms the tarski strategy in PVS. Particularly, the advantage of our tactic becomes greater as the problems become more complex, which can be attributed to the fact that our tactic has much better worst-case computational complexity (polynomial vs. exponential in the number of polynomials).
In the case of general multivariate problems, the CAD procedure is doubly exponential while Tarski’s quantifier elimination procedure is non-elementary in the number of variables [2, Chapter 11]). When limited to univariate problems, the CAD procedure degenerates to root isolation and sign determination on a set of univariate polynomials, which is of polynomial complexity in the number of polynomials and their degree bound [2, Chapter 10]). In comparison, Tarski’s quantifier elimination procedure, even when limited to univariate problems, is still exponential in the number of polynomials [7].
In addition, it is worth noting that as the problems become more complex (e.g., ex6 and ex7 in Fig. 3), certificate checking becomes the bottleneck factor of our tactic (especially for universal problems). This indicates that, despite the fact that certificate searching is much harder than certificate checking, the Mathematica implementation is still much more efficient than our verified certificate-checking procedure. This leaves much room for future optimisations.
Our work has also been greatly inspired by Cyril Cohen’s PhD thesis [6], within which a quantifier elimination procedure has been built upon the Sturm–Tarski theorem and real algebraic numbers formalised within the Coq theorem prover. However, our goals and approaches are very different.
Cohen’s work is part of a large project that has formalised the Feit–Thompson theorem (odd order theorem) in Coq [15], and focuses more on theoretical developments than we do. For example, they proved the Sturm–Tarski theorem to construct an RCF quantifier elimination procedure in the spirit of Tarski’s original method, which has important theoretical properties but is not practical as a proof procedure. Moreover, he has formalised arithmetic on real algebraic numbers and shown that they form a real closed field via resultants. We have not formalised resultants at all. Our sign determination algorithm uses the Sturm–Tarski theorem, which is significantly more efficient in practice than using resultants. On the other hand, as it was unnecessary for our proof procedure, we have not proved in Isabelle that the real algebraic numbers form a real closed field. In general, compared to his work, ours stresses the practical side over the theoretical. Fundamentally, we want to build procedures to solve non-trivial problems in practice.
Decision procedures based on Sturm’s theorem have been implemented in Isabelle and PVS before [14, 26]. Their core idea is to count the number of real roots within a certain (bounded or unbounded) interval. Generally, they can only handle formulas involving a single polynomial, so they are not complete for first-order formulas (unlike our tactic and the tarski strategy in PVS).
Assia Mahboubi [22] has implemented the executable part of a general CAD procedure in Coq, but as far as we know, the correctness proof for her implementation is still ongoing. This is also one of the reasons for us to choose the certificate-based approach rather than directly verifying an implementation.
There are other methods to handle nonlinear polynomial problems in theorem provers, such as sums of squares [17], which is good for multivariate universal problems but is not applicable when the existential quantifier arises, and interval arithmetic [18, 34], which is very efficient for some cases but is not complete. These methods and ours should be used in a complementary way.
## 9 Discussion and Applications
One of our driving motivations is the integration of MetiTarski with Isabelle. MetiTarski [1] is a first-order theorem prover for real number inequalities involving transcendental functions such as $$\sin$$, $$\tan$$ and $$\exp$$. It can automatically prove formulas like
\begin{aligned}&\forall x \in (0,1.25).\, \tan (x)^2 \le 1.75 \times 10^{-7} + \tan (1)\tan (x^2)\\&\forall x > 0.\, \frac{1-e^{-2 x}}{2x(1-e^{-x})^2} -\frac{1}{x^2} \le \frac{1}{12}\\&\forall x \in (0,1).\, 1.914 \frac{\sqrt{1+x}-\sqrt{1-x}}{4+\sqrt{1+x}+\sqrt{1-x}} \le 0.01+\frac{x}{2+\sqrt{1-x^2}}. \end{aligned}
The main idea behind MetiTarski is to approximate transcendental functions by polynomial or rational function bounds, and then solve the formula by a combination of a resolution theorem proving and an external Real Closed Field (RCF) decision procedure (QEPCAD, Mathematica or Z3). MetiTarski is a version of Joe Hurd’s Metis prover [19], modified to include arithmetic simplification and integration with RCF decision procedures, along with many other refinements.
Applications of MetiTarski include verification problems arising in air traffic control [13] and analogue circuit designs [11]. As some of the applications are safety critical, it is natural to consider to integrate MetiTarski with an existing interactive theorem prover, whose internal logic can be used to ensure the correctness of MetiTarski’s proofs. Besides, the automation provided by MetiTarski is generally useful to interactive theorem provers.
MetiTarski has been integrated with the PVS theorem prover [28] as a trusted oracle [12]. The authors state that the automation introduced by MetiTarski for closing sequents containing real-valued functions considerably outperforms existing tactics in PVS. However, this tactic should not be used in a certification environment, where external oracles are not allowed.
Our eventual goal is to integrate MetiTarski into the Isabelle/HOL theorem prover. Isabelle can verify purely logical inferences (in fact, it contains an internal copy of the Metis theorem prover), and the third author has just formalised most of the bounds of transcendental functions used by MetiTarski [30]. The primary remaining hurdle is the RCF decision procedure, and the work presented here is the first step towards it.
Finally, let us say a bit about how our work might be generalised to multivariate problems. In doing so, we plan to continue our certificate-based approach, as we are unlikely to implement a verified internal CAD procedure comparable in efficiency to a state-of-the-art implementation. It is still not obvious to us where the clear separation between search and verification should be in the multivariate case, but we have already made some progress:
• The bivariate sign determination procedure based on recursive application of the Sturm–Tarski theorem described in our previous work [21] can be easily generalised to a multivariate one (i.e., a procedure to decide the sign of a multivariate polynomial at real algebraic points), which can be then used to efficiently certify purely existential multivariate formulas over reals.
• Our recent formalisation of Cauchy’s residue theorem [20] can be used to certify a key theorem used in general CAD: that the complex roots of a polynomial continuously depend on its coefficients.
## 10 Conclusion
We have described our work of building a procedure for first-order univariate polynomial problems in Isabelle/HOL. Compared to existing tactics among proof assistants, noticeable features of our tactic are
• It is based on univariate cylindrical algebraic decomposition (CAD).
• It sceptically integrates efficient external solvers in a certificate-based way, so that its soundness solely depends on Isabelle’s logic (and code generation machinery) rather than the external solvers.
This is made possible by certificate-based approaches to real root isolation and sign-determination for evaluating polynomials at real algebraic points. As much of the novelty in our work is motivated by practical efficiency considerations, we have performed experiments comparing our procedure with another real algebraic proof procedure, the tarski method in PVS. By making use of efficient external solvers, our procedure is shown to empirically outperform this other method by substantial margins. We believe this adds further impetus to the certificate-based methods for a wide variety of formal proof procedures.
Certificate-based methods can be compared on the basis of how much mathematics and computation are required both to find and check their certificates. For example, to convert a Positivstellensatz certificate into a HOL-Light proof of a universal theorem, Harrison’s sums-of-squares tactic only requires simple sign-based reasoning and rational arithmetic, while in our case, we need more mathematics (e.g., real algebraic numbers and the Sturm–Tarski theorem) and more computation (especially for the universal case). A good certificate design needs to balance the difficulty of the formalisation effort and verified computation required to check the certificates with the efficiency improvements offered by offloading the construction of the certificates to high-performance external tools.
## Footnotes
1. 1.
2. 2.
As our tactic is computationally intense, our procedure makes use of the proof by reflection technique [16].
3. 3.
Besides the application described in this section, the Cauchy index also plays a critical role in the Routh–Hurwitz theorem. Interested readers may consult [32, Chapter 10, 11] for historical notes.
4. 4.
In fact, their tactic does not handle arbitrary boolean expressions like ours, but we believe this should not be too hard to overcome.
5. 5.
6. 6.
## Notes
### Acknowledgements
We thank Florian Haftmann for helping with code generation for our procedure. We are also grateful to the anonymous referees for their constructive suggestions.
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Thiemann, R., Yamada, A.: Algebraic numbers in Isabelle/HOL. Archive of Formal Proofs (2015). http://isa-afp.org/entries/Algebraic_Numbers.shtml. Formal proof development | 12,082 | 46,232 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-39 | longest | en | 0.881847 |
https://bestincentivetours.com/safe-places/how-long-does-a-bullet-travel.html | 1,659,958,136,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570793.14/warc/CC-MAIN-20220808092125-20220808122125-00251.warc.gz | 146,794,903 | 17,034 | # How Long Does A Bullet Travel?
If you’re looking to hit a target at a greater distance, the National Rifle Association recommends setting your angle of elevation to be around 30 degrees above the horizontal. The National Rifle Association estimates that a bullet fired from a 9 millimeter pistol, which is the most popular firearm on Guns.com, may travel up to 2,130 yards, which is equivalent to nearly 1.2 miles.
Where can I find out how far a bullet is capable of traveling? According to Paskiewicz, it has a range of around 1.5 miles at an altitude of 12,000 feet. There are a lot of things that can affect where a bullet goes, such as the wind, obstructions, the weight of the bullet, and its trajectory.
## How fast does a bullet travel?
Different calibers, which refer to the diameter of the bullet, are available, and caliber is, to some extent, what determines how quickly a bullet travels.When it leaves the barrel, a bullet of a caliber 22 lr travels at a speed between 1200 and 1750 feet per second (820 to 1200 miles per hour).The velocity of the larger.50 caliber bullet will be anywhere between 2800 and 3150 feet per second (1900 to 2100 mph).
## How far does a 40 caliber bullet travel?
The muzzle velocity of a 40-caliber bullet will determine how far it travels after leaving the barrel.The majority of rounds with a caliber of 40 have a velocity that ranges between 1200 and 1400 feet per second (FPS).When launched from a height of five feet above the ground and under ideal conditions, the bullet is capable of traveling up to 6,600 feet.The 45 caliber bullet is another common choice for handgun ammunition.
## How far does a 22 pistol bullet travel?
How far does a bullet fired from a.22 caliber pistol travel?The distance traveled by a bullet fired from a handgun with a caliber of 22 mm will also be determined by a number of other factors.In any case, the greatest range of the 22-caliber weapons is one mile.On the other hand, when this occurs, a rifle is utilized to fire the bullet.If you’re using a pistol, you’ll need to go closer to your target.
## Why do Bullets travel further when fired from longer barrels?
When it comes to the distance that bullets travel, those that are fired from long barreled firearms travel at a faster rate and cover more ground. Therefore, the velocity of the bullet directly correlates to its range. The reason behind this is that it requires some time for air resistance to catch up to a quicker bullet.
We recommend reading: When Is The Best Time To Travel To Bali?
## What is the fastest a bullet will travel?
The speed of the quickest bullet is calculated to be 2600 feet per second.The most fascinating aspect of this information is that a bullet may move twice as fast as sound.Is there a bend in the path of the bullet?When traveling over shorter distances, it moves in a straight line; but, when traveling over relatively longer distances, a number of variables may influence whether or not it moves in a straight line.
## How far can a bullet from a rifle travel accurately?
When this is taken into consideration, the distance might be anywhere from a few hundred yards to several miles.Low velocity rifle rounds like the.22 LR may be fired accurately out to around 50 to 200 yards, while subsonic or barely supersonic pistol rounds can be fired correctly out to about the same distance.The majority of centerfire rifle rounds have rather precise ranges of two thirds of a mile or farther, sometimes even further.
## How far does a bullet travel went shot into the air?
This is especially true when the conditions outside are ideal and the shot is taken at an angle that is absolutely horizontal. Therefore, the correct answer to the question of how far a bullet fired from a 308 rifle can go is between 800 and 4600 yards. However, if you shoot while the projectile is in the air, it will go a greater distance—anywhere from 3.5 to 4.5 miles. | 865 | 3,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-33 | longest | en | 0.907185 |
https://healthunlocked.com/headway/posts/131379339/puzzles-week-3 | 1,477,137,944,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718957.31/warc/CC-MAIN-20161020183838-00104-ip-10-171-6-4.ec2.internal.warc.gz | 853,916,076 | 36,137 | How old?
A friend wrote you the following letter:
"Our baby girl Nat was born last year in March. If you multiply her age by those of her older sister Kat and brother Pat, the product is 36. If you add all of their ages together, you'lll get a sum of 13. Everyone is healthy and happy. Write back soon!"
What are the ages of Nat, Kat, and Pat?
3 hints if you need them, gogogo.
### 13 Replies
• Nat isn't 1 yet
Kat is 9
Pat is 4
I think, Janet x
• Afraid not, sorry Janet! Nat a has to be a component in the maths here, so if she's zero or say 6 and a half or anything like that it wouldn't work out as 13...
• If she was born in March 2013 she must be...
• My brains gone into meltdown now so I'll look at it later xxxxxx Janet
• No right so if Nat is 1 then Kat and Pat must be twins and are both 6 years old xxxx Janet
• Well done Janet
I was making it way more complicated eg if Nat was born March 2013 she was 9 months old, so I was trying to work it out in months then convert to years. I got very lost
• I get to think more clearly now. Nat must be at 1. Pat and Kat must be 6. 1x6x6=36. 1+6+6=13.
I admit though, I did have a quick look at Janet's answer.
Janet truely worked this one out :).
• Better late than never...
Nat can only be 1, so that makes pat and kat 6 year old twins.
• Hey look I'm right, but I could have known Janet would've pipped me to this one.
• Sorry Barny, I get really competitive with puzzles,was a member of Mensa years ago, and I've been very worried my brain injury might affect that working of my brain so I work very hard at maintaining it. I'm going to leave the next few alone, keep the answers to myself xxxxx Janet
• Don't be Janet, I'm not upset or anything and I wasn't trying to be the first to answer. Actually I'm quite impressed that I'm on par with someone who was a member of Mensa lol
• The thing is Barny, it's practice with puzzles and thinking abstractly, not going for the obvious.
When I was in primary school many moons ago!, our form teacher used to give us an IQ test every week so I got lots of practice and learned a lot of the abstract ways they set questions. Xxxxx Janet
• IF YOU STILL WANT TO SOLVE THE PUZZLE DON'T SCROLL DOWN!
Blah blah blah blah blah blah 1+6+6 is correct! Well done. | 605 | 2,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-44 | longest | en | 0.985391 |
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## Karnataka 2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.1
Part – A
2nd PUC Basic Maths Mathematical Logic Ex 6.1 One Mark Questions and Answers
Question 1.
Symbolise the following propositions:
(1) 3x = 9 and x<7
(ii) 33 + 11 ≠ 3 or 8 – 6 = 2
(iii) If two numbers and equal then their squares are not equal.
(iv) If oxygen is a gas then gold is a compound
(v) y + 4 ≠ 4 ore is not a vowel
(i) Let p:3x = 9, q = x < 7
Given in symbols is p ∧ q
(ii) Let p:33 = 11 = 3, q= 8 – 6 = 2
Given is ~p ∨ q
(iii) Let p: Two numbers are equal, q: Squares are equal then given is p → ~q
(iv) Let p: Oxygen is a gas
q: Gold is a compound
Then given is p → q
(v) P: y + 4 = 4, q: e is a vowel given proposition is ~p v ~q
Part – B
2nd PUC Basic Maths Mathematical Logic Ex 6.1 Two or Three Marks Questions and Answers
Question 1.
if p, q and r are propositions with truth values F, T and F respectively, then find the truth values of the following compound propositions:
(i) (~p → q) ∨ r
(ii) (p ∧ ~q) → r
(iii) p → (q → r)
(iv) ~(p → q) ∨~(p ↔ q)
(v) (p ∧ q) ∨ ~ r
(vi) ~(p ∨ r) → ~q
(i) (~p → q)∨ r
(~F →T) ∨ r
(T →T) ∨ F
T ∨ F
= T
(ii) (p ∧~q) → r
(F∧ ~T) → F
(F ∧F) → F
F →F
= T
(iii) p → (q → r)
F →(T →F)
F →(T →F)
= T
(iv) ~(p → q) ∨~(p↔ q)
~(F →T) ∨ ~(F↔T)
~T ∨ ~F
F ∨ T = T
(v) (p ∧ q) ∨ ~ r
(F ∧ T) ∨ ~ r
F ∨ T
= T
(vi) ~(P ∨ r) → ~ q
~(F ∨ F) ∨ ~ T
~F → ~ T
T → F
= F
Question 2.
(1) If the compound proposition “(p → q) ∧ (p ∧ r)” is false, then find the truth values of p, q and r.
(ii) If the compound proposition p→ (q ∨ r) is false, then find the truth values of p, q and r.
(iii) If the compound proposition p → (~q ∨ r) is false, then find the truth values of p, q and r.
(iv) If the truth value of the propositions (p ∧ q) → (r ∨ ~s) is false, then find the truth values of p, q, rand s.
(i) Given (p → q) ∧ (p ∧ r) is false
(a) Case 1: p → q is true & p ∧ r is false
p is T q is T & p is T &ris F
p = T, q = T, r = F
Case 2(a): p = F, q = T p ∧ r= F → p = F, r = F
p = F, q = T, r = f
(b): (p →q) is F & par is true
T → F = F
T ∧ T is T
P=T, q = F, r=T
Case 3: (p → q) is F & (p ∧ r) is false
T → F = F
F ∧ F = F
F ∧ T = F
F ∧ F = F .
∴ p = T, q = F, r= F.
(ii) Given p → (q ∨ r) is false
T → F = F
∴ p = T & q ∨ r is false =
F ∨ F= F
∴ p = T, q = F & r= F
(iii) Given p → (q ∨ r) is false
Then T → F= F
∴ P = T, ~ q ∨ r= F
F ∨ F = F
∴ q = T, q = T, r = F.
(iv) Given (p ^ q) → (r ∨ ~s) is false
We know that T → F = F
∴ p∧q = T and r ∨ ~s = F is false
T∧ T = T
F ∨ F= F is false
∴ p = T, q = T, r = F, S = T | 1,139 | 2,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-33 | latest | en | 0.695889 |
https://www.unitconverters.net/time/month-synodic-to-second.htm | 1,713,610,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817576.41/warc/CC-MAIN-20240420091126-20240420121126-00858.warc.gz | 928,124,035 | 3,242 | Home / Time Conversion / Convert Month (synodic) to Second
# Convert Month (synodic) to Second
Please provide values below to convert month (synodic) to second [s], or vice versa.
From: month (synodic) To: second
### Month (synodic) to Second Conversion Table
Month (synodic)Second [s]
0.01 month (synodic)25514.4384 s
0.1 month (synodic)255144.384 s
1 month (synodic)2551443.84 s
2 month (synodic)5102887.68 s
3 month (synodic)7654331.52 s
5 month (synodic)12757219.2 s
10 month (synodic)25514438.4 s
20 month (synodic)51028876.8 s
50 month (synodic)127572192 s
100 month (synodic)255144384 s
1000 month (synodic)2551443840 s
### How to Convert Month (synodic) to Second
1 month (synodic) = 2551443.84 s
1 s = 3.9193494456848E-7 month (synodic)
Example: convert 15 month (synodic) to s:
15 month (synodic) = 15 × 2551443.84 s = 38271657.6 s | 302 | 850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.672961 |
https://capitalsportsbet.com/ZuluCode5/how-to-interpret-sports-betting-lines.html | 1,568,913,608,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573561.45/warc/CC-MAIN-20190919163337-20190919185337-00206.warc.gz | 401,108,431 | 10,615 | In theory, sportsbooks don't care about the outcome of a game, although for those of you who bet on the Steelers (-5.5) last season versus the Chargers and saw a game winning TD returned by S Troy Palumalu as the game expired reversed, thus negating a seven point victory and putting the final at 11-10, you might think otherwise, but this is how it's supposed to be.
Here you can see that the Rams are +3.5, while the Cowboys are -3.5. So for this example the Cowboys are 3.5 point favorites, while the Rams are underdogs of 3.5 points. If you were to bet on St Louis you would need them to lose by 3 or fewer points or just win the game outright. If you were to bet on Dallas you would need the Cowboys to win by 4 or more points.
In most cases, the favorite will be the team with a negative moneyline (in some cases both teams can have a negative moneyline if they are both closely matched). A line of -160 means that you would have to bet \$160 to win your base amount of \$100. A team with a moneyline of -130 wouldn't be favored nearly as strongly as a team with a moneyline of -330.
Something that all of our pro bettors on staff preach is that there is no reason to make anything more complicated than it needs to be. Specifically, they are referring to the awful trend of aspiring sports bettors thinking that you must make more complex bets in order to make a living betting sports. The reason we say it’s an awful trend is because this couldn’t be further from the truth.
```Armed with the knowledge of how to remove vig, it's now possible to prevent yourself from making the same mistakes that the majority of bettors make. Most bettors understand the importance of line shopping (i.e. comparing the lines and odds at different bookmakers and betting sites). However, if they fail to also understand how moneylines and vig work, then they're probably going to make wagers where they think there's positive expected value (+EV), even though there's not.
```
Let's look at a different option. We bet the Cowboys +3, and the final score is Chargers 21, Cowboys 19. Even though the Cowboys lost the game, we still win our bet because they lost by less than three points. How much are we going to get paid on this bet? Well, we look in the parenthesis and see we will get paid at -120. This means that for every \$100 we bet, we will get paid \$83.33 in profit.
A point spread (or line) is a tool used by sportsbooks to attract wagers on both sides of a game. The line is most commonly used in football and basketball games. Because it’s rare for two teams in a pro sports game to be completely evenly-matched, one team will have an advantage, another will be seen as the underdog. The point spread is the handicap offered to the underdog to level the playing field, so to speak.
Futures are wagers on long term events which generally won’t have a result for many weeks or even months. The most popular type of futureswager is placed on the outright market which basically means, who will win a league’s championship? FanDuel also offer many other markets outside of the championship, usually including divisional and conference wagering. The odds on futures change as the event gets closer and more is known about the teams. Very often odds will also be updated as the tournament or league progresses.
You have three choices for the three betting options: Home, Away, or Draw (tie). The result of the game is decided after regulation play (90 minutes plus injury time). Overtime, the Golden Goal rule, and penalty kicks are not taken into consideration for soccer bets unless otherwise stipulated. You can usually bet on a winner or advancement (including OT & shootouts) but with different odds would be given.
Almost all point spread wagers are paid out at moneyline odds of -110. This is almost even money minus the percentage that is taken for the sportsbook's cut known as the vig. Sometimes you will see a bit of variation in the payout odds, but for the most part, you should expect to see -110. If you don't see the payout numbers posted but just the point spread, you can most likely assume that you are to interpret that as being paid out at -110. If you're ever curious, though, just ask for clarification or look at your betting slip.
This arrangement tells us a lot: which team is home (listed on bottom, in this case the Seattle Seahawks), which team is the underdog (listed with a plus sign next to their name, in this case the New England Patriots), we know which team is the favorite (listed with a minus sign next to their name, in this case the Seattle Seahawks), and we know the point spread (2.5 points).
Above, you can see several numbers to the right of both teams. These all represent the different lines that are available on the San Francisco vs. Los Angeles game. The first set of numbers for both teams is the point spread, the second set is the moneyline, and the third set is the over/under (a.k.a. totals). We'll explain each of these lines more in-depth below.
There's another reason to bet the underdogs on the moneyline as well. If your handicapping has made you feel very strongly that a poor team is due for a big win then the moneyline allows you to profit much more handsomely from your conclusion than a point spread bet does. The moneyline, then, is a powerful situational tool for people who closely follow the NBA.
Had you initially bet \$100 on the Mavericks, you would walk away with a profit of \$90.91. Now, if you bet on the Mavericks, you will see a profit of only \$76.92. On the other side of the bet, an earlier bet on the Magic would have paid you \$105. Now that same \$100 bet will get you \$125 if the Magic win the game. Ideally, this will entice more people to bet on the Magic and the action on the Mavericks to slow down.
You could even take it a step further and take the next rectangle down and bet Liverpool +1. This means that Liverpool can tie or win by any amount of goals and you win your bet. As Liverpool is a huge favorite, you won't be paid very well at all for this bet, but you can still turn a profit when you are right. You would be paid at 1 to 10 which means you would get \$1 for every \$10 you bet. If you bet \$100, you would get a \$10 profit on this bet.
This used to be impossible or extremely time-consuming when only land based sportsbooks and casinos existed. You would have to drive hundreds of miles if you wanted to get to another casino and then hope they still had the different line. This would cost you travel money as well as time. Online sportsbooks make this extremely easy now. You can check several sportsbooks lines on a game within a matter of seconds or minutes. It doesn't cost you any additional money, just a few minutes of your time and can have a huge impact on the outcome of your bets.
Buying points: Changing the point spread in order to favor your chances of winning. The odds here are proportionally lowered. Buying points allows the option of moving the point-spread on totals in your favor when betting on football or basketball (college and pro). You’re able to move the line up to 2 points. For every ½ point you want to buy, you must risk 10% (10 cents) in extra juice (except when buying on/off of a 3 in NFL).
You can bet the money line option in every single sport that is offered up. It is the simplest form of betting and it is also the primary way to bet sports in which a point spread isn’t available (think hockey or baseball). Money line wagers are also available in football and basketball, but the point spread wager is much more popular. It is also used in tennis, golf, boxing, MMA, cricket, table tennis, and any other sport you can think of that has a winner at the end of the game.
While we aren't exactly sure at which dollar amounts or what formulas sportsbooks use to determine when they shift the lines, we do know why they do it. It is their attempt to minimize their risk as much as possible and guarantees sportsbook profit. Lines will also move if something major happens (like Lebron breaking his leg or something) so keep an eye out for this. Ultimately, the shift in the line is done for the exact same reason to keep the same amount of money on both sides of the game. As you'll see in the strategy and tips section, shifting lines do present some interesting opportunities for sports bettors.
For example, in a cricket match a sports spread betting firm may list the spread of a team’s predicted runs at 340 – 350. The gambler can elect to ‘buy’ at 350 if they think the team will score more than 350 runs in total, or sell at 340 if they think the team will score less than 340. If the gambler elects to buy at 350 and the team scores 400 runs in total, the gambler will have won 50 unit points multiplied by their initial stake. But if the team only scores 300 runs then the gambler will have lost 50 unit points multiplied by their initial stake.
There are four elements to the moneyline bet that you can see here. The first column is just an identifier of which bet is which for the sportsbook. When you place your bets, you can tell the sportsbook you want to bet on the Eagles to win or that you want to take bet 055. This number has nothing to do with the actual game and is just a code for the sportsbook to keep their bets organized.
Understanding how a moneyline wager pays isn’t simple but it’s not very complicated. That said, it might take running through a few examples before fully grasping the payouts. Moneylines for football and basketball games are often tied to the point spread. When a game has a large point spread it usually has a large moneyline. Both are separate bets but are shown together in a sports wagering app screen and in a brick and mortar sportsbook.
# --Fractional odds are most commonly found in racing. A 10/1 payout should be read "\$10 paid for every \$1 wagered." When the bigger number is on the left, you will find that bet is normally an underdog in the race. Also note, however, that in case such as "Who will win the Super Bowl in the NFL?" you will see all the teams listed as "underdogs"… i.e. paying at least 2/1 (some up to 300/1 or more).
Single day matchups are wagers on the complete 18-holes for that day. Holes played as part of a completion from the previous day's round, and playoff holes are not included in Single Day matchups. The full 18 holes will be considered in determining the outcome of the bet even if they are played on consecutive days. If both members of the matchup do not complete the full 18 holes, all bets are "No Action". If both players end the 18 holes in a tie, the bet is considered a "No Action" wager.
Remember earlier when we said that most point spread bets in basketball pay out at -110? Well, this is where the vig is located. Sportsbooks will work to get equal amounts of money on both sides of a game and make their money off of the vig. If they are successful in doing so, it does not matter to them who wins the game. For example, let's look at our earlier example. Here are what the odds would look like at the sportsbook:
In most cases, the favorite will be the team with a negative moneyline (in some cases both teams can have a negative moneyline if they are both closely matched). A line of -160 means that you would have to bet \$160 to win your base amount of \$100. A team with a moneyline of -130 wouldn't be favored nearly as strongly as a team with a moneyline of -330.
Before we had the options of wagering on future bets, parlays, teasers, alternative lines, Asian lines, prop bets and each-way, there was one betting option that reigned supreme. It was the money line bet. From a non-gambling perspective, winning a game in any sports will make a team happy. However, depending on the score, that win may not please bettors. That’s because the point spread betting option has taken over as the popular option, leaving the money line far behind. If you are the kind of person who bets on your favorite team each and every game, this is the bet for you. There is nothing worse than watching your team win the game, but lose you money by not covering the point spread.
When the point spread was invented in Chicago by Charles McNeil the money line took a backseat. When two unevenly matched teams played, the playing field was leveled by having the favorite give points (for example Chicago Bears –7) while the underdog got points (Minnesota Vikings +7). No matter which team the bettor took the bettor would always risk \$110 to win \$100. The extra \$10 needed to win \$100 is called the juice or the vig, it is basically the house’s or the bookie’s take. It’s 10-percent of the bet so it would take \$33 to return \$30 and \$440 to return \$400 etc. (winning bettors get the vig back).
You see, people too frequently get caught up on their win-loss rate, which actually has no real effect on their bottom line. You can win more bets than you lose and still be losing money. On the other side of things, you can lose more bets than you win and be wildly profitable at sports betting. It all comes down to successfully finding value and pouncing on it when you see it.
By this point, you should be feeling pretty well versed in all things NBA betting. We’ve walked you through how to use our free expert picks, where to place your bets, the strategies you need that are specific to the NBA, and the different types of bets you have at your disposal. Whether or not you become a successful NBA sports bettor now is up to you. If you study this material, do your research, and put in some hard work, you can be on the road to crushing the books in no time. We wish you the best of luck and are always here if you ever need any additional help.
Changes to the lineup for a game will have a big effect on the moneyline as well as any other bets you’re looking to place on that game. If a superstar is suddenly out, it’s going to have a big change on where the money is coming in, which will inevitably cause a big shift in the line. Sometimes the sportsbook will even adjust the line preemptively if they anticipate a large change in where the money is going to be coming in on a particular game.
Additionally, we’ll discuss line movement, how the casino profits (important for you to understand), and moneyline betting strategies that can help you crush the books. These strategies will range from basic to advanced, so even the most seasoned of sports bettors should expect to get some value from this. Feel free to skip to a specific section if you came here for specific information. If you’re newer or it’s been a while since you’ve bet, we highly recommend reading this guide from top to bottom, as the sections will build on knowledge from previous sections.
```Unless otherwise stated all handicaps listed on DraftKings site are to be calculated based on the result from the start of the listed period to the end of the specified period. It is however customary that for certain handicap odds in specific sports (Asian Handicap in Soccer), only the outcomes obtained from the time of bet placement until the end of the listed timeframe will be taken into consideration, thus disregarding any points scored before the time the bet was placed and accepted. Any odds with these characteristics will be clearly displayed on site and highlighted in the user’s Bet History with the score at the time of bet placement.
```
Proline will set a point spread for each game of either -0.5 / +.05, -1.0 / +1.0, or -1.5 / +1.5. If you choose -0.5 the team must win by 1 goal or more to win your wager. If you choose +0.5 your team must win or tie the game. Note that OLG Proline includes the shoot-out for point spreads so there are never any ties for the -0.5 / +0.5 spreads. If you choose the -1.0 spread your team must win by 2 or more goals to win the bet. If they win by 1 goal exactly it is a push. If you choose the +1.0 spread your team must win the game by 1 or more goals. If they lose by 1 goal it is a push. If you choose the -1.5 spread your team must win by 2 or more goals. If you choose the +1.5 spread your team must win the game, tie, or lose by 1 goal to win the bet.
82 games per year means a lot of opportunities to wager on your favorite teams and Sportsbetting.ag sees that opportunity and gives you what you want to bet on. Sometimes this means having spreads for big games a day or two in advance, especially if both teams are off the day before the big matchup. They take care of their NBA bettors... give them a shot.
In most football games there is a favorite and an underdog. Very occasionally there are games where the two team are completely evenly matched, but for the most part one team is favored over the other to win. With point spreads, the idea is to create an even money proposition when betting on the game. So the favorite has to win by at least a certain number of points for a wager on them to be successful, and the underdog has to lose by no more than a certain number of points for a wager on them to be successful. The bigger the gap in quality between the two teams, the bigger the point spread.
There are some major differences between the two that all punters should be aware of. One of the great things about betting online is the offering of betting tutorials. With these, the newest punters can learn all about the different types of bets, how to place them and what odds they carry for winning. For those that are looking to enjoy amazing betting options and superb odds, there are many trusted and respected online casinos in canada that offer sportsbook action and awesome chances to collect payouts.
There is no such thing as a half point in sports, but there is in sports betting! The half point ensures that a side will win and that the match will not end in a push, where the spread equals the actual difference in points between the two teams. In a push all bettors get their money back, which is no good for the oddsmaker! Half points also give oddsmakers more control over lines, allowing them to set more specific values.
Before we had the options of wagering on future bets, parlays, teasers, alternative lines, Asian lines, prop bets and each-way, there was one betting option that reigned supreme. It was the money line bet. From a non-gambling perspective, winning a game in any sports will make a team happy. However, depending on the score, that win may not please bettors. That’s because the point spread betting option has taken over as the popular option, leaving the money line far behind. If you are the kind of person who bets on your favorite team each and every game, this is the bet for you. There is nothing worse than watching your team win the game, but lose you money by not covering the point spread.
```The 2-way moneyline is what most North American bettors would simply refer to as “the moneyline”. This is one of the most common wagering options where the user bets which side will win the game straight up. (A draw or tie results in a push with the 2-way moneyline.) The term is sometimes highlighted during soccer betting to differentiate from the 3-way moneyline - a more popular option with the draw added as a wagering option.
```
It is a very good idea to shop around to find the best line when you are betting on sports. Books may offer slightly different lines, and you might be able to gain a point or half a point in your favor on certain markets. Consider an NBA game between the Heat and the Lakers – one firm has the Heat 9.5 point favorites and another have the Heat as 8.5 point favorites. You back the Heat with the second firm and they win by exactly nine points. Here you’ve gone from a loss to a win simply by shopping around, and in the long run this will make a big difference to your bottom line.
Let's use this formula to calculate the implied probability of the Celtics winning their game against the Grizzlies. We know the odds are -240, which means we'd have to risk \$240 for a total potential return of \$340 (the initial stake plus the \$100 winnings). So the calculation here is \$240 divided by \$340. This gives us an implied probability of 0.7059.
If you want to bet on football, then you have plenty of options. There are not only lots of games you can bet on, there are also lots of different types of wagers you can place. Point spreads and totals are the most popular types, by quite a distance, and many football bettors stick solely to those. This isn't really the ideal approach, as some of the other wagers can be very useful in the right circumstances.
Doc's Sports is offering \$60 worth of handicapper picks absolutely free - no obligation, no sales people - you don't even have to enter credit card information. You can use this \$60 credit any way you please for any handicapper and any sport on Doc's Sports Advisory Board list of expert sports handicappers. Click here for more details and take advantage of this free \$60 picks credit today .
If your sports betting experience consists mostly of office pools during March Madness or a casual wager between you and a friend while you watch the Super Bowl, the transition to serious sports betting means learning how to read betting lines. The biggest difference between making the kind of casual bets mentioned above and placing wagers with online sportsbooks or at brick-and-mortar bookshops is the use of sports betting lines. Casual wagers usually involve each person in the bet picking one team to win, then wagering an equal amount, say \$20 or \$30. Professional bookmakers, online sports betting exchanges, and sports betting facilities in casinos have a more complex system for offering wagers on sporting events, in part to ensure profit on the part of the book, and in part to present a standardized representation of odds.
A teaser is a bet that alters the spread in the gambler's favor by a predetermined margin – in American football the teaser margin is often six points. For example, if the line is 3.5 points and bettors want to place a teaser bet on the underdog, they take 9.5 points instead; a teaser bet on the favorite would mean that the gambler takes 2.5 points instead of having to give the 3.5. In return for the additional points, the payout if the gambler wins is less than even money, or the gambler must wager on more than one event and both events must win. In this way it is very similar to a parlay. At some establishments, the "reverse teaser" also exists, which alters the spread against the gambler, who gets paid at more than evens if the bet wins. | 5,025 | 22,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-39 | latest | en | 0.974305 |
https://statisfaction.wordpress.com/2010/07/16/using-graphic-cards-to-perform-statistical-computation/ | 1,531,950,169,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00367.warc.gz | 768,331,841 | 21,660 | # Statisfaction
## Using graphic cards to perform statistical computation
Posted in Geek by Pierre Jacob on 16 July 2010
Hi!
For my first post I’m going to blog about a trendy topic in computational statistics: the increasingly common use of graphic cards to perform any kind of scientific computation (see here) and in particular statistical computation.
The fun thing about graphic cards (fun in a weird and definitely nerd way) is that they were built to display pictures (obviously), mainly for movies and video games. These cards became more and more powerful in the 90s and until now, helping the games to achieve an always higher level of photorealism. But I’m not sure their early developers realized how much they could be useful in scientific computation.
In the last years it has become clear that, for a limited amount of money, say 300\$, you can get much more computational power if you buy a GPU (a graphic card processor) than a CPU (a standard processor). This gain can be of order 100 or even more!… meaning a program that takes several hours on a standard CPU can be run in less than a minute on a GPU. Of course it doesn’t work for any computation (otherwise GPU would simply replace CPU and there would be nothing to blog about).
Compared to a CPU, a GPU is extremely good at doing parallel computation. It means that it can do hundreds of little tasks at the same time, where a standard CPU with two cores can do… two tasks, basically. Algorithms that need to do a lot of independent little tasks can therefore be parallelized, other algorithms cannot.
• For example, if you have a vector $(x_0, x_1, \ldots x_n)$ and want to take the log of each component of this vector, you can do it in parallel, because you don’t need to know the result of $log(x_i)$ in order to compute $log(x_j)$, for any $i \neq j$. Similarly, evaluating a likelihood of a vector of independent variables can also be parallelized.
• On the contrary, suppose you want to compute Fibonacci numbers, which are defined by $F_0 = F_1 = 1$ and the recurrence relation: $F_n = F_{n-1} + F_{n-2}$. Then you obviously need to compute the n-th term before computing the n+1-th term. The program doing this computation would therefore be sequential. Any algorithm requiring a recurrence is thus hard to parallelize, and that includes for instance the Metropolis-Hastings algorithm. Though I’m working on using parallel computation to improve Metropolis-Hastings based estimation… maybe more about this in a future post.
My PhD focuses on Sequential Monte Carlo methods (SMC), also known as Particle Filters, which happen to be very easy to parallelize. Hence my interest in the topic! Instead of waiting hours for my computations to finish, I could have the results in (near) real-time. But then I would have no excuse for playing table football
If you are interested, there are plenty of links to check. Owning a NVIDIA device, I use CUDA and its python counterpart, PyCUDA. Apparently you can also use CUDA in R with this. There is also a MATLAB plugin. An alternative is the open standard called OpenCL, that allows you to program NVIDIA and ATI devices (those two being the main graphic card designers), so that’s probably the best solution, or at least will be in the future. In the statistical literature, there has been recent articles, like this one in the SMC context. This website from the University of Oxford gives lots of links as well, and this one too from Duke University. Programming on a GPU already interests a lot of people, namely those who have intensive computations to perform (financial engineers, physicists, biologists…), I hope you’ll find it useful too.
Stay tuned for more exciting posts!…
### 7 Responses
1. Julyan said, on 16 July 2010 at 12:15
I like the surreptitious link to table football!
2. Aledine said, on 3 September 2010 at 03:50
Interesting and many thanks for the links. I am wondering how much computational gain can be achieved by running several video-cards in SLI or Crossfire mode and overclocking them.
• pierrejacob said, on 3 September 2010 at 19:09
Good question, I don’t know! Theoretically two GPUs would allow twice as many cores, and twice as much GPU memory. However the bandwidth between the “normal” memory and the GPUs is probably unchanged so filling the whole GPU memory would take twice as long. Anyway multiple GPUs are totally supported by CUDA C library so it should provide a significant gain.
3. […] Metropolis-Hastings (IMH) algorithm that can highly benefit from parallel processing units (see this previous post for an introduction on parallel […]
4. […] it’s my temporary hobby (see here for instance), I’ve coded the Box-Muller transform with pyCUDA to take advantage of the GPU. The result is […]
5. […] Using graphic cards to perform statistical computation […]
6. […] of graphics processing units in statistics, a quickly expanding area that I’ve blogged about here. Xian and I are going to talk about Parallel IMH and Parallel Wang Landau. We’ll be able to […] | 1,135 | 5,050 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-30 | latest | en | 0.943106 |
https://math.stackexchange.com/questions/2635036/induced-maps-in-local-homology-and-isotopies | 1,685,276,647,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00162.warc.gz | 456,086,731 | 33,989 | # Induced maps in local homology and isotopies
Let $f,g: M \to M$ be two continuously isotopic homeomorphisms, where $M$ is an orientable manifold, not necessarily compact. We pick an orientation on $M$, where by an orientation I mean a continuous (in the sense explained here) choice of generators for the local homology groups $H_n(M, M - \{x \} )$, where $n = dim \, M$.
I want to prove that if $f$ is orientation preserving in the sense mentioned above, then $g$ must also be orientation preserving. The analogous result for diffeomorphisms and smooth isotopies is easy to prove, since we can use the continuity of $sgn \, (d_xf_t)$. However, I'm having some difficulty in the topological case:
First, it's not clear to me how to prove (assuming it's true, as in the smooth case) that the fact that $f$ is orientation preserving at a point $x \in M$ implies that it is orientation preserving at every point. That is, I'd like to prove that if for some $x$ in $M$ and $f_*: H_n(M, M - \{ x\}) \to H_n(M, M - \{ f(x)\})$ we have that $f_*([M]_x) = [M]_{f(x)}$, then the same holds for every point in $M$ (here $[M]_x$ is our selected generator for $H_n(M, M-\{ x\})$).
Second, I'd have to show that $f_*([M]_x) = [M]_{f(x)}$ for some $x$ implies $g_*([M]_x) = [M]_{g(x)}$. If it were the case that for every $t \in [0,1]$, $f_t(x) = g(x) = f(x)$ then it would be easy, since we'd have $f_* = f_{t*} = g_*: H_n(M, M - \{ x\}) \to H_n(M, M - {f(x)}) = H_n(M, M - \{ g(x)\})$ by using the maps induced in relative homology by the $f_t: (M, M - \{ x\}) \to (M, M-\{ f(x)\})$.
Since that doesn't happen when $f(x) \neq g(x)$, it's not as simple (at least for me!) as looking at the induced maps and using homotopy invariance like above.
Any hints or comments would be appreciated. Thanks!
First, it's not clear to me how to prove (assuming it's true, as in the smooth case) that the fact that $f$ is orientation preserving at a point $x \in M$ implies that it is orientation preserving at every point. That is, I'd like to prove that if for some $x$ in $M$ and $f_*: H_n(M, M - \{ x\}) \to H_n(M, M - \{ f(x)\})$ we have that $f_*([M]_x) = [M]_{f(x)}$, then the same holds for every point in $M$ (here $[M]_x$ is our selected generator for $H_n(M, M-\{ x\})$).
This is true as long as $M$ is connected. For any $x\in M$, there is some open set $U$ containing $x$ and an element $[M]_U\in H_n(M,M-U)$ such that for all $y\in U$, the image of $[M]_U$ in $H_n(M,M-\{y\})$ is the generator $[M]_y$ chosen by our orientation. Shrinking $U$, we may assume that there is also such a class $[M]_{f(U)}\in H_n(M,M-f(U))$. We may further shrink $U$ to assume $U$ is an open ball in some neighborhood of $x$ homeomorphic to $\mathbb{R}^n$ so that the map $H_n(M,M-U)\to H_n(M,M-\{y\})$ is an isomorphism for each $y\in U$ (and similarly for $f(U)$) and so the elements $[M]_U$ and $[M]_{f(U)}$ are unique.
We then have that for any $y\in U$, $f$ sends $[M]_y$ to $[M]_{f(y)}$ iff $f_*:H_n(M,M-U)\to H_n(M,M-f(U))$ sends $[M]_U$ to $[M]_{f(U)}$. The latter condition does not depend on the point $y\in U$. Thus if $f$ is orientation-preserving at $x$, it is orientation-preserving in a neighborhood of $x$, and similarly if it is orientation-reversing. So the sets of points where $f$ is orientation-preserving and where $f$ is orientation-reversing are an open partition of $M$. If $M$ is connected, one of these sets must be all of $M$.
Second, I'd have to show that $f_*([M]_x) = [M]_{f(x)}$ for some $x$ implies $g_*([M]_x) = [M]_{g(x)}$.
It suffices to show that $f_*([M]_x) = [M]_{f(x)}$ implies $(f_t)_*([M]_x) = [M]_{f_t(x)}$ for all sufficiently small $t$. Fixing a neighborhood $V\cong\mathbb{R}^n$ of $f(x)$, there is an open neighborhood $U$ of $x$ and an $\epsilon>0$ such that $f_t(U)\subseteq V$ for all $t\in[0,\epsilon]$. We may as well assume $\epsilon=1$, $U$ is our entire domain, and $V$ is our entire codomain. That is, we may as well assume we have a homotopy consisting of open embeddings $f_t:U\to\mathbb{R}^n$ for each $t\in [0,1]$, and wish to show $f_0$ is orientation-preserving at $x$ iff $f_1$ is orientation-preserving at $x$.
But now we can modify our maps $f_t$ so that we can look at just one point: let $f_t'(y)=f_t(y)-f_t(x)$. Note that $f_t'$ is orientation-preserving at $x$ iff $f_t$ is orientation-preserving at $x$, since it differs by composition with a translation, which is orientation-preserving on $\mathbb{R}^n$. We also have $f_t'(x)=0$ for all $t$. So, by considering the induced maps $H_n(U,U-\{x\})\to H_n(\mathbb{R}^n,\mathbb{R}^n-\{0\})$, we see that $f_0'$ is orientation-preserving at $x$ iff $f_1'$ is orientation-preserving at $x$. It follows that $f_0$ is orientation-preserving at $x$ iff $f_1$ is orientation-preserving at $x$, as desired.
[In this argument I am using the fact that $\mathbb{R}^n$ has only two different orientations, so that when we orient the codomain $\mathbb{R}^n$ in the same way as our given orientation on $V$, we have either the standard orientation or its opposite. This guarantees that translation is orientation-preserving on $\mathbb{R}^n$, for our chosen orientation. The fact that $\mathbb{R}^n$ only has two orientations is because it is connected, using the first part of the answer above.]
• Thanks, that was very detailed and helpful! It had actually occurred to me that the isotopy could be modified like that when $M = \mathbb{R}^n$, but I didn't realize I could use that locally in the general case... Now I know! Thanks again. Feb 4, 2018 at 6:21 | 1,816 | 5,545 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-23 | latest | en | 0.940163 |
https://tutorialhorizon.com/algorithms/fizz-buzz-challenge-java-implementation/ | 1,702,288,644,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00320.warc.gz | 643,346,981 | 10,169 | Beginner
# Fizz Buzz Challenge - Java Implementation
Objective: Write a program that prints the numbers from 1 to 100. But for multiples of three print "Fizz" instead of the number and for the multiples of five print "Buzz". For numbers which are multiples of both three and five print "FizzBuzz"
Output:
```1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14,
Fizz Buzz, 16, 17, Fizz, 19, Buzz, Fizz, 22, 23, Fizz,….
```
Approach:
• Iterate from 1 to 100.
• If number is multiple of 3, print ‘fizz’.
• If number is multiple of 5, print ‘buzz
• If number is multiple of 3 and 5, print ‘fizz Buzz’
• If none of the above is true, print the number itself.
Java Code:
Output:
`1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, Fizz Buzz,16, 17, Fizz, 19, Buzz, Fizz, 22, 23, Fizz, Buzz, 26, Fizz, 28, 29, Fizz Buzz,31, 32, Fizz, 34, Buzz, Fizz, 37, 38, Fizz, Buzz, 41, Fizz, 43, 44, Fizz Buzz,46, 47, Fizz, 49, Buzz, Fizz, 52, 53, Fizz, Buzz, 56, Fizz, 58, 59, Fizz Buzz,61, 62, Fizz, 64, Buzz, Fizz, 67, 68, Fizz, Buzz, 71, Fizz, 73, 74, Fizz Buzz,76, 77, Fizz, 79, Buzz, Fizz, 82, 83, Fizz, Buzz, 86, Fizz, 88, 89, Fizz Buzz,91, 92, Fizz, 94, Buzz, Fizz, 97, 98, Fizz, Buzz` | 504 | 1,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | longest | en | 0.880773 |
https://pepeitalianstreetfood.com/pizza/often-asked-how-big-is-a-dominos-large-pizza-2.html | 1,631,907,283,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00349.warc.gz | 527,172,927 | 10,176 | # Often asked: How Big Is A Domino’s Large Pizza?
## How big are Dominos large pizzas?
A Large Domino’s pizza is 13.5 inches, with 10 glorious slices of cheesy goodness.
## How many slices are in a 14 inch pizza from Dominos?
Figure Out the Slices Per Size Large pizzas are 14 inches in diameter and will offer approximately 10 slices.
## What size pizzas does Domino’s have?
At Domino’s, we offer all the traditional pizza sizes — small, medium, and large — plus an extra large option. It helps when you’re ordering to know that Domino’s pizza sizes vary by crust type. For example, our Gluten Free Crust is available only as a small pizza, making it the perfect personal pizza.
## How big is a 12 inch pizza from Dominos?
Small Pizza: 8-10 inches with 6 slices. Medium Pizza: 12 inches with 8 slices. Large Pizza: 14 inch with 10 slices. Extra- large Pizza: 16-18 inch with 12 slices.
## Are 2 Medium Pizzas bigger than a large?
A large pizza is almost always a better deal than two mediums. When you increase the width of your pizza it actually adds to the total size of your pie exponentially. For example, a 16-inch large might seem twice as big as an 8 inch small but it’s actually four times as much pizza.
You might be interested: Can You Eat Pepperoni Pizza When Pregnant?
## How many large pizzas do I need for 6 adults?
A Medium 12″ inch Pizza is normally cut into 8 slices and serves 3-4 people. A Large 14″ inch Pizza is normally cut into 8 or 10 slices and serves 3-5 people. An Extra Large 16″ inch Pizza is normally cut into 6 or 12 slices and serves 5- 6 people. An 18″ inch Pizza is normally cut into 6 or 12 slices and serves 6 -7
## How many large pizzas do I need for 10 adults?
10 People = 4 Pizzas. 15 People = 6 Pizzas. 20 People = 8 Pizzas. 30 People = 12 Pizzas.
## How many people will a large pizza feed?
Most large pizzeria pizzas, measuring 14 to 16 inches in diameter, feed approximately four to eight people, depending on how many slices each person eats. Large pizzas typically contain eight to 12 triangular slices or 10 to 12 square slices.
## How much bigger is a 14 inch pizza?
Pizza Comparison (old version)
Diameter (in) Area (sq in) Times bigger than 10in
12 113 1.4
14 154 2.0
16 201 2.6
18 254 3.2
## How much is a large 2 topping pizza at Dominos?
Get a Large 2-Topping Pizza at Domino’s for \$5.99.
## How much bigger is a 12 inch pizza than a 10 inch pizza?
The amount of multiplication means that small changes in the diametre of the pizza lead to much bigger changes in the area. A 10 – inch pizza, for example, is 78 square inches, with a 12 – inch pizza a comparative 113 square inches. A 14- inch pizza comes in at 153 inches, so roughly the same size as 10 -inchers. | 723 | 2,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.926005 |
http://math.stackexchange.com/questions/260909/sum-i-1n-1-left-dfraca-ia-n-ia-n-right-geq-c-2nn-1 | 1,467,101,625,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396538.42/warc/CC-MAIN-20160624154956-00084-ip-10-164-35-72.ec2.internal.warc.gz | 211,936,419 | 19,224 | # $\sum_{i=1}^{n-1} \left|\dfrac{a_ia_{n-i}}{a_n}\right| \geq C_{2n}^n-1$
Given that the equation $$p(x)=a_0x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n=0$$ has $n$ distinct positive roots, prove that
$$\sum_{i=1}^{n-1} \left|\dfrac{a_ia_{n-i}}{a_n}\right| \geq C_{2n}^n-1$$
I had tried to calculate $P'(x)$ but can't go further. Please help me. Thanks
-
What is $C^n_{2n}$? The central binomial coefficient? – Mike Spivey Dec 17 '12 at 20:05
I guess $C_n^p={n\choose p}$. – Mercy King Dec 17 '12 at 20:07
use Cauchy-Schwarz inequality. – Yimin Feb 26 '13 at 23:17
This doesn't make sense. If we replace all the coefficients by $\lambda a_k$, where $\lambda\gt0$, then $$p(x)=\lambda a_0x^n+\lambda a_1x^{n-1}+\dots+\lambda a_{n-1}x+\lambda a_n=0$$ and yet $$\sum_{i=1}^{n-1}\left|\frac{\lambda a_i \lambda a_{n-i}}{\lambda a_n}\right|=\lambda\sum_{i=1}^{n-1}\left|\frac{a_i a_{n-i}}{a_n}\right|$$ can be made as small as we wish. Did you mean $a_n^2$ in the denominator? – robjohn Mar 9 '13 at 6:48
Even $a_n^2$ in the denominator doesn't work. For any given $p$ and $\lambda\gt0$, we have another $$p_\lambda(x)=a_0x^n+\lambda a_1x^{n-1}+\dots+\lambda^{n-1}a_{n-1}x+\lambda^na_n=0$$ whose roots are $\lambda$ times the roots of $p$ (hence positive and distinct), yet $$\sum_{i=1}^{n-1}\left|\frac{\lambda^ia_i\lambda^{n-i}a_{n-i}} {\lambda^{2n}a_n^2}\right|=\frac1{\lambda^n}\sum_{i=1}^{n-1}\left|\frac{a_ia_{n-i}}{a_n^2}\right|$$ can be made any size we wish. I think these types of scaling can be discounted if the denominator is $a_0a_n$. – robjohn Mar 9 '13 at 7:23
## 2 Answers
The statement in the question is false (as I mention in comments), but $(3)$ seems possibly to be what is meant.
For any positive $\{x_k\}$, Cauchy-Schwarz gives $$\left(\sum_{k=1}^nx_k\right)\left(\sum_{k=1}^n\frac1{x_k}\right)\ge n^2\tag{1}$$ Let $\{r_k\}$ be the roots of $p$, then $$\left|\frac{a_1a_{n-1}}{a_0a_n}\right| =\left(\sum_{k_1}r_{k_1}\right)\left(\sum_{k_1}\frac1{r_{k_1}}\right) \ge\binom{n}{1}^2$$ $$\left|\frac{a_2a_{n-2}}{a_0a_n}\right| =\left(\sum_{k_1<k_2}r_{k_1}r_{k_2}\right)\left(\sum_{k_1<k_2}\frac1{r_{k_1}r_{k_2}}\right) \ge\binom{n}{2}^2$$ $$\left|\frac{a_3a_{n-3}}{a_0a_n}\right| =\left(\sum_{k_1<k_2<k_3}r_{k_1}r_{k_2}r_{k_3}\right)\left(\sum_{k_1<k_2<k_3}\frac1{r_{k_1}r_{k_2}r_{k_3}}\right) \ge\binom{n}{3}^2$$ $$\vdots\tag{2}$$ Summing $(2)$ yields $$\sum_{i=1}^{n-1}\left|\frac{a_ia_{n-i}}{a_0a_n}\right|\ge\binom{2n}{n}-2\tag{3}$$ If we include the end terms, we get the arguably more aesthetic $$\sum_{i=0}^n\left|\frac{a_ia_{n-i}}{a_0a_n}\right|\ge\binom{2n}{n}\tag{4}$$
Note that $(3)$ and $(4)$ are sharp. If we cluster roots near $1$, we will get coefficients near $(x-1)^n$, for which the sums in $(3)$ and $(4)$ are equal to their bounds.
-
Hint:
1. Represent $\dfrac{a_k}{a_0}$ in terms of roots, and try to figure out the relationship between $\dfrac{a_k}{a_0}$ and $\dfrac{a_{n-k}}{a_0}$.
2. Use Cauchy-Schwarz Inequality.
3. Use the equality $\sum_{p\ge 0} C_n^p C_n^{n-p} = C_{2n}^n$.
- | 1,309 | 3,024 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2016-26 | longest | en | 0.699076 |
https://keolistravelservices.com/how-much-does-a-basketball-weigh-in-grams/ | 1,652,728,700,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00754.warc.gz | 407,087,356 | 4,982 | Beginners in the human being of basketball could think that all basketballs sweet the same. However, the ball can have various weights, escape on particular factors. So just how much go a basketball weigh?
A usual basketball weighs about 21 to 22 ounces or around 1.3 come 1.5 pounds. The waiting in the basketball is practically irrelevant come the weight because it just adds around 0.48 ounces or 0.03 pounds. The overall weight that a basketball additionally depends on the products used in the production.
Take note that men’s and women’s basketballs have different sizes and weights. Keep analysis to know more about these differences.
You are watching: How much does a basketball weigh in grams
Table that Contents
## Basketball Height and also Weight Chart
Different basketballs come in differing sizes, circumferences, and weight. The table below will provide you a rapid preview the the assorted basketballs being used.
SizeCircumferenceWeight
Size 116 inches8 ounces
Size 322 inches10 ounces
Size 425.5 inches14 ounces
Size 527.5 inches17 ounces
Size 628.5 inches20 oz
Size 729.5 inches22 oz
## How lot Does the Basketball weigh in the NBA?
The weight of a 29.5-inch main NBA basketball v a 75-centimeter one is 624 grams or 1.4 pounds. Moreover, the ball should be inflated to in between 7.5 and 8.5 pounds per inch.
## How lot Does the Basketball sweet in the WNBA?
The WNBA basketball is slightly smaller than the one supplied in the NBA. It should have actually a maximum of 29 inches through a weight of no more than 20 ounces but not less than 18 ounces.
## How much Does the Basketball sweet in the NCAA?
The basketball provided in the nationwide Collegiate athletic Association (NCAA) has actually the same size and weight together the version used in the NBA. However, the women’s NCAA division uses a 20-ounce basketball through no sports in weight, unequal the one in the WNBA.
## How lot Does the Basketball weigh in FIBA?
The men’s division of the global Basketball Federation (FIBA) needs its basketball to have actually a weight that’s not less than 20 ounces (567 grams) however no much more than 23 ounces (650 grams). As for this association’s women’s division, they usage a basketball that’s has actually a weight of no less than 18 ounces (510 grams) yet no more than 20 ounces (567 grams).
See more: Photoshop: Where Is The Dodge Tool In Photoshop, : Photoshop
## How much Does the Basketball weigh in the Youth League?
The weight of the basketball of youth leagues generally depend top top the place of the association. The north American Basketball Youth organization uses a 14-ounce (400-gram) basketball. On the other hand, the FIBA Youth organization use a basketball with the 16 to 17-ounce (465 come 495-gram) range.
## How much Does a Mini Basketball Weigh?
Mini basketball hoop sets have tendency to vastly differ in size and also weight, making their accompanied balls vary considerably in those aspect. Still, numerous of the mini basketballs tend to sweet 10 ounces (280 grams). | 685 | 3,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-21 | latest | en | 0.940247 |
http://mathhelpforum.com/geometry/89594-circle-geometry.html | 1,527,145,362,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865928.45/warc/CC-MAIN-20180524053902-20180524073902-00496.warc.gz | 191,560,262 | 9,621 | 1. ## Circle Geometry
Hey guys
Two perpendicular chords AB and CD of a circle, centre O, intersect at E. Prove that $\displaystyle \angle AOD$ + $\displaystyle \angle BOC$ = 180 degrees.
2. Hello xwrathbringerx
Originally Posted by xwrathbringerx
Hey guys
Two perpendicular chords AB and CD of a circle, centre O, intersect at E. Prove that $\displaystyle \angle AOD$ + $\displaystyle \angle BOC$ = 180 degrees.
In $\displaystyle \triangle ACE$:
$\displaystyle \angle E = 90^o$ (given)
$\displaystyle \Rightarrow \angle ACD+\angle BAC = 90^o$ (angle sum of triangle)
But $\displaystyle \angle AOD = 2\angle ACD$ (angle at centre = twice angle at circumference)
and $\displaystyle \angle BOC = 2\angle BAC$ (angle at centre = twice angle at circumference)
$\displaystyle \Rightarrow \angle AOD + \angle BOC = 180^o$ | 238 | 824 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-22 | latest | en | 0.667122 |
https://tophomeworkanswers.com/mathematics/which-graph-models-the-function-fx-%E2%88%9242x/ | 1,606,820,411,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00543.warc.gz | 536,100,999 | 10,875 | # Which Graph Models The Function F(X) = −4(2)X?
7223 users searched for this homework answer last month and 93 are doing it now, let’s get your homework done.
This Top Homework Answer is High School level and belongs to the Mathematics subject.
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## Question
Which graph models the function f(x) = -4(2)x?
Answer:Option C is correct.Step-by-step explanation:The exponential function is of the form: where a is initial value and b is the base.*If base b>1(i) For positive value of a.If x increases then, f(x) tends to positive infinity.(ii)For negative value of a.if x increases, then f(x) tends to negative infinity.*if 01From the above definition,as we increase x , f(x).Y-intercept ( plug x =0 to solve for y)Substitute the value of x =0 in given equation;Y-intercept (0, -4)Therefore, the only correct option is C because if we increase the value of x, the function f(x) tends to negative infinity and also it cut the y-axis at y = -4.
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If you have more homework to do you can use the search bar to find the answer to other homework: 300 have done it today and 119 in the last hour. | 360 | 1,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-50 | latest | en | 0.853555 |
https://cognitivecardiomath.com/blog/page/10/ | 1,695,322,343,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506029.42/warc/CC-MAIN-20230921174008-20230921204008-00165.warc.gz | 212,709,664 | 27,571 | ### Playing Fraction War: A Fraction Card Game for Upper Elementary and Middle School Math
Easy-prep Card Game to Practice Comparing Fractions This is a repost from 2013, transferred from my previous blog:-) Some students finally got to play ‘Fraction
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Memory Wheel Templates for a Favorite End of School Year Activity Memory Wheel templates are a low-prep, easy way to structure one of my favorite
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### Playing The Factor Game in Middle School Math
Finding The Factor Game Again! I was looking through my middle school math folders on my computer, and came across a document called “The Factor
Welcome to Cognitive Cardio Math! I’m Ellie, a wife, mom, grandma, and dog ‘mom,’ and I’ve spent just about my whole life in school! With nearly 30 years in education, I’ve taught:
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http://www.scienceprojectchennai.com/555-timer-ic-basics.php | 1,566,806,825,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331228.13/warc/CC-MAIN-20190826064622-20190826090622-00017.warc.gz | 320,379,183 | 9,015 | # Ajantha College and School Students Science Projects materials
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## 555 Timer IC
#### Ramp Generator Circuit-using 555 Timer IC
We know that if a capacitor is charged from a voltage source through a resistor, an exponential waveform is produced while charging of a capacitor from a constant current source produces a ramp. This is the idea behind the circuit. The circuit of a ramp generator using timer 555 is shown in figure. Here the resistor of previous circuits is replaced by a PNP transistor that produces a constant charging current.
Circuit diagram.
Charging current produced by PNP constant current source is
i C = Vcc-V E / R E
where V E = R 2 / (R1 + R2) * V CC + V BE
When a trigger starts the monostable multivibrator timer 555 as shown in figure, the PNP current source forces a constant charging into the capacitor C. The voltage across the capacitor is, therefore, a ramp as illustrated in the figure. The slope of the ramp is given as
Slope, s = I/C | 273 | 1,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-35 | longest | en | 0.842041 |
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 6 Mathematics5.2 Graphs
Graphs are used to represent data. Graphs help understand data better. The commonly used graphs are line plots, pie charts, bar graphs, pictographs, scatter plot, stem and leaf graph, plot and whisker plots. Example: Which of the graphs uses a circle? Answer: Pie chart Directions: Answer the following questions. Also write at least 5 examples of your own.
Q 1: What is a line plot used for?a graph comparing data using picturesto show change over timeto compare datacompares different sets of data Q 2: A circle graph is used to show the relationship of different parts to a whole. It is usually divided by fractions or percentages.truefalse Question 3: This question is available to subscribers only! Question 4: This question is available to subscribers only! | 226 | 985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-30 | latest | en | 0.866401 |
http://seniorresident.com/2016/08/how-to-retire-early/ | 1,566,810,331,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331485.43/warc/CC-MAIN-20190826085356-20190826111356-00383.warc.gz | 169,683,234 | 19,040 | # How to Retire Early 9
The primer for this post is here: Retiring Early
We’re going to go through some of the math you need to understand in order to retire early.
Let’s start with this:
### How much money do you need a year during your retirement to last from 55 until ~ 90? (from my prior post)
Is it 60k? Then you need ~\$2 million.
Is it 80k? Then you need between \$2.5 million to \$2.75 million.
Is it 100k? Then you need \$3.25 million to \$3.5 million.
### Let’s try to calculate the two extremes of \$2 million and \$3.5 million.
Now this will vary for some of you depending on how long your residency/fellowship training is. However, if you started work at 30, you have 25 years and if you started at 35, you have 20 years of compound interest in order to earn you nest egg. Let’s also assume a 6% return to be more conservative.
### Let’s look at \$2 million first.
The handy dandy compound interest calculator helps us here again.
Just input 1, 5000, 20, 6% and 1 to start with.
If you put away \$5000 a month (\$60,000 a year), you will have \$2,207,138.68 in 20 years.
If you put away \$5000 a month (\$60,000 a year), you will have \$3,291,875.01 in 25 years.
If you put away \$3250 a month (\$39,000 a year), you will have \$2,139,720.26 in 25 years.
So there you have it, in order to retire early and have a ~\$2 million nest egg, you need to put about \$60,000 away a year for 20 years (starting at age 35) or \$39,000 a year for 25 years (starting at age 30).
### Now let’s talk about the \$3.5 million:
The handy dandy compound interest calculator helps us here again.
Just input 1, 7000, 20, 6% and 1 to start with.
If you put away \$8000 a month (\$96,000 a year), you will have \$3,531,419.96 in 20 years.
If you put away \$8000 a month (\$96,000 a year), you will have \$5,266,997.44 in 25 years.
If you put away \$5500 a month (\$66,000 a year), you will have \$3,621,062.08 in 25 years.
So there you have it, in order to retire early and have a ~\$3.5 million nest egg, you need to put about \$96,000 away a year for 20 years or \$66,000 a year for 25 years.
### That seems pretty doable right?
This calculation assumes you have no debt and no other high costs to attend to (such as a mortgage or car payment). You will be living on \$60,000 to \$100,000 a year, some of which will probably be taxed because at least a portion of your nest egg probably came from a 401k/403b account which is pre-tax.
Additionally, these calculations do not take inflation into account, and you should probably try to live on less than what you expected:
There are a ton of inflation calculators around, but I like this one because it shows a nice graph and lets you select your expected inflation rate.
If you assume 2.5% inflation, then \$60,000 in 2016 is worth ~ \$98,317 in 2036 and \$111,237 in 2041 (20 year and 25 year retirement from today respectively).
This means that you would need > \$100,000 a year in 2036 and 2041 to approximate a similar lifestyle to \$60,000 a year in 2016.
If we mess around with the numbers and assume a 2.5% inflation rate, then we figure out that:
\$35,000 in 2016 will be worth \$57,352 in 2036
\$35,000 in 2016 will be worth \$64,888 in 2041
### Wait… what?!
Inflation also compounds. You need your own compound interest to offset it.
If you calculate for a \$60,000 a year nest egg and save your \$2 million dollars at the age of 55, then you will need to live off of the equivalent of \$35,000 a year. This is STILL doable assuming you live somewhere with a low cost of living, own your own low maintenance house, and own your own cars. However, you will not be taking any lavish vacations anywhere.
Saving ~\$3.5 million for the equivalent of a \$100,000 a year is more reasonable because that is ~\$60,000 a year equivalent in 2036 and little less than that in 2041.
### Ok, so what should I do then?
There have been studies on how much money a person needs to “be happy” over which there isn’t much of a difference. This was researched previously to be the round number of \$75,000 a year back in 2010, out of Princeton. Note that this \$75,000 number is a household number.
Using our inflation calculator again (with real values this time, not predictions).
\$75,000 in 2010 will be worth \$81,866 in 2016
However, Huffington Post did another follow-up to this in 2014 where they tried to break things down state by state. This intuitively makes sense because “\$75,000 happiness” is not the same as Mississippi (\$65,850) versus Hawaii (\$122,175). This of course is based on cost-of-living, but does not take into account retirement benefits like certain pensions/social security being taxed differently in different states. However, that is a whole other post for another time. (Spoilers: There is a reason why everyone retires to Florida.)
So how much is a \$82,000 a year (2016) equivalent nest egg in 2036? Our inflation calculator (2.5%) says:
\$82,000 in 2016 will be worth \$134,367 in 2036
\$82,000 in 2016 will be worth \$152,023 in 2041
That’s a lot of money to be withdrawing every year. However, that is a pretty comfortable retirement in my opinion.
Remember the retirement calculator from my prior post? BankRate? Let’s fire it up again:
If you have \$4,5000,000 saved and withdraw \$12,000 a month, assuming a 1% return. You will last the 35 years and die at 90 with \$348,252 left.
So you would need to have between \$4 million and \$5 million dollars to last the 35 years.
Of course, this number is different if you need it to last a shorter amount of time… which is why retiring at 55 is so much more difficult than retiring at 65.
Having enough money for 25 years of retirement is entirely different from saving enough for 35 years.
### Other things to consider:
You will get some form of Social Security (probably), but it won’t be much.
Expect medical costs to increase and long term care costs to increase, and they are already pretty expensive in 2016. Reference
No one can tell you for sure how much your health care and/or long term care will cost. All you can do is make a conservative estimate.
The farther you are from retirement and the longer your nest egg has to last for makes it very difficult to predict costs. These numbers will need to be worked and reworked probably every year. So just saying you want to save \$3 million dollars and retire at 55 is a reasonable plan… but it’s still only a plan.
### So, you need to ask yourself… How much do I need?
Is it \$3 million? \$4 million? \$5 million?
#### What is your number?
Mine is \$5 milllion. That is the amount I am saving toward by the time I hit 65.
Will I get there? I hope so.
I’ll talk more about my long term plan later.
### TL;DR
Retiring Early is difficult… but not impossible.
\$60,000 today is not \$60,000 when you retire.
Inflation compounds too.
“Happiness” was \$75,000 back in 2010. Probably a reasonable starting point for a comfortable retirement.
What is your number?
-Sensei
Agree? Disagree? Questions, Comments and Suggestions are welcome.
You don’t need to fill out your email address, just write your name or nickname.
Share this:
### About Sensei
A young attending physician trying to navigate the mine field that is life after medical school...
## 9 thoughts on “How to Retire Early”
• Physician on FIRE
My number is \$3 million in retirement savings. That will be 40 years of expenses at our current spending level as a family of four.
Other numbers I’m aiming for:
\$100,000 or more in each of the two boys’ 529 funds.
Zero debt. [Done] Two properties in low cost of living areas are fully paid off.
45 (Age by which I’ll likely be retired from clinical medicine)
I’m anticipating working full time another 3 years or so plus perhaps an additional year or so overseas (New Zealand and/or Australia). And of course, I will continually evaluate the plan, reserving the right to change course.
Cheers!
-PoF
• Future Proof, MD
My number is \$5 million. That’s including anticipated costs of starting/raising a family, future education costs and travel expenses – I would like to be able to take the family on 2-3 international trips/year. Of course, as circumstances change, so will my number. But right now that sounds about right. I hope to reach it by age 55. It sounds quite ambitious I know, but I think it’s not entirely out of reach.
• Passive Income M.D.
Again, I think about my retirement number differently than most people. I think of it in terms of “how much passive income do I need monthly.” If I can build up a portfolio of rental property, dividend stocks, and other passive income cash-flowing ventures to hit “X” amount monthly, then I’m good right then and there. Sometimes I think that number is \$15,000 a month, sometimes it’s \$20,000. (I live in one of the cities with the highest cost of living in the US.) I think I can get there in 10-15 years.
That’s not to say I don’t contribute to my 401k. I contribute the max for the tax benefits, but I consider it gravy.
Would love to hear what you guys think about this…
• Sensei Post author
Hey dude, I forgot to reply to this comment.
The idea of passive income from rental property, dividend stocks, and other ventures to hit a monthly income goal is admirable. However, I am not sure many doctors have the risk tolerance or ambition to do such things. Just “being a doctor” is enough stress for most doctors.
That aside, having sources of passive income would be a boon to retiring early. Any way to achieve a recurring monthly passive income is of course favorable. I would imagine that many bloggers (physicians or not) would like to have this kind of income generation. However, this is not achievable for most people.
I think that the most efficient form of “passive income” is compound interest. ie. Getting your first million in index funds is the hardest, then it gets easier to make the 2nd and the 3rd, and so on. Of course, if you can acquire other forms of passive income while still getting to your first million in index funds, then of course, that is all the better.
Of course, that is easier said than done and requires some degree of risk on your part… | 2,574 | 10,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-35 | latest | en | 0.93397 |
http://www.jiskha.com/display.cgi?id=1305771745 | 1,498,196,769,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320003.94/warc/CC-MAIN-20170623045423-20170623065423-00486.warc.gz | 550,962,239 | 3,775 | # math
posted by on .
Find all the points having an x-coordinate of 9 whose distance from the point (3, -2) is 10.
(9, 6), (9, -10)
(9, 13), (9, -7)
(9, -12), (9, 8)
(9, 2), (9, -4)
• math - ,
So ll the points whose distance from the point (3, -2) is 10 would be the circle
(x-3)^2 + (y+2)^2 = 100
when x = 9
36 + (y+2)^2 = 100
(y+2)^2 = 64
y + 2 = ± 8
y = 6 or y = -10
looks like choice #1 | 176 | 398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-26 | latest | en | 0.886125 |
https://www.vedantu.com/question-answer/if-a-is-an-idempotent-matrix-satisfying-left-class-12-maths-cbse-5edf37cdd6b8da423ce2df66 | 1,721,349,922,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00563.warc.gz | 881,969,341 | 30,045 | Courses
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# If A is an idempotent matrix satisfying, ${{\left( I-0.4A \right)}^{-1}}=I-\alpha A$ where ‘ I’ is the unit matrix of same order as that of ‘A’, then the value of |9$\alpha$| is equal to?
Last updated date: 17th Jul 2024
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Hint: Use the properties of matrix for solving this problem, also use the property of Idempotent matrix i.e. ‘If ‘X’ is an idempotent matrix then, ${{X}^{2}}=X$’ which will work as a key point is the solution.
We will write the given equation first,
$\therefore {{\left( I-0.4A \right)}^{-1}}=I-\alpha A$
We will multiply both sides by $\left( I-0.4A \right)$ therefore we will get,
$\therefore \left( I-0.4A \right)\times {{\left( I-0.4A \right)}^{-1}}=\left( I-0.4A \right)\times \left( I-\alpha A \right)$……………………………… (1)
To proceed further in the solution we should know the property of matrix which is given below,
Property:
$A\times {{A}^{-1}}=I$
Where, A be any matrix of the order $m\times n$
‘I’ be the identity matrix of the order $m\times n$
If we observe the equation (1) we say that in the term $\left( I-0.4A \right)$ there are some operations performed. As matrix ‘A’ is multiplied by 0.4 therefore it is going to give a matrix at the end and as given in the problem both ‘A’ and ‘I’ have same order therefore the whole term $\left( I-0.4A \right)$is going to give us a matrix at the end.
Therefore if we use the property given above in equation (1) we will get,
$\therefore I=\left( I-0.4A \right)\times \left( I-\alpha A \right)$
Now we are going to multiply the brackets on the right hand side of equation, therefore we will get,
$\therefore I={{I}^{2}}-\alpha \times A\times I-0.4\times A\times I+0.4\times \alpha \times {{A}^{2}}$………………………………………. (2)
To proceed further in the solution we should know the properties of matrix given below,
Properties:
${{I}^{2}}=I$
$A\times I=A$ Provided that the given matrix and identity matrix should have the same order.
By using above properties we can write equation (2) as,
$\therefore I=I-\alpha \times A-0.4\times A+0.4\times \alpha \times {{A}^{2}}$
By cancelling ‘I’ from both sides we will get,
$\therefore 0=0-\alpha \times A-0.4\times A+0.4\times \alpha \times {{A}^{2}}$
$\therefore 0=-\alpha \times A-0.4\times A+0.4\times \alpha \times {{A}^{2}}$
As ‘A’ is an Idempotent matrix therefore we can use the property of Idempotent matrix given below to solve the equation,
Property:
If ‘X’ is an idempotent matrix then, ${{X}^{2}}=X$
By using the above property we can write the above equation as,
$\therefore 0=-\alpha \times A-0.4\times A+0.4\times \alpha \times A$
$\therefore 0=-\alpha \times A+0.4\times \alpha \times A-0.4\times A$
By taking $\alpha \times A$ common we will get,
$\therefore 0=\alpha \times A\left( -1+0.4 \right)-0.4\times A$
Now, A can be cancelled out therefore we can write,
$\therefore 0=\alpha \left( -0.6 \right)-0.4$
By shifting 0.4 on the left hand side of the equation we will get,
$\therefore 0.4=\alpha \left( -0.6 \right)$
$\therefore \dfrac{0.4}{-0.6}=\alpha$
$\therefore \alpha =\dfrac{0.4}{-0.6}$
$\therefore \alpha =-\dfrac{4}{6}$
$\therefore \alpha =-\dfrac{2}{3}$
Multiplying by 9 on both sides of the equation we will get,
$\therefore 9\alpha =9\left( -\dfrac{2}{3} \right)$
Taking the modulus on both sides of the equation we will get,
$\therefore |9\alpha |=\left| 9\left( -\dfrac{2}{3} \right) \right|$
$\therefore |9\alpha |=\left| -3\times 2 \right|$
$\therefore |9\alpha |=\left| -6 \right|$
If we remove the modulus sign from the right hand side then we will get,
$\therefore |9\alpha |=6$
Therefore the value of $|9\alpha |$ is equal to 6.
Note: Do remember that the value of $A\times I$ is ‘A’ only when both the matrices have the same orders and if they are of different order then$A\times I\ne A$. This might confuse students during exams so always read the conditions carefully. | 1,289 | 3,959 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-30 | latest | en | 0.847879 |
https://www.auraauro.com/2018/02/ | 1,717,028,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00600.warc.gz | 558,978,773 | 38,640 | ## scratch programming for teaching physics concepts
I started to learn a scratch program software to create interactive model for learning. Then I got a task to create a model for: How pressure of gas change with respect to temperature and volume to container: I chose to represent it using Scratch. Initially, I had a problem on bouncing of gas molecules with same angle when it hit the wall of container. Then sanjeev helped me to solve it. In the model, I scaled the pressure in range of 0 to 150 and temperate in range of 0 to 300. But it didn’t give exact value of pressure for given temperature ( I will figure out it soon), because main goal of this model is show how pressure (no. of hit in wall by gas molecules) is affect by temperature and volume. I shared the model in scratch.mit.edu website. click this link to see the model https://scratch.mit.edu/projects/205711792/
## First digit on clock
– Logeshwari
In electronics, initially I displayed my name using Seven Segment Display(SSD). I had already posted that in My First Blog. After that I wanted to learn more on SSD. So I was asked to make a digital clock. I have completed the first digit of the seconds. I used two JK flip flops (HD74LS76A), decoder (SN74LS47), a SSD (Common Anode) and Signal Generator. I first made the connections of JK flip flop on a breadboard. The input is clock signal which is generated from a signal generator of Frequency 1Hz. The combination of the two flip flops acts like a counter. Once the circuit is connected, I tested the output of the flip flop using an ocilloscope. I got the waveforms like this.
Then I took the decoder and connected the outputs to the input pins of the decoder. Decoder (converts the binary information to 2n outputs) is connected to the seven segment display. While doing decoder I was able to recall Karnaugh-Map which is the logic behind it. The output of the decoder is fed to the seven segment display. Once everything is done then it counts from zero to fifteen (f-represented in binary).
Since the one’s digit of the clock counts only upto nine and after that it should reset to zero. For resettting the digit to zero I used NAND gate. Binary representation of nine is 1001. When the Most Significant Bit and the Least Significant Bit are high(1) then it should reset to zero.
I have completed the first step only, but I feel happy that I learnt and understood what I did. I acknowledge Sanjeev, Sundhar and Ranjith for guiding me.
## Sinthanai Sangamam with around 600 teachers(Azim Premji Foundation)
Saranya
Sinthanai Sangamam is happened on 21/02/2018. It is a new kind of experience with around 600 teachers from different places. I have presented a paper about STEM land. The great thing is I got an opportunity to share about, my journey with the children in STEM land.The wonderful thing is that I am in STEM land. Why I came to STEM land is, I want to learn more about electronics that is why I came to STEM. I have learned about electronics what I have learnt that came as a few product that is inverter, display my name using 7-segment display, mobile charger, solar powered charger etc..Then I learned some of the programming from students that is scratch and geo gebra these program will be used in mathematics. These things I have explained in very shortly. In Sinthanai Sangamam everyone shared their experience with the children. How they interact with the children, what method they are following in their school, what kind of projects their students done and how we can connect to science and maths in our real life. The one main thing what I learnt is there should not a gap between children and teacher.
VLSI Layout
In VLSI layout course I learned to design a gates using NAND. Then I have checked the NAND gate truth table with the help of bread board, battery and multimeter. I learned to use magic tool, how to extract and how to simulate the layout.
I learned the cross section of CMOS and what colour coding we have to use for the each components.
## Parul Kudawla to STEM land
Poovizhi, Pratap
We met Parul when we presented our paper (“Fostering responsibility in learning in rural schools) in TISS Mumbai. She is a management trainee in BPCL. She wanted to learn more about STEM land so she visited STEM land last Sunday. She came with her parents.
She saw all the Montessori materials we used and asked how they were used and for what purpose. Then she was keen in understanding how children learn multiplication tables using Vaughn cube. We explained her the pronunciation and the corresponding numbers. She tried finding it herself and was happy that she understood it. She also shared it with her parents. She learned about the 3D printer and designed a 3D object in Blender. We showed her some projects made by children in scratch and Geogebra.
Her parents were trying to solve some cast puzzles. We all played dimension games and were explaining about how STEM land works and what we do there. Parul enjoyed STEM land and wanted to support us on our research work.
## Oven wall EBD with children
– Ranjith
I’m back after two week gap, I notice lot of things happened here in two weeks.
I worked on an EBD (Education by Design) project with children in Isai Ambalam School. They had built a red mud and brick oven, but are concerned that in the rainy season the red sand get eroded by water. They want build a short wall to protect the oven from flow of the water during the rains.
Now they are in the process of cleaning and breaking and making the floor plain to raise a wall from there. It was amazing experience with children and they are eager to finish it soon.
In this week, I learned many more things including
• A tool called magic used for VLSI layout. I did the layout of an Invertor, NAND gate. Then I learned how to create a hierarchy and import a NAND gate layout and use it to create XOR gate by wiring using different layers of metals to avoid a short circuit.
## Geogebra, EBD & Goanimate
– Saranya
I learned to use Geogebra and have used it to solve a linear equation. This worked for me and it was new to me. It is increasing my interest in mathematics.Now I am eager to solve many equations. Whatever we enter an equation the tool puts the waveform for it. For example, I have taken 3x+2y=0 is a linear equation. From that diagram we can easily identify that is linear equation. This was surprising for me.
I worked on EDB (Education by Design) with students at Isai Ambalam school.The pond which they have built before was cracked by roots of old trees .So because of that students want to reinforce the pond.Now they are in process of adding iron rod and concrete item so that the pond won’t get break again.It was wonderful experience with them.Is very good to see like that students they were full of energy.
I also made a small animated video with Goanimate.com it was great idea. It gave an idea of how I can convey something that feels more interactive. It was fun and awesome because whatever we want convey to the students we can convey by making an animation video. It was my first experience with goanimate.I made a video about multiplication story.
Saranya Multiplication Story by SanjeevRanganathan on GoAnimate
Here are some snapshots in case you cannot see the flash content above.
(I.e) there are 3 children, if I gave 4 bananas to each of them how many bananas did I give in all? The answer also I have shown.The answer is 12(add there number of bananas each have and we will get the answer easily).
## Projects for Raman awards – Isaiambalam
Raman Young Science Innovator Award is a contest for students studying from III to X standards grouped into three. Students from IIIto IV, V to VIII and IX to X. We heard about this contest and started registering for it on 31st January. The registered students got their login credentials through their mail after 31st January. We submitted their projects on the last date on 10th February (extended to 12th February). Eight students submitted their innovation projects from Isai ambalam School. They are Ayush Jena, Suresh and Vasanth Kumar from class VII. Balamurugan, Kaviya, Sabari anandh and Yuvasri from class VI and Sharani from class V.
Ayush Jena (Class VII)
Ayush chose the topic buoyancy. He made an animation using scratch which explains the buoyancy of ships.
Suresh (Class VII)
He made his project on conductors and non-conductors.His main objective of the project is to find if the material is a conductor or non-conductor.He made a circuit and placed in inside a box.If the material is a conductor then the light will glow. If the light doesn’t glow then the material is not a conductor(non-conductor).The materials he used to make this project are wires, bulb, battery, box and iron.
Vasanth Kumar (Class IX)
Vasanth Kumar chose the topic Plant Kingdom. He took banana and he made an animation using scratch.In his animation he explained the benefits of bananas, scientific names and different things about it.
Balamurugan (Class VI)
He chose the topic density. He took a bowl of water and filled water upto 100 ml. Then he put some stones in the bowl until the water level rises to 110 ml. Then he took the stones in the bowl and weighed them with a weighing machine. It was 100 gm, which is equal to the rise in the water level.
Kaviya (Class VI)
Kaviya made her project in convection. She experimented by using coloured ice cubes. She was able to see the colour of the ice cubes get sink in the water. She observed that the speed of the speed of the ice cubes dissolving in water.
Sabari Anandh (Class VI)
Sabari chose the topic Plant Kingdom.He made his project using scratch. He selected neem, squirel, mamgo and described their important characteristics. Then he explained what is the connection between them using scratch.
Yuvasri (Class VI)
Yuvasri chose the topic melting point. She tested the melting point of ice cubes. She understood why the ice cubes melts when heated and what happens when heated.
Sharani (Class V)
Sharani chose the topic density. She took honey, vegetable oil, soup oil, coloured water and syrup. She poured all the liquid one by one glass bottle. Honey settles down because it has density greater than one. Similarly she formed five layers in the bottle.
## VLSI course
We are conducting a basic VLSI course. Around 16 participants from in and around Auroville are coming in to learn. We are only focusing on Digital layout and understanding basic concepts.
Course outline:
Day 1:
Give an introduction to CMOS technology and fabrication process. Give an intro to CMOS Transistors and it’s working. Introduce people to stick diagrams.
• Introduction to VLSI technology.
• Introduction on CMOS transistors and working.
• Demo on LT spice simulation tool.
• Introduced CMOS Inverter and ask participants to simulate an inverter in LT spice.
• Showed a video on fabrication process and manufacturing a chip.
• Introduced stick diagram and how it helps to get started with layout. Participants drew a stick diagram for an Inverter.
Day 2:
• Introduced magic layout tool.
• Helped them to login to our central server through VNC. This helped them to avoid installing magic in their laptops. We haven’t figured out a way to install in windows so far, but Magic sort of works fine in Ubuntu platform.
• Showed top view and cross section of a CMOS transistor. This helped them to understand the fact that layout is all about top view of the devices.
• Participants laid out an invereter using magic.
• Once that was completed we showed them to extract and create a netlist. Netlist helps us to understand the connection and cross check whether the layout we laid out is correct or not.
Day 3:
• Introduced NAND gate
• Laying out NAND gate. (Note: We gave a circuit diagram for a NAND gate)
• Extracting NET list and with the same netlsit draw the schematic usind the netlist.
• Whoever completes should layout a NOR gate and repeat STEP 2 and 3.
Day 4:
• Introduced to gates. (OR, AND, INVERETER, NAND, NOR and EXOR)
• Understand the truth table for all gates.
• Introduced few rules like Involution law, Idempotency law and De Morgan’s law
• Task 1: Using NAND gate create OR, AND, INVERETER and EXOR. Draw schematic and derive the logic for the same.
• Task 2: Build those gates using DM74S00N and bread board and test whether their logic were appropriate.
Day 5:
• Intorduced Hierarchy. Showed how to import models that can be used to create complex layouts.
• Task: Layout an XOR gate importing the NAND gate built in the previous classes
Day 6:
• Introduced IRSIM
• Task 1: Test different layouts with IRSIM
• Task 2: Whoever didn’t complete XOR should complete XOR and test it out using IRSIM
Day 7:
We wanted people to see a real layout and at the end of the course, layout a micro processor.
• Showed them a layout of a real chip.
• There are three main blocks necessary for a micro processor. ALU unit, Memory and Counter. We used full adder for ALU and Flip flops for every bit for a memory and counter.
• Task: Layout a micro processor
## Goanimate, MIT App inventor and Stykz
During the activity class in Udavi school children were working on different applications like Goanimate, MIT Appinventor, Stykz and scratch. Muralidharan anad Nirmal was working on Goanimate. Murali made a video called help others. He made an animation of a man sailing on the boat searching for treasure but then on his way he finds a man in an island asking for help then he goes and rescues him from there. Aravindh built a timer app using MIT Appinventor. Jeeva worked on scratch and programmed a game. Yuvan was working on Stykz and learnt to insert figures in it. It was interesting to see them working on their own on different activities. They were so curious to learn new things and enjoyed learning. Nirmal learnt Go animate and was animating a video to represent a geometrical concept during his math class.
## Raman Award Projects – Udavi School
Raman Young Science Innovator Award is a contest for students studying from III to X standards grouped into three. Students from III to IV, V to VIII and IX to X. We heard about this contest and started registering for it on 31st January. The registered students got their login credentials through their mail after 31st January. We submitted their projects on the last date on 10th February (extended to 12th February). 11 students from Udavi School chose their topics and submitted their Innovation projects. They are Abitha, Amsavalli, Hariharan, Kabilan, Punidhavel, Suriya and Vignesh from IX standard. Baranidharan from VIII and Arvindh, Pranav and Vishal from VII standard. The topics were based on their standards.
Abitha (Class IX)
Abitha chose the topic refraction. She studied about refraction in her academic books, so she thought of making it in practical. She made her project using GeoGebra. GeoGebra is a Dynamic Mathematics Software (DMS) for teaching and learning mathematics. She used two mediums. Both of them are air. The incident ray and the refracted ray has the same angle. The refraction index of the second medium increases then the angle decreases.
## 2.Amsavalli (class IX)
Amsavalli did a project on reflection. She did her project using GeoGebra.When the incident ray falls on the mirror and the ray gets reflected. This is the reflected ray. She showed the angles of the reflection in an animated form.
## 3.Hariharan (Class IX)
Hariharan chose the topic frequency. He made his project using Arduino. He used an ultrasonic sensor which will detect the object. He wrote his program in Arduino. When any object is placed in front of the ultra sonic sensor it makes beep sound. When the object is far from the sensor the volume becomes low. Based upon the distance of the object to the sensor, the sound will differ.
## 4.Kabilan (Class IX)
Kabilan made a microscope using a mobile phone and lens of laser light. He placed the object on the lense and it shows the enlarged image of it in the mobile. He made this microscope, because when he was using a microscope the light hurt him. So he wanted to make a microscope which does’t hurt. He tested the microscope by seeing the microscopic view of leaves and insects.
## 5.Punidhavel (Class IX)
Punidhavel made a piano using resistors, basar and battery. He made his connections in a breadboard. He used resistors and buzzors to make sound. After completing the project he made the sa re ga ma tune.
## 6.Suriya (Class IX)
Suriya made a pie chart using GeoGebra about the natural gases in the atmosphere.
## 7.Vignesh (CLass IX)
Vignesh chose the topic frequency. The materials which Vignesh used are ocilloscope, microphone and 341 tuning fork. He measured the frequency of tuning fork with the help of microphone and oscilloscope. By using this we can also measure the heart beat frequency.
## 8.Bharanidharan (Class VIII)
He made an inverter using Arduino. The materials he used are Arduino, Breadboard, voltage regulator, resistor(10 k ohm), 9V battery, transformer, wires and laptop. He made a program to blink the Led at 50Hz. Then he dumped the program in the Arduino. He connected it to a CRO and saw the square waveform. Then he connected the remaining circuits on the breadboard. He used the voltage source from which nine volt DC input is fed to the transformer. The output of the transformer is 230 volt AC. After that he connected a 5 Watts bulb to the ouput. Eventually the bulb glowed.
## 9.Arvindh (Class VII)
He made his project on slow combution explaining the ignition temperature.
## 10.Pranav (Class VII)
Pranav made a steam engine.He took two wooden pieces and on each edge he made a hole and put a screw. He placed a tin above the wooden pieces and stuck it with glue. He made a hole on top and one side of the tin. He broke the pen into two and inserted each piece in the two holes. Then he kept a motor which was connected to another wooden piece. A Led was also connected to this wooden piece. When fire is lit up, it goes through the tin and it makes the turbine to rotate.
## 11.Vishal (Class VII)
Vishal made a project on combution. He made a small stove using two tins. First he took the small tin and made holes then the bigger tin and made holes in it. The smaller tin is placed inside the bigger tin. A vessel is placed on the tin. Using small wooden pieces he made fire. He poured water in the vessel, which got heated quickly.
Students made their project with their full involvement and felt happy after submitting their innovation project. | 4,218 | 18,578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-22 | latest | en | 0.930542 |
https://howthingsworks.com/hw/learning-how-many-cups-is-8-oz-dry-orzo-80437/ | 1,653,193,863,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543797.61/warc/CC-MAIN-20220522032543-20220522062543-00131.warc.gz | 371,128,865 | 12,072 | ## How many cups is 8 oz of dried pasta?
2 cups
You can also do it by weight. Generally 8 ounces of short pasta (like macaroni) is about 2 cups. Having said all that, what’s generally listed on the packages/boxes of most dry pasta as the serving size is 2 ounces.
## How many cups is 6 oz of dry orzo?
The simplest way to prepare orzo is to boil it in a pot of salted water and drain it as you would any other pasta. It cooks to al dente in about 8 to 10 minutes. 1 cup raw orzo (about 6 oz) will give you about 2 cups cooked.
## How many cups is a dry orzo?
2 cups
Soup Shapes
BARILLA PASTA – PRODUCT YIELDS
Soup ShapesDRY PASTA – 2-OUNCE SERVINGCUPS UNCOOKED PASTA PER PKG
Ditalini1/3 cup2-1/4 cups
Orzo1/4 cup2 cups
Pastina1/3 cup2 cups
## How many cups is 2 ounces of dry orzo?
What does 2 ounces of dry pasta look like? It depends on the shape.
SOUP SHAPES.
BARILLA PASTA – PRODUCT YIELDS Orzo 1/3 cup 4/5 cup 2-1/2 cups
## How much is 8 oz of fettuccine?
Dry pasta guidelines
Type of PastaUncooked AmountCooked Amount
Fettuccine8 oz.3-1/4 cups
Linguine8 oz.4 cups
Medium Shell3 cups/8 oz.4 cups
Rigatoni3 cups/8 oz.4 cups
Jun 17, 2019
## How do you measure dry pasta in cups?
To measure out 1 cup of dry pasta, use your palm as a guide (fill a closed fist). This works best for smaller noodle shapes like macaroni or rigatoni. Save a soda bottle. While the hole in a pasta spoon may vary, the size of a soda bottle opening will always be the same.
## How many cups is 2 oz dry?
Two ounces of dry pasta is equal to 1/2 cup dry, which boils up to be 1 1/2 cup cooked, or a heaped 1 cup.
## How much is 2 oz in Cup?
2 oz = 0.25 cups
You may also be interested to know that 1 oz is 1/8 of a cup.
## How many cups of uncooked pasta is 16oz?
Generally 8 ounces of short pasta (like macaroni) is about 2 cups. So a 1 pound box of dry pasta (16 ounces = 4 cups dry) cooks up to about 8 cups.
## Is a dry cup 8 oz?
For dry measurements, the rules change. Because dry ingredients vary greatly in weight, you can’t rely on the same conversion. For example, 1 cup of all-purpose flour weighs 4.5 ounces, not 8 ounces. On the other hand, 1 cup of chocolate chips weighs a little over 6 ounces.
## How many dry ounces is 3 cups?
Dry/Weight Measure
Ounces
10 tablespoons plus 2 teaspoons2/3 cup5.2 ounces
12 tablespoons3/4 cup6 ounces
16 tablespoons1 cup8 ounces
32 tablespoons2 cups16 ounces
## Is 1 cup dry the same as 1 cup liquid?
Technically, yes. They both measure the same amount of volume. 1 cup in a dry measuring cup is the same as 1 cup in a liquid measuring cup.
## How do you measure dry ounces?
The convention in the US is this: If a dry ingredient is listed in ounces, it’s a unit of weight and should be measured on a scale. If a wet ingredient is listed in ounces, it’s fluid ounces and should be measured in a wet measuring cup.
## What measurement is dry cup?
A liquid measuring cup like the one below measures 1 cup or 8 FLUID ounces of liquid. A dry measuring cup is specific to dry ingredients such as flour and sugar – a cup of flour is not 8 ounces. In fact a cup of sugar weighs 7 ounces. If it is confectioners sugar ( powder sugar) it is 4.5 ounces.
## What size is an 8 oz cup?
Overall Dimensions:
Top Diameter:3 1/8 Inches. Bottom Diameter:2 1/4 Inches. Height:3 1/2 Inches. Capacity:8 oz.
## How many cups is 12 dry ounces?
12 fl oz = 1.5 cups, the formula which we derived earlier for converting fluid ounces to cups is not true for flour.
## How many cups is 4 dry ounces?
Dry measures
3 teaspoons1 tablespoon1/2 ounce
8 tablespoons1/2 cup4 ounces
12 tablespoons3/4 cup6 ounces
32 tablespoons2 cups16 ounces
64 tablespoons4 cups32 ounces
## How many cups are Oz?
0.12500004 cups
How many cups in an ounce? 1 fluid ounce is equal to 0.12500004 cups, which is the conversion factor from ounces to cups.
## How many cups is 7 oz dry?
7 oz = 0.875 cups
Thus, you can divide 7 by 8 to get the same answer.
## How much is 10 dry ounces in cups?
10 oz = 1.25 cups
You may also be interested to know that 1 oz is 1/8 of a cup. Thus, you can divide 10 by 8 to get the same answer.
## Does 4 ounces equal 1 cup?
Volume Equivalents (liquid)*
8 tablespoons1/2 cup4 fluid ounces
12 tablespoons3/4 cup6 fluid ounces
16 tablespoons1 cup8 fluid ounces
2 cups1 pint16 fluid ounces
## How many ounces are in a cup of dry ingredients?
On average, one dry cup is equal to 6.8 US dry ounces. One cup equals 16 tablespoons equals 8 ounces equals. 5 pounds equals 221.23 grams.
## How many dry ounces are in a dry quart?
37.23 oz
How many ounces in 1 quart? Liquid: There are 32 fluid ounces in 1 quart. Dry: There is 37.23 oz in 1 quart.
## How many dry ounces are in a cup of sugar?
One US cup of granulated sugar converted to ounce equals to 7.05 oz. | 1,389 | 4,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-21 | latest | en | 0.90047 |
https://www.maasaimarathon.org/category/keeper/ | 1,596,488,840,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735833.83/warc/CC-MAIN-20200803195435-20200803225435-00434.warc.gz | 757,140,041 | 130,660 | Categories
## Marathon Pace Keeper
His research highlighted a discrepancy between men’s and women’s marathon results, showing that women run at a slightly higher percentage of their half marathon race-pace during a full marathon pace keeper marathon. which means that this marathon predictor is gender-specific, as well as being based on a large number of race performances. 2. In order to finish a marathon in 4 hours, your average pace needs to be 9 min, 9. 62 sec per mile, or 9:9. 62 minutes per mile. example calculation 2 : how long will it take me to run a half marathon if my average pace is 10 minutes, 30 seconds per mile?. Aug 09, 2017 · marathon pace represents a realistic goal pace if you are training for a marathon; if you are not training for a marathon, you can use marathon pace to interject some moderately paced days and variety into your easy runs. threshold (tempo) pace. commonly, threshold pace is described as “between 10k and half marathon race pace. ”. A pacemaker or pacesetter, sometimes informally called a rabbit, is a runner who leads a middleor long-distance running event for the first section to ensure a fast time and avoid excessive tactical racing. pacemakers are frequently employed by race organisers for world record attempts with specific instructions for lap times.
## Running Pace Chart Race Pace Runners World
Determine how fast your pace should be if you have a certain finish time for a desired distance or race. for example, find out what pace you need to keep to run a 28-minute 5k or a sub-2:00 half marathon. determine what your pace was for your training run around the neighborhood or track. for example, find out how fast your pace was for that 46. Feb 26, 2019 · knowing what your pace is will help make all of your runs feel better. finding your perfect pace means feeling comfortable and relaxed. check out our top tips for how you can find your pace. For example, find out what pace you need to keep to run a 28-minute 5k or a sub-2:00 half marathon. determine what your pace was for your training run around the neighborhood or track. for example, find out how fast your pace was for that 46-minute 5-mile training run. determine the distance you ran. For example, find out what pace you need to keep to run a 28-minute 5k or a sub-2:00 half marathon. determine what your pace was for your training run around the neighborhood or track. for example, find out how fast your pace was for that 46-minute 5-mile training run. determine the distance you ran.
## How To Pace A Marathon Myprocoach
Pacemaker (running) wikipedia.
## How To Train For A Half Marathon For Beginners Shape
A similar case occurred in the 1994 los angeles marathon when veteran marathoner paul pilkington was paid to set a fast pace then drop out. when the elite athletes failed to follow his pace, he kept going, ultimately winning \$27,000 and a new mercedes marathon pace keeper to the surprise of the expected favourites. that year, the l. a. marathon was the national. Here’s a handy marathon pacing chart to determine your needed pace in minutes per mile. pacing is based on the marathon standard distance of 26 miles and 385 yards, or approximately 26. 22 miles. the important numbers to look at for your goal time is the avg min/mile and 13. 1 and 20 mile split times based on your race strategy. Kipchoge, an eight-time major marathon winner and three-time olympic medalist, pounded his chest twice as he crossed the finish line in vienna’s leafy prater park, where the majority of the run. The number of saves a keeper makes is not always the best indication of their ability because those playing regularly for struggling clubs will invariably face more shots. it is no surprise then to.
Marathon pace represents a realistic goal pace if you are training for a marathon; if you are not training for a marathon, you can use marathon pace to interject some moderately paced days and variety into your easy runs. threshold (tempo) pace. commonly, threshold pace is described as “between 10k and half marathon race pace. ”. But a pace difference of just five or 10 seconds per mile in the first half of a marathon could make the difference between hanging on and falling apart in the second half. so choosing an appropriate time goal, which in turn gives you an appropriate target pace, is very important.
Running pace calculator calculate your finish time for popular race distances (5k, 10k, 10 mile, half marathon, marathon) based on your expected pace. 3 min 4 min 5 min 6 min 7 min 8 min 9 min 10 min 11 min 12 min 13 min 14 min 15 min 16 min 17 min 18 min 19 min 20 min 21 min 22 min 23 min 24 min 25 min 26 min 27 min 28 min 29 min 30 min. Heads up: there’s one fartlek workout in this half-marathon training schedule. after doing a warm-up, you’ll run 1 minute at your goal pace, then recover for 1 minute at marathon pace. keep matching your effort to recovery interval 1:1 while working through a pyramid: 1 minute, 2 minutes, 3 minutes, 2 minutes, 1 minute. According to running coach susan paul, typically runners add 30 seconds per mile to their goal race pace to as much as 2 minutes per mile. how much you choose to slow down is up to you, but.
The easiest marathon pace chart on the web. How these charts help: our pace charts show what time a given pace will produce for six common race distances: 5k, 5 miles, 10k, 10 miles, half marathon, and marathon. charts are available for marathon pace keeper pace. Mar 04, 2019 · how these charts help: our pace charts show what time a given pace will produce for six common race distances: 5k, 5 miles, 10k, 10 miles, half marathon, and marathon. charts are available for pace. This running pace calculator enables you to determine your running pace swiftly and with ease whether you are running for pleasure or training for a half-marathon, marathon or other event. use the calculator to figure out your pace per yard, mile, meter or kilometer, and view your splits in any of distance measurement.
Knowing what your pace is will help make all of your runs feel better. finding your perfect pace means feeling comfortable and relaxed. check out our top tips for how you can find your pace. More marathon pace keeper images. | 1,436 | 6,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-34 | latest | en | 0.931261 |
https://math.stackexchange.com/questions/1008978/why-open-interval-in-formal-definition-of-limit-at-infinity | 1,713,710,369,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00591.warc.gz | 349,547,416 | 36,449 | # Why Open Interval In Formal Definition Of Limit At Infinity
The formal definition of limit at infinity usually starts with a statement requiring an open interval. An example from OSU is as follows:
Limit At [Negative] Infinity: Let $f$ be a function defined on some open interval $(a, \infty)$ [$(-\infty, a)$]. Then we say the limit of $f(x)$ as $x$ approaches [negative] infinity is $L$, and we write $\lim_{x\to[-]\infty} f(x) = L$ if for every number $\epsilon$, there is a corresponding number $N$ such that $\left|f(x) - L\right| < \epsilon$ whenever $x > N\;[x < N]$.
But, I think it is also valid when the interval is half-open as in the following: Let $f$ be a function defined on some right-open interval $[a, \infty)$ [left-open interval $(-\infty, a]$]. Then we say the limit of $f(x)$ as $x$ approaches [negative] infinity is $L$, and we write $\lim_{x\to[-]\infty} f(x) = L$ if for every number $\epsilon$, there is a corresponding number $N$ such that $\left|f(x) - L\right| < \epsilon$ whenever $x \geq N\;[x \leq N]$.
So, why the formal definition of limit at infinity does not start with a statement requiring a half-open interval, which is more general?
Is it because people want to match it with the formal definition of limit? I understand that the formal definition of one-sided limit requires an open interval because that is necessary to define a limit at a point $a$. But, such requirement does not exist for limit at infinity, and therefore, why the more general version of half-open interval is not used in the formal definition of limit at infinity?
There is not really a difference between both approaches: If $f$ is defined on the open interval $(-\infty,a)$, then we may as well consider the restriction to the closed interval $(-\infty,a-1]$ and similarly vice versa. The reason that open interval may be preferred is that the limit requires $f$ to be defined on a topological neighbourhood of $\infty$. A neighbourhood of $\infty$ is a set that contains an open set containing $\infty$ and the basic open sets are open intervals. So the the following definition might be considered "best", but I'm afraid it is way less intuitive for the learner:
Limit At Infinity: Let $f\colon A\to\mathbb R$ be a function where $A$ is a punctured neighbourhood of $\infty$ in the two-point compactification of $\mathbb R$. Then we say ...
It's great to ask what hypotheses in a theorem or definition are really germane. What is important is that $f$ is defined on some interval $(a,\infty)$ for sufficiently large $a$ so that when we write down the $\epsilon-\delta$ definition, we don't have to worry about pathologies.
It's not more general. If your function is defined on a closed interval $[a,\infty)$ then it's also defined on an open interval $(b,\infty)$ where $b > a$. And the values of the function on the interval $[a,b]$ don't matter at all when it comes to the limit of the function at $+\infty$.
The question of whether the limit of $f(x)$ as $x\to +\infty$ exists is basically the same as asking whether $f$ can be extended from $(a,\infty)$ to $(a,+\infty]$ (see Wikipedia on the extended real line) such that it's continuous at $+\infty$. | 813 | 3,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-18 | latest | en | 0.850777 |
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IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies. Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..
Q1. Which condition is not satisfied by an ideal solution?
• mix H=0
• mixmixV=0
• mix S=0
• Obeyance of Raoult’s law
Solution
Q2.When 0.004 M Na2 SO4 is an isotonic acid with 0.01 M glucose, the degree of dissociation of Na2 SO4 is
• 75%
• 50%
• 25%
• 85%
Q3. The correct relationship between the boiling points of very dilute solutions of AlCl3(t1) and CaCl2(t2), having the same molar concentration, is
• t1=t2
• t1>t2
• t2>t1
• t2≥t1
Solution
Q4. The osmotic pressure of a sugar solution at 24℃ is 2.5 atm. The concentration of the solution in mole per liter is
• 10.25
• 1.025
• 1025
• 0.1025
Solution
Q5.The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of the theoretical value of water and alcohol mixture. Hence the mixture shows
• The solution is highly saturated
• Positive deviation from Raoult's law
• Negative deviation from Raoult's law
• Nothing can be said
Solution
Q6. The freezing point among the following equimolal aqueous solutions will be highest for
• C6 H5 NH3 Cl (aniline hydrochloride)
• Ca(NO3 )2
• La(NO3 )3
• C6 H12 O6 (glucose)
Solution
Q7.Which aqueous solution has minimum freezing point?
• 0.0 1M NaCl
• 0.005 M C2 H5 OH
• 0.005 M MgI2
• 0.005 M MgSO4
Solution
Q8.The osmotic pressure of blood is 7.40 atm at 27℃. The number of mol of glucose to be used per liter for an intravenous injection that is to have the same osmotic pressure as blood is
• 0.3
• 0.2
• 0.1
• 0.4
Solution
Q9.Which is not a colligative property?
• Lowering of vapour pressure
• Freezing point
• Osmotic pressure
• Elevation in boiling point
Solution
Q10. Semi-permeable membrane is chemically
• Copper ferrocyanide
• Copper ferricyanide
• Copper sulphate
• Potassium ferrocyanide
Solution
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2021-10-02, 16:01 #1 petrw1 1976 Toyota Corona years forever! "Wayne" Nov 2006 Saskatchewan, Canada 10100001111012 Posts Is it feasible to define a simple P-1 probability "approximator"? I know there is a very robust accurate P-1 probability calculator here. But I'd like to find a simplified version that is still "pretty close"; and that can be used as a single formula or function. At first glance it does not seem like it should be very hard. There are only 4 variables: - Exponent - Current TF Bit level - Bound 1 - Bound 2 However, it has eluded my synapses. I track all my P1 work for my Under-2000 sub-project in an Excel spread sheet (sorry, I'm a geek from the 80's). A big part of this project is determining whether is it more efficient to find the remining required factors with TF or with P-1. (Or more accurately how much more of each and in what order?) Where I determine I need more P-1, the bigger question is: "How much more?" Or more specifically what are the optimal bounds that will produce the desired number of factors with the least total work effort. My vision is to have a spreadsheet where I: (I actually already have this ... but step 2 below is inefficient and poorly done) 1. Download the current P-1 work from here 2. Add a column with a "neat and tidy" P-1 estimator formula for the current success rate. 3. Add another few columns with proposed bounds, success rates and work effort 4. Determine which new bounds will generate the required number of factors with the least total work effort. I can easily complete steps 1, 3 and 4. I just have not had any luck creating a useful simpler estimator. Thanks
2021-10-02, 16:31 #2 axn Jun 2003 10100111011112 Posts You can write a macro for the full blown function in your spreadsheet tool of choice. Probably the easiest way (in the long run). The actual C code can be extracted from P95 source for reference. Last fiddled with by axn on 2021-10-02 at 16:32
2021-10-02, 17:04 #3
petrw1
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
120758 Posts
Quote:
Originally Posted by axn You can write a macro for the full blown function in your spreadsheet tool of choice. Probably the easiest way (in the long run). The actual C code can be extracted from P95 source for reference.
Thanks...I did that for the prior version.
The problem I had then was being unable to (or not knowing how to) have the macro trigger for every row in the spreadsheet as soon as it was populated.
Having to run it manually for every row (or at least every different row) was time consuming.
2021-10-03, 02:31 #4 axn Jun 2003 14EF16 Posts Write it as a function that takes all the 4 parameters and then returns a double. Then you just put it as a formula in a cell. Are you using Excel (in Windows) or something else for your spreadsheet?
2021-10-03, 03:20 #5
petrw1
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
3×11×157 Posts
Quote:
Originally Posted by axn Write it as a function that takes all the 4 parameters and then returns a double. Then you just put it as a formula in a cell.
That's exactly what I am trying to do .
Quote:
Are you using Excel (in Windows) or something else for your spreadsheet?
Yes
2021-10-03, 03:42 #6 axn Jun 2003 23·233 Posts Ok. So you need a .xslm file, where you insert a Module and define a public function. Code: Public Function Pminus1Probability(e as Long, tf as Integer, B1 as Long, B2 as Long) As Double and then put a formula in the cell as, for example, =Pminus1Probability(A1, B1, C1, D1)
2021-10-03, 05:24 #7
petrw1
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
3×11×157 Posts
Quote:
Originally Posted by axn Ok. So you need a .xslm file, where you insert a Module and define a public function. Code: Public Function Pminus1Probability(e as Long, tf as Integer, B1 as Long, B2 as Long) As Double and then put a formula in the cell as, for example, =Pminus1Probability(A1, B1, C1, D1)
Yes, I understand the process...my problem is converting this into an excel Function/macro.
Having looked at this source and after a few poor attempts at converting this version into a macro I've come to realize that for my purposes I don't need accuracy to many decimal points; 2 would be enough.
That is why I'd be content with a much simpler "reasonable" formula.
Thanks
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1 Leverage and Capital Structure Ross Chapter 16 Spring Leverage Financial Leverage Financial leverage is the use of fixed financial costs to magnify the effect of changes in EBIT on EPS. Fixed financial costs can be, for instance, interest payments and dividends on preferred shares. 2
2 10.1 Leverage Financial Leverage Let T denote the tax rate, let I denote interest expense and let NS denote the number of shares outstanding. Then earnings per share (EPS) are given by EPS = (1 T )(EBIT I). NS Leverage Financial Leverage At a given EBIT level, how do percentage changes in EBIT translate into percentage changes in EPS? EPS EPS (1 T )(EBIT I) NS ( (1 T )(EBIT I) NS = (1 T )(EBIT I) NS = (1 T )(EBIT EBIT) (1 T )(EBIT I) = = EBIT (1 T )EBIT EBIT EBIT (1 T )(EBIT I) EBIT EBIT I EBIT EBIT ) 4
3 10.1 Leverage Financial Leverage That is, when EBIT increases by 1%, the percentage increase in EPS is EBIT EBIT I = EBIT EBIT I. This the degree of financial leverage (DFL) at base level EBIT. 5 Types of Capital Assets Debt & Equity NWC Long-Term Debt (D) Fixed Assets Equity (E) 6
4 Capital Structure Theory Modigliani and Miller s propositions: Proposition I: The market value of a firm is constant regardless of the amount of leverage that it uses to finance its assets. Proposition II: The expected return on a firm s equity is an increasing function of the firm s leverage. 7 Capital Structure Theory: M&M Proposition I The value of a firm is given by the present value of all the cash flows its assets are expected to generate in the future. The value of a firm is equal to the value of its assets. Unlevered Firm: V U = E U Levered Firm: V L = D + E L. 8
5 Capital Structure Theory: M&M Proposition I M&M Proposition I states that V U = V L. Why? Consider an all-equity firm with value V U = E U. Suppose there existed a way to finance this firm s assets with debt and equity such that V L = D + E L > V U. 9 Capital Structure Theory: M&M Proposition I An arbitrageur could buy α shares of the above firm, place them in a trust and sell debt and equity claims against these shares in proportions such that making then a riskless profit. α(d + E L ) > αe U, 10
6 Capital Structure Theory: M&M Proposition I Similarly, someone could buy all of the firm s shares for E U and modify the firm s capital stucture to have V L = D + E L > E U and then resell the firm for a riskless profit of V L V U. 11 Capital Structure Theory: M&M Proposition I In a frictionless market, this arbitrage oppotunity would lead to an increase in the firm s unlevered equity to the point where for any level of D and E L. V U = E U = D + E L = V L 12
7 Capital Structure Theory: M&M Proposition I with Taxes Consider an unlevered firm, denoted U, that expects constant earnings before interest and taxes, denoted EBIT, forever. Each period, if the corporate tax rate is T, shareholders receive and the government receives (1 T )EBIT T EBIT. 13 Capital Structure Theory: M&M Proposition I with Taxes Let E U = V U denote the present value of the payment (1 T )EBIT forever. Let G U denote the present value of the payment T EBIT forever. Let k 0 denote the firm s WACC when unlevered. Then E U = V U = (1 T )EBIT k 0. 14
8 Capital Structure Theory: M&M Proposition I with Taxes Consider a levered firm, Firm L, with the same EBIT as U, but with a perpetual debt issue D with coupon rate i. Interest payments are tax exempt. Shareholders receive (1 T )(EBIT id) each period forever, bondholders receive id each period forever, and the government receives T (EBIT id) each period forever. 15 Capital Structure Theory: M&M Proposition I with Taxes Each period, the total cash flow to shareholders and bondholders of Firm L is (1 T )(EBIT id) + id = (1 T )EBIT + TiD. The value of the levered firm is then the sum of two perpetuities, i.e. (1 T )EBIT forever and TiD forever. 16
9 Capital Structure Theory: M&M Proposition I with Taxes As before, the present value of (1 T )EBIT forever is V U. Discounting the tax shield cash flows TiD at the bonds coupon rate i, their present value is TiD i = T D, and thus the value of the levered firm is V L = V U + T D. 17 Capital Structure Theory: Financial Distress In a world with uncertainty, however, increasing D also increases the risk of bankruptcy. Financial distress creates some costs. Business risk is not affected by the level of debt but has an impact on the firm s capability to meet in financial obligations. Financial risk is directly affected by the firm s level of debt. 18
10 The Optimal Capital Structure The value of the levered firm can also be obtained using V = (1 T )EBIT k a, where k a is the firm s cost of capital. 19 The Optimal Capital Structure Suppose we have D D+E k e k d WACC (k a ) 0% 13.00% 8.00% 13.00% 20% 13.20% 8.20% 12.20% 40% 13.80% 8.80% 11.80% 50% 14.25% 9.25% 11.75% 60% 14.80% 9.80% 11.80% 80% 16.20% 11.20% 12.20% 20
11 The Optimal Capital Structure Then the value of the firm V = k a is minimized. (1 T )EBIT k a, is maximized when 21 The Optimal Capital Structure Suppose EBIT = 1,000 and T = 40%. Then D D+E k e k d WACC (k a ) V = (1 T )EBIT k a 0% 13.00% 8.00% 13.00% % 13.20% 8.20% 12.20% % 13.80% 8.80% 11.80% % 14.25% 9.25% 11.75% % 14.80% 9.80% 11.80% % 16.20% 11.20% 12.20%
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# Reflections
## Transformations that turn a figure into its mirror image by flipping it over a line.
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Reflections
Scott looked at the image below and stated that the image was reflected about the -axis. Is he correct? Explain.
### Watch This
First watch this video to learn about reflections.
CK-12 Foundation Chapter10ReflectionsA
Then watch this video to see some examples.
CK-12 Foundation Chapter10ReflectionsB
### Guidance
In geometry, a transformation is an operation that moves, flips, or changes a shape to create a new shape. A reflection is an example of a transformation that takes a shape (called the preimage) and flips it across a line (called the line of reflection) to create a new shape (called the image).
You can reflect a shape across any line, but the most common reflections are the following:
• reflections across the -axis: values are multiplied by -1.
• reflections across the -axis: values are multiplied by -1.
• reflections across the line : and values switch places.
• reflections across the line . and values switch places and are multiplied by -1.
#### Example A
Describe the reflection shown in the diagram below.
Solution: The shape is reflected across the y-axis. Let’s examine the points of the shapes.
Points on
Points on
In the table above, all of the -coordinates are multiplied by -1. Whenever a shape is reflected across the y-axis, it's -coordinates will be multiplied by -1.
#### Example B
Describe the reflection of the purple pentagon in the diagram below.
Solution: The pentagon is reflected across the x-axis. Let’s examine the points of the pentagon.
Points on
Points on
In the table above, all of the -coordinates are the same but the -coordinates are multiplied by -1. This is what will happen anytime a shape is reflected across the x-axis.
#### Example C
Describe the reflection in the diagram below.
Solution: The shape is reflected across the line . Let’s examine the points of the preimage and the reflected image.
Points on
Points on
Notice that all of the points on the preimage reverse order (or interchange) to form the corresponding points on the reflected image. So for example the point on the preimage is at (-1, 1) but the corresponding point on the reflected image is at (1, -1). The values and the values change places anytime a shape is reflected across the line .
#### Concept Problem Revisited
Scott looked at the image below and stated that the image was reflected across the -axis. Is he correct? Explain.
Scott is correct in that the preimage is reflected about the -axis to form the translated image. You can tell this because all points are equidistant from the line of reflection. Let’s examine the points of the trapezoid and see.
Point for Point for
All of the -coordinates for the reflected image are the same as their corresponding points in the preimage. However, the -coordinates have been multiplied by -1.
### Vocabulary
Image
In a transformation, the final figure is called the image.
Preimage
In a transformation, the original figure is called the preimage.
Transformation
A transformation is an operation that is performed on a shape that moves or changes it in some way. There are four types of transformations: translations, reflections, dilations and rotations.
Reflection
A reflection is an example of a transformation that flips each point of a shape over the same line.
### Guided Practice
1. Describe the reflection of the pink triangle in the diagram below.
2. Describe the reflection of the purple polygon in the diagram below.
3. Describe the reflection of the blue hexagon in the diagram below.
1. Examine the points of the preimage and the reflected image.
Points on
Points on
Notice that all of the -coordinates of the preimage (purple triangle) are multiplied by -1 to make the reflected image. The line of reflection is the -axis.
2. Examine the points of the preimage and the reflected image.
Points on
Points on
Notice that all of the -coordinates of the preimage (image 1) is multiplied by -1 to make the reflected image. The line of reflection is the -axis.
3. Examine the points of the preimage and the reflected image.
Points on
Points on
Notice that both the -coordinates and the -coordinates of the preimage (image 1) change places to form the reflected image. As well the points are multiplied by -1. The line of reflection is the line .
### Practice
If the following points were reflected across the -axis, what would be the coordinates of the reflected points? Show these reflections on a graph.
1. (3, 1)
2. (4, -2)
3. (-5, 3)
4. (-6, 4)
If the following points were reflected across the -axis, what would be the coordinates of the reflected points? Show these reflections on a graph.
1. (-4, 3)
2. (5, -4)
3. (-5, -4)
4. (3, 3)
If the following points were reflected about the line , what would be the coordinates of the reflected points? Show these reflections on a graph.
1. (3, 1)
2. (4, -2)
3. (-5, 3)
4. (-6, 4)
Describe the following reflections:
### Vocabulary Language: English
Coordinate Plane
Coordinate Plane
The coordinate plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. The coordinate plane is also called a Cartesian Plane.
Geometric Patterns
Geometric Patterns
Geometric patterns are visual patterns of geometric figures that follow a rule.
Image
Image
The image is the final appearance of a figure after a transformation operation.
perpendicular bisector
perpendicular bisector
A perpendicular bisector of a line segment passes through the midpoint of the line segment and intersects the line segment at $90^\circ$.
Perpendicular lines
Perpendicular lines
Perpendicular lines are lines that intersect at a $90^{\circ}$ angle.
Preimage
Preimage
The pre-image is the original appearance of a figure in a transformation operation.
Reflection
Reflection
A reflection is a transformation that flips a figure on the coordinate plane across a given line without changing the shape or size of the figure.
Transformation
Transformation
A transformation moves a figure in some way on the coordinate plane.
Rigid Transformation
Rigid Transformation
A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure. | 1,482 | 6,659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 2, "texerror": 0} | 4.875 | 5 | CC-MAIN-2016-07 | latest | en | 0.812969 |
https://www.freetechtrainer.com/how-to-create-dummy-variables-using-ifelse-statement-in-r-with-an-example/ | 1,721,214,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00792.warc.gz | 667,436,704 | 17,880 | # How to Create Dummy Variables Using ifelse Statement in R with an Example
A dummy variable is a type of variable that represents a categorical variable as a numerical variable that takes on one of two values: 1 or 0. In this tutorial, I’ll show a step-by-step process how to create dummy variables using ifelse statement in R with an Example.
Let’s create a sample dataframe with three variables id, age, marital_status, salary and we want to predict salary by age and marital_status. In the example of the tutorial, I’ll use the following sample data frame:
```>df <- data.frame( id = c("001", "002", "003"), age = c(35, 20, 40), marital_status = c("Married", "Single", "Divorced"), salary = c(41000, 21000, 42000) ) ```
Now, let’s run the below code to see the variable types of the dataframe.
>str(df)
```'data.frame': 3 obs. of 4 variables: \$ id : chr "001" "002" "003" \$ age : num 35 20 40 \$ marital_status: chr "Married" "Single" "Divorced" \$ salary : num 41000 21000 42000 ```
To use marital_status as a predictor variable in a regression model, we have to convert it into a dummy variable. Since, it is currently a character variable that can take on three different values (“Single”, “Married”, or “Divorced”).
Now, we can use the ifelse() statement in R, to define dummy variables and then define the final data frame we’d like to use to build the regression model:
``` >marital_status_single <- ifelse(df\$marital_status== 'Single', 1, 0) >marital_status_married <- ifelse(df\$marital_status== 'Married', 1, 0) >marital_status_divorced <- ifelse(df\$marital_status== 'Divorced', 1, 0) ```
Re-creating the dataframe with the newly created dummy variables
```>ddf <- data.frame(id = df\$id, age = df\$age, salary = df\$salary, marital_status_Single = marital_status_Single, marital_status_Married = marital_status_Married, marital_status_Divorced = marital_status_Divorced ) ```
To see the newly created final data frame we have to execute the following command:
>str(ddf)
```'data.frame': 3 obs. of 6 variables: \$ id : chr "001" "002" "003" \$ age : num 35 20 40 \$ salary : num 41000 21000 42000 \$ marital_status_Single : num 0 1 0 \$ marital_status_Married : num 1 0 0 \$ marital_status_Divorced: num 0 0 1 ```
In this tutorial, I tried to show how to create dummy variables using ifelse statement in R with an example. Hope you have enjoyed the tutorial. If you want to get updated, like the facebook page https://www.facebook.com/LearningBigDataAnalytics and stay connected. | 686 | 2,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.702004 |
https://brainmass.com/physics/acceleration/uniform-circular-motion-centripetal-acceleration-37351 | 1,701,730,364,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00163.warc.gz | 171,500,805 | 6,457 | Purchase Solution
# Uniform Circular Motion and Centripetal Acceleration
Not what you're looking for?
A car with a constant speed of 83.0 km/h enters a circular flat curve with a radius of curvature of 0.400 km. If the friction between the road and the car's tires can supply a centripetal acceleration of 1.25 m/s^2, does the car negotiate the curve safely? Justify the answer.
##### Solution Summary
The expert examines uniform circular motion and centripetal acceleration.
##### Solution Preview
v = 83 kmph = 83*5/18 m/s = 23.055 m/s
r = 0.400 km = 400 m
Centripetal ...
##### Introduction to Nanotechnology/Nanomaterials
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Some short-answer questions involving the basic vocabulary of string, sound, and water waves.
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Test your knowledge of moon phases and movement.
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How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz. | 292 | 1,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.856601 |
http://www.mental-arithmetic.co.uk/Number-Bonds-to-100.htm | 1,495,694,584,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608004.38/warc/CC-MAIN-20170525063740-20170525083740-00288.warc.gz | 579,739,402 | 2,848 | # Number Bonds to 100 Worksheets
Click here to return to the main worksheet index
Click here for our other Year 2 Worksheets.
A number bond is just a maths fact that a pair of numbers when added together make a given total. For example the number bonds for 10 are 0+10=10, 1+9=10; 2+8=10, 3+7=10, 4+6=10, 5+5=10 etc, and the number bonds for 20 are
0+20 = 20, 1+19 = 20, 2+18 = 20 etc.
Number bonds like times tables are something which a child should know instantly without needing to think - therefore lots of practice by play and worksheets are essential to memorise these number relationships.
Number bonds help children to understand how numbers work - that a number can be made up of parts, and the sum of the parts is the whole. Number bonds are also a good introduction to the fact that subtraction is the inverse of addition - A child learning that 4 add 6 equals 10 is also learning that 10 minus 6 is 4 and that 10 minus 4 is 6.
Try our Number Bonds to 10 Worksheets and Number Bonds to 20 Worksheets if you have not already.
NEW Click here to try our new free *Number Bonds PDF Quiz Generator* with which you can generate your own printable PDF worksheets and quizzes to meet your exact needs. Set the number bond to be tested to any value from 5 to 999, select the number of questions required from 10-99, and click to generate your bespoke worksheet.
Number Bonds to 100 Worksheet 1Number Bonds to 100 Worksheet 2Number Bonds to 100 Worksheet 3Number Bonds to 100 Worksheet 4Number Bonds to 100 Worksheet 5Number Bonds to 100 Worksheet 6Number Bonds to 100 Worksheet 7Number Bonds to 100 Worksheet 8Number Bonds to 100 Worksheet 9
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© 2010-2015 Mental-Arithmetic.co.uk - All Rights Reserved | 468 | 1,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-22 | longest | en | 0.85543 |
https://math.stackexchange.com/questions/1240973/expected-value-and-variance-of-a-max-randomized-stocks | 1,718,267,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00257.warc.gz | 363,864,475 | 36,323 | # Expected Value and variance of a max randomized stocks
Hey guys I have been working on a probability and expected value/variance problem and the problem is:
Each day the price of a stock in the market is a random number between 0 and 1 independently of its value in previous days. A record breaking price occurs when the price on the current day exceeds all previous prices. What is the expected number of record breaking prices in n days? What is the variance of the number of record breaking prices in n days? (Hint: If I throw k points at random in [0,1] what is the chance that the ith point I throw is the largest among those k points? Given that the ith point is the largest among those k points, is its distribution different from uniform in [0, 1]?)
My approach was: first day Expected v = 1*1*(1 choose 1) = 1 second day expected v = first day Expected V + P(second price>first price) * 2(2 choose 1) but then from third day, it becomes way too complicated.
I know I should be using the integral to find Expected value of continuous variable, but then I would need to know the probability of each cases, which I have no idea of figuring it out.
Intuitively I know that the probability of getting the max value gets significantly lower as days go on, and I would assume it's on the line of a log function. But again, this is just a guess.
Can someone please lend a hand? thank you!
• What's the distribution on the random number between 0 and 1? Uniform?
– Newb
Commented Apr 18, 2015 at 21:24
• It's not necessarily uniform I don't think.
– PGod
Commented Apr 18, 2015 at 21:25
• It's got to be uniform. Otherwise the question is impossible, right? Commented Apr 18, 2015 at 21:33
• Yes they should be considered uniform. !
– user232537
Commented Apr 18, 2015 at 22:04
For $i=1,2,...,n$, let $X_i=1$ if there is a record set on day $i$, and $X_i=0$ if there isn't a record. For $X_i=1$, we must have that the value on day $i$ is the highest of the first $i$ days. So $P(X_i=1)=\frac1i$. So $E[X_i]=\frac1i$.
Since expectation is linear we have $E[\sum_{i=1}^n X_i]=\sum_{i=1}^n E[X_i]=1+\frac12+\cdots+\frac1n$, and this gives the expected number of records over $n$ days.
This confirms your idea that the number of records is roughly logarithmic in $n$.
• It seems that the key question for the variance is what is $E[X_i X_j]$ for $i \neq j$, since $E[X_i^2]$ is just $E[X_i]$ again.
• @Ian I think you're right, since the $X_i$s aren't independent. Commented Apr 19, 2015 at 2:18 | 701 | 2,501 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-26 | latest | en | 0.943254 |
jintian93.github.io | 1,607,027,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141732696.67/warc/CC-MAIN-20201203190021-20201203220021-00264.warc.gz | 358,770,224 | 8,523 | pytorch自己实现一个CrossEntropy函数
This article was original written by Jin Tian, welcome re-post, first come with https://jinfagang.github.io . but please keep this copyright info, thanks, any question could be asked via wechat: jintianiloveu
理论(theory)
$$CrossEntropy = \sum{(ylog(y’) + (1-y)log(1-y’))}$$
[[1, 0, 0],
[0, 1, 0]]
[[545, 54, 2],
[232, 54, 546]]
1. 先计算softmax, 然后计算log;
2. 计算 $yi*log(yi)$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 import torch import numpy as np # label = torch.Tensor([ # [1, 2, 3], # [0, 2, 1], # ]) label = torch.Tensor([0, 2, 1]).long() fc_out = torch.Tensor([ [245, 13., 3.34], [45., 43., 37.], [1.22, 35.05, 1.23] ]) def one_hot(a, n): b = a.shape[0] c = np.zeros([b, n]) for i in range(b): c[i][a[i]] = 1 return np.array(c) def softmax(a): return [np.exp(i)/np.sum(np.exp(i)) for i in a] def cross_entropy_loss(out, label): # convert out to softmax probability out_list = out.numpy().tolist() out1 = softmax(out_list) print(out1) out2 = torch.softmax(out, 0) print(out2) # [0, 2, 1] -> [[1, 0, 0], [0, 0, 1], [0, 1, 0]] # onehot label and rotate label_onehot = one_hot(label, 3) loss = np.sum(out1 * label_onehot.T) print(loss) loss = torch.nn.CrossEntropyLoss() lv = loss(fc_out, label) print(lv) lv = cross_entropy_loss(fc_out, label) print(lv) | 557 | 1,389 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-50 | latest | en | 0.478189 |
https://www.unitconverters.net/speed/meter-hour-to-millimeter-second.htm | 1,720,939,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00149.warc.gz | 919,156,951 | 3,361 | Home / Speed Conversion / Convert Meter/hour to Millimeter/second
# Convert Meter/hour to Millimeter/second
Please provide values below to convert meter/hour [m/h] to millimeter/second [mm/s], or vice versa.
From: meter/hour To: millimeter/second
### Meter/hour to Millimeter/second Conversion Table
Meter/hour [m/h]Millimeter/second [mm/s]
0.01 m/h0.0027777778 mm/s
0.1 m/h0.0277777778 mm/s
1 m/h0.2777777778 mm/s
2 m/h0.5555555556 mm/s
3 m/h0.8333333333 mm/s
5 m/h1.3888888889 mm/s
10 m/h2.7777777778 mm/s
20 m/h5.5555555556 mm/s
50 m/h13.8888888889 mm/s
100 m/h27.7777777778 mm/s
1000 m/h277.7777777778 mm/s
### How to Convert Meter/hour to Millimeter/second
1 m/h = 0.2777777778 mm/s
1 mm/s = 3.6 m/h
Example: convert 15 m/h to mm/s:
15 m/h = 15 × 0.2777777778 mm/s = 4.1666666667 mm/s | 304 | 798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.293987 |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-1-section-1-2-combining-functions-shifting-and-scaling-graphs-exercises-page-20/62 | 1,585,958,151,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518767.60/warc/CC-MAIN-20200403220847-20200404010847-00396.warc.gz | 988,220,156 | 12,418 | ## University Calculus: Early Transcendentals (3rd Edition)
The equation of the stretched graph is $$y=3\sqrt{x+1}$$
For $c\gt1$, the graph is scaled: - $y=cf(x)$ - stretched - vertically - factor of $c$ - $y =\frac{1}{c}f(x)$ - compressed - vertically - factor of $c$ - $y=f(cx)$ - compressed - horizontally - factor of $c$ - $y=f(\frac{x}{c})$ - stretched - horizontally - factor of $c$ $$y=\sqrt{x+1}$$ The graph is STRETCHED - VERTICALLY - by a factor of $3$. Therefore, the equation of the new graph is $$y=3\sqrt{x+1}$$ | 173 | 526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-16 | longest | en | 0.780156 |
http://www.mathcs.emory.edu/~cheung/Courses/554/Syllabus/2-disks/pointer-swizzling4.html | 1,513,204,737,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948531226.26/warc/CC-MAIN-20171213221219-20171214001219-00718.warc.gz | 408,729,158 | 2,567 | ### Un-pin blocks efficiently
• Solving problem 2: store information to help us perform unpinning on a block
• Recall problem 2:
• When we want to release (clear) a block/record in memory that is being referenced through a virtual memory address:
The block can only be released after all referencing virtual addresses have been unswizzled:
• Pre-requisite to the solution:
• We must maintain information on the locations of all referencing pointers to a given database address
Example:
The location information is necessary to perform the unswizzling operation on all referencing virtual memory addresses
• Assumption:
• Length of a database address 2 × Length of a virtual memory address
Graphically:
• If length of a database address < 2 × Length of a virtual memory address, we will
allocate extra bytes in the database address field to make its length equal to 2 × length(virtual address)
Reason for doing this:
• We use the first half of the (memory) space of the database address:
To store the swizzled virtual memory address of the database object in memory
• We use the second half of the (memory) space of the database address:
• The linked list data structure used to store information on the locations of all referencing pointers to a given database address:
Notice that:
• We add an extra field into the translation table:
• The Linked List field contains the start of the linked list
``` Before swizzling: +---------------------+---------------------+ | database address of some database object | +---------------------+---------------------+ After swizzling: +---------------------+---------------------+ | Vir Mem Addr of Obj | next field of list | +---------------------+---------------------+ ```
• Unpinning information: How the linked list is constructed
• Initially: the memory is empty
• Read a block with a certain record into memory:
We enter the database address and its corresponding virtual memory address into the translation table
The linked list is initialized to an empty list
• We read in a block containing a database address of the block:
• We read in a another block containing a database address of the block:
• Unpinning a block
• Algorithm to unpin a record located at virtual memory address x:
• Find the entry for x in the translation table
(Use a brute force search in the translation table or use the hash table technique discussed above)
• Traverse the linked list and unpin every virtual address field found in the list:
Before:
After: | 512 | 2,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-51 | latest | en | 0.836711 |
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# RE: Concrete Tank Design Help
• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Concrete Tank Design Help
• From: "Jim Kestner" <jkestner(--nospam--at)somervilleinc.com>
• Date: Fri, 27 Sep 2002 10:30:55 -0500
Mark:
At interior wall conditions, I generally use standard hooks on the vertical
wall dowels and continuous steel in the base slab. Check CRSI to make sure
that the vertical dowels have enough embedment to be fully developed with a
standard hook. If not, use smaller bars at closer spacing or thicken the
base slab to provide the required embedment.
At exterior walls, I typically use a full lap splice both in the vertical
wall and the base slab. I require the same dowels on both wall faces since
tanks are generally hydrostatically tested before backfilling, so large
forces can occur in each direction. I lap both dowels with the bottom layer
of steel in the base slab.
You do not want to lap an inside bar in the wall with a top bar in the base
slab. This is an "inside to inside" connection that if stressed will pop out
the corner. Extend those bars to the outside faces and hook.
How much fixity to use at the base is a matter of judgement. In reality, no
connection or its resiting elements can provide perfectly pinned or fixed
conditions.
You could do a sophisticated analysis by a finite element model in 3D with
compression springs to model the soil. This may be easy or difficult
depending on what software package is used.
A simpler anaylsis would be to design the wall for the maximum steel given
by each analysis (pinned or fixed) whichever is greater. Then provide dowels
to fix the base. Design the base to resist this overturning as if it is the
base of a retaining wall. Another analysis that you could use is a beam on
elastic foundation (Enercalc)with an end moment and vertical load. You may
have to thicken the base out a distance away from the wall to resist these
forces. The outward forces (caused by hydrostatic tesing) are harder to
resist by the base so I would probably consider this condition pinned. In
other words, conservatively bound the problem and design for the maximum
forces produced by either (pinned or fixed) analysis, whichever is greater.
I hope this helps!
Jim K.
-----Original Message-----
From: Mark Seifried [mailto:mseifried(--nospam--at)ggjengineers.com]
Sent: Friday, September 27, 2002 7:11 AM
To: seaint(--nospam--at)seaint.org
Subject: RE: Concrete Tank Design Help
My question was a general one for all tank walls: interior, exterior, and
*What about preventing the transfer of slab moment into the walls due to the
fixed condition between the wall and slab?
*Partial fixity cracks with 90d hooks; at what plane in the joint would
adequate tension development end? It would seem that this point would not
occur until the bend in the hook, wouldn?t this moment dissipate as it has
traveled through the slab?
*I have run analysis of structural models with many different degrees of
fixity at the wall base, what percentage moment transfer could a 90d hook
pinned joint provide along the base of the wall? Zero?
Thank you for the response. I was beginning to think my question was too
easy, but I have looked through many ACI & PCA design guides that have not
touched on this subject. This list server has given provided tremendous
knowledge within many areas over the last year, especially since I?ve only
been out of school a couple of years.
Mark Seifried, EIT
From: Sherman, William [mailto:ShermanWC(--nospam--at)cdm.com]
Sent: Thursday, September 26, 2002 10:59 AM
To: 'seaint(--nospam--at)seaint.org'
Subject: RE: Concrete Tank Design Help
Is the wall to slab connection at an interior wall or exterior wall? If at
an exterior wall, is there exterior soil load which will put the exterior
face in tension at the base of wall?
At interior walls, the wall dowels may use standard hooks into the base
slab, assuming continuous reinforcement in the base slab.
At an exterior corner between a wall and base slab, and assuming moment
transfer with tension at both faces, the outside vertical wall reinforcement
should be lap spliced to the bottom slab reinforcement and the inside
vertical wall reinforcement may be doweled with a standard hook. A standard
hook at the outside of the corner does not provide adequate continuity if in
tension. (At any plane cut thru the joint, you should have continuity of the
outside tension reinforcement. If you just provide a standard hook on such a
bar, you may get cracking in the joint due to partial restraint without
Also, the slab reinforcement must be able to resist the moment transferred
from the wall.
William C. Sherman, PE
CDM, Denver, CO
Phone: 303-298-1311
Fax: 303-293-8236
email: shermanwc(--nospam--at)cdm.com
-----Original Message-----
From: Mark Seifried [mailto:mseifried(--nospam--at)ggjengineers.com]
Sent: Wednesday, September 25, 2002 8:29 AM
To: seaint(--nospam--at)seaint.org
Subject: Concrete Tank Design Help
My question deals with the transfer of moment from the base of a tank wall
into the tank slab. I know that this connection can be assumed closer to
pinned of fixed depending on the reinforcement detailing.
But, if this connection is assumed close to fixed, what kind of
reinforcement is necessary to complete this transfer of moment into the
slab? I typically see 90d standard hooks used as dowels at the same spacing
as the vertical wall rebar... Is this tension development length provided by
the hook adequate for transfer? Or does the standard hook EF result in more
of a pinned connection since the length of rebar parallel with the bottom
slab rebar is very small?
Thank you in advanced for your help or any direction you can provide? Mark
Seifried, EIT
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* send to the list is public domain and may be re-posted | 1,837 | 7,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-17 | latest | en | 0.898196 |
https://www.racecar-engineering.com/articles/tech-explained-chassis/ | 1,723,423,064,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00719.warc.gz | 732,883,574 | 40,011 | We’ve covered the journey of forces from the tyre contact patch through the wheels and suspension into the springs and dampers, and now we arrive at their final destination, the chassis.
In this penultimate article of the racecar vehicle dynamics series, tyre dynamics, suspension, kinematics and spring-damper systems all come together and interact with the chassis to complete the puzzle of vehicle performance.
Chassis Modes
In the vehicle dynamics world, we refer to chassis modes. Modes are combinations of wheel deflections that produce a particular form of chassis displacement. Traditionally these modes are discussed concerning road inputs, but the concept is also used to illustrate chassis displacements caused by longitudinal and lateral forces generated whilst driving.
There are four displacement modes: Heave, Pitch, Roll, and Warp. For brevity, I’ll summarise the modes in the table below.
Heave is an interesting mode in motorsport due to its relationship with ride height and underbody aerodynamics. With huge aerodynamic loads forcing the chassis towards the track surface, spring stiffnesses must be specified very high to maintain the chassis in the optimal window of its aero map; this introduces some significant compromises when those loads aren’t present.
Roll receives a lot of attention in chassis design due to the transient effects on tyre contact patch loads and the dynamic impact on wheel camber.
When designing a racecar platform, it’s common to have a roll gradient (°/g of lateral acceleration) target, which is agreed upon by aerodynamicists and the suspension and kinematics team to ensure the car works together as a unit. There’s no magic number to roll or pitching targets. To understand why roll happens, I introduce the much-discussed concept of the roll centre.
The roll centre is the application point of tyre forces to the chassis – its vertical distance from the CoM is the source of roll moment (or roll torque) acting upon the CoM; this is simple leverage.
As you see from the illustration, the resultant tyre forces at the contact patches of a particular axle have both a vertical and horizontal component. The magnitude of each is determined by the roll centre’s vertical height from the track surface. The vertical component of these forces is the jacking force and acts to lift the chassis, which is uncomfortable for a driver and impacts weight transfer by raising the CoM.
It’s desirable to keep the roll centre as low as possible to maximise the horizontal component, but lowering the roll centre too much creates a large moment arm between the CoM, though, so there’s a balance to be made here.
Maintaining a low roll torque also means less roll stiffness is required from the system – returning full circle to the tyre, which finds itself in more favourable conditions. You should now start to see the origins of a vehicle dynamicists desire to keep CoM height low!
There is a direct analogy to this in the side view where the roll centre concept can be applied to the Pitch centre. Here, it’s anti-squat and anti-dive geometry that controls the location of pitch and dive centres, as described in the kinematics article.
Roll and pitch also influence aerodynamic performance by changing the proximity of the underfloor and aerodynamics components to the track surface and introducing the chassis rake angle. Warp is of less significance as it’s not stimulated by any specific chassis accelerations and responds to road inputs. Warp is coupled to the roll mode.
Mode Decoupling
Treatment of these chassis modes is significant in vehicle dynamics because each mode has a different impact on dynamic situations. With different responses, we should manage them separately.
In an ideal situation where freedom to design for optimal chassis control is given, each mode would be controlled by a separate spring-damper system – providing different natural frequencies and different damping rates for each mode. Standard passive suspension systems like those on GT cars are challenging to realise that due to vehicle packaging.
Systems that act to decouple each chassis mode from one another are called mode-decoupling systems. The anti-roll bar contributes towards this by separating spring rates in heave and pitch from a roll in its simplest form. Designers can start to get creative with their approaches on prototype and open-wheel formula cars where there’s more design freedom.
Spring and damping rates suitable to support the heave mode would be way too high in roll, where they will introduce very high variations in contact patch loading and a fall in average grip levels during cornering.
Commonplace in LMP and F1/IndyCar style chassis are suspension designs that tackle this problem by separating heave and pitch damping from roll damping by introducing a heave spring and damper. It provides the proper support for the chassis under high aero loading to maintain the aero map whilst providing a supple enough platform for track compliance in other modes.
Passively decoupling the pitch mode is the tricky one. It usually requires hydraulic linking the front and rear suspension systems to introduce some additional damping to the movement. Porsche had some success with this in their 919 LMP1 car with the FRIC (Front Rear Interconnected Control) system.
Chassis Balance
One of the most critical variables controlling the chassis is the Centre of Mass (CoM) location relative to the contact patches. That can be the difference between a great car and a terrible one.
The chassis balance is often assessed subjectively, but it does have very objective roots. In the most basic terms, chassis balance describes which axle loses grip first and leads to under or oversteer. The physics behind balance forms one of the fundamental equations of cornering, namely the concept of yaw moment equilibrium.
For this article, it’s enough to understand that with all else being equal (i.e. the exact tyre, spring-damper and suspension kinematics on all four wheels), the significant input into chassis balance is the longitudinal CoM location relative to the wheelbase of the racecar.
One can easily deduce the longitudinal CoM position from the static front-to-rear weight distribution of the chassis. With a 50:50 weight distribution, the CoM is precisely between the two axles. Dynamically this is an ideal scenario as it means the tyres on the front and rear axles are operating at the same slip angles and work equally.
With a forward weight distribution, the front tyres are operating at larger slip angles than the rears. From the tyre dynamics article, you’ll remember that once a tyre reaches its optimum slip angle, the lateral force it generates starts to fall, in this case leading to understeer. The opposite is true for oversteer.
As the race car in this situation is grip limited by the overactive axle, it’s harmful for maximum lateral acceleration and tyre wear, so significant efforts are made to package the CoM as centrally as possible during the design phase. One can adjust balance through the springs and dampers, but this addresses the symptom rather than the cause. A fundamentally balanced chassis is always the target.
Weight Transfer
Now that’s understood, we move on to vertical CoM positioning. Starting from the most elementary situations of longitudinal acceleration – a driver applies some throttle, and the car accelerates.
Due to inertia, the race car’s mass resists acceleration, which the tyres feel as a shift of weight from the front wheels to the rear wheels. The extent of this weight transfer is a function of the vertical location of the chassis CoM from the track surface and the vehicle’s wheelbase. It’s directly proportional to the magnitude of the acceleration.
In a rear-drive car with this simplified scenario, weight transfer is nothing but positive as it will result in an increase of grip at the driven tyres. Still, a high CoM has definite drawbacks in other handling scenarios.
In cornering, weight transfer is not helpful as it means the inside tyres become unloaded, while the outside tyres become heavily loaded. This time, the extent of the weight transfer is proportional to the vertical CoM height, track width and lateral acceleration.
As you saw in the tyre dynamics article, grip levels increase with vertical loads, but the CoF falls, so in the end, this means that the grip gained by the outside wheels is less than the grip lost by the inside wheels – overall lateral acceleration suffers.
Spring stiffness nor suspension geometry can change this; it’s a hard fact. Weight transfer is undoubtedly something to be minimised, which is why you see and hear such emphasis on keeping the CoM as low as possible during the design process.
Moments of inertia
The inertia of the chassis is crucial in roll and pitch responses, and it’s also an essential influence in the yaw response of the chassis. The idea of considering the vehicle’s mass as concentrated at the CoM is an excellent assumption in many static cases. Still, it doesn’t account for the distribution of that mass within the chassis and dynamic instances.
The closer the mass is located to the CoM (which, for argument’s sake, we will assume to be the axis of rotation), the lower rotational inertia the chassis displays. That ultimately means that the yawing moments generated by the chassis generate higher yaw accelerations and result in a sharper dynamic response.
Generally, on circuit racing cars, locating heavy powertrain components close to the centre of mass is standard practice to keep the yaw inertia as low as possible.
The result is a subjectively sharp response to steering input and positive adjustability at the grip limit for the driver, with an objectively measured improvement in time response to yaw moments for the chassis. That generates more significant yaw accelerations and translates to reduced lap times.
What I hope has been communicated throughout the series thus far is that a holistic approach to racecar performance is required when vehicle dynamics is concerned. Ultimately, the main bulk of the car lays at the chassis, so all other pieces in the puzzle work in symbiosis to accelerate it around the race track as rapidly as possible.
By locating the main masses strategically to optimise balance, weight transfer, and yaw accelerations, interaction with the suspension system provides a platform for all other pieces in the puzzle to create the equation of performance.
The next and final article in the series will focus on data acquisition. Stay close! | 2,063 | 10,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.905283 |
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• 1. to BACK TO SCHOOL NIGHTand SOUTHAMPTON!
• 3. SPELLING
STORYTOWN SPELLING PROGRAM
WEEKLY PRE AND POST TESTS
HOMEWORK MON. – THURS.
FINAL GRADE: Weekly tests, accuracy on written assignments/reading journal, performance in Word Work
• 4. StoryTown and JGB Program
CSI: COMPREHENSION STRATEGY INSTRUCTION
Teach and reinforce skills related to reading instruction (i.e. comparing, summarizing, etc.)
Students will use anthology stories, books, short stories, and informational text
• 5. 100 Most Misspelled Words
OTHER WORD WORK: FOCUS ON VOCABULARY, WORKING WITH SPELLING WORDS, USING WORD WALL
Nifty Thrifty Fifty
• 6. WRITTEN COMPOSITIONWith Mrs. Thompson
Emphasis on:
Language mechanics, parts of speech, writing process and application, paragraph and composition format, poetry, figurative language, language review and paragraph editing
Starting with Touchtone Texts Unit to analyze the author’s craft.
Students write about topics that interest them.
• 7. HANDWRITING
Gradual immersion in class handwriting practice with emphasis on:
Cursive letter connections, formations and neatness
• 8. Reference Book, Journal, and Study Link
Homework: Daily Monday-Thursday
Daily review routine of math facts
Daily questioning forum with student/student problem solving
Skills:
Number Theory, Estimation and Computation, Geometry, Division, Fractions, Decimals, Percents, Exponents and Negative Numbers, Fractions and Ratios, Coordinates, Area, Volume and Capacity, Algebra, Volume, and Probability
Extension using Games, Activities, and Projects
MATHEMATICS
EVERYDAY MATHEMATICS
• 9. Environmental Issues
Astronomy and Weather
Energy Around Us
Elements of Life
Family Life
• 10.
• Geography Review: Investigations
• 11. Cultural Groups: Encounters
• 12. Colonial Life: Progressions
• 13. The American Revolution
• 14. The Constitution
• 15. African American Culture
Social Studies(With Mrs. Grabau)
• 16. Expect nightly homework in Spelling, Math, and ReadingHomework planner checked daily Monday Folders
• 17. COMMUNICATION
Homework Planner
Teacher/Parent Notes
E-MAIL:
Online:
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https://thecuomofoundationmonaco.org/articles/binomial-probability-examples-5d1bad | 1,611,823,393,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704839214.97/warc/CC-MAIN-20210128071759-20210128101759-00380.warc.gz | 604,887,483 | 12,925 | +377 97 77 01 66 info@fondationcuomo.mc
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A ball is chosen at random and it is noted whether it is red. Solution: (a) The repeated tossing of the coin is an example of a Bernoulli trial. Example 2: For the same question given above, find the probability of: Solution: P (at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1). × (½)2× (½)3 P(x=2) = 5/16 (b) For at least four heads, x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5) Hence, P(x = 4) = 5C4 p4 q5-4 = 5!/4! Each question has four possible answers with one correct answer per question. Hence. There are two possible outcomes: true or false, success or failure, yes or no. n = 10, p=0.15, q=0.85, x=4. The 0.7 is the probability of each choice we want, call it p. The 2 is the number of choices we want, call it k. And we have (so far): = p k × 0.3 1. 6 times, a ball is selected at random, the color noted and then replaced in the box.What is the probability that the red color shows at least twice?Solution to Example 7The event "the red color shows at least twice" is the complement of the event "the red color shows once or does not show"; hence using the complement probability formula, we writeP("the red color shows at least twice") = 1 - P("the red color shows at most 1") = 1 - P("the red color shows once" or "the red color does not show")Using the addition ruleP("the red color shows at least twice") = 1 - P("the red color shows once") + P("the red color does not show")Although there are more than two outcomes (3 different colors) we are interested in the red color only.The total number of balls is 10 and there are 3 red, hence each time a ball is selected, the probability of getting a red ball is $$p = 3/10 = 0.3$$ and hence we can use the formula for binomial probabilities to findP("the red color shows once") = $$\displaystyle{6\choose 1} \cdot 0.3^1 \cdot (1-0.3)^{6-1} = 0.30253$$P("the red color does not show") = $$\displaystyle{6\choose 0} \cdot 0.3^0 \cdot (1-0.3)^{6-0} = 0.11765$$P("the red color shows at least twice") = 1 - 0.11765 - 0.30253 = 0.57982. eval(ez_write_tag([[300,250],'analyzemath_com-large-mobile-banner-2','ezslot_13',701,'0','0']));Example 880% of the people in a city have a home insurance with "MyInsurance" company.a) If 10 people are selected at random from this city, what is the probability that at least 8 of them have a home insurance with "MyInsurance"?b) If 500 people are selected at random, how many are expected to have a home insurance with "MyInsurance"?Solution to Example 8a)If we assume that we select these people, at random one, at the time, the probability that a selected person to have home insurance with "MyInsurance" is 0.8.This is a binomial experiment with $$n = 10$$ and p = 0.8. Hence if you When we replace in the formula: Interpretation: The probability that exactly 4 candies in a box are pink is 0.04. Example 1A fair coin is tossed 3 times. There are two parameters n and p used here in a binomial distribution. According to the problem: Number of trials: n=5 Probability of head: p= 1/2 and hence the probability of tail, q =1/2 For exactly two heads: x=2 P(x=2) = 5C2 p2 q5-2 = 5! the probability of getting a red card in one trial is $$p = 26/52 = 1/2$$The event A = "getting at least 3 red cards" is complementary to the event B = "getting at most 2 red cards"; hence$$P(A) = 1 - P(B)$$$$P(A) = P(3)+P(4) + P(5)+P(6) + P(7)+P(8) + P(9) + P(10)$$$$P(B) = P(0) + P(1) + P(2)$$The computation of $$P(A)$$ needs much more operations compared to the calculations of $$P(B)$$, therefore it is more efficient to calculate $$P(B)$$ and use the formula for complement events: $$P(A) = 1 - P(B)$$.$$P(B) = \displaystyle {10\choose 0} 0.5^0 (1-0.5)^{10-0} + {10\choose 1} 0.5^1 (1-0.5)^{10-1} + {10\choose 2} 0.5^2 (1-0.5)^{10-2} \\ = 0.00098 + 0.00977 + 0.04395 = 0.0547$$$$P(\text{getting at least 3 red cards}) = P(A) = 1 - P(B) = 0.9453$$. When you throw the dice 10 times, you have a binomial distribution of n = 10 and p = ⅙. × (½)4× (½)1 = 5/32. $$S = \{ (H H H) , \color{red}{(H H T)} , \color{red}{(H T H)} , (H T T) , \color{red}{(T H H)} , (T H T) , (T T H) , (T T T) \}$$Event $$E$$ of getting 2 heads out of 3 tosses is given by the set$$E = \{ \color{red}{(H H T)} , \color{red}{(H T H)} , \color{red}{(T H H)} \}$$In one trial ( or one toss), the probability of getting a head is$$P(H) = p = 1/2$$and the probability of getting a tail is$$P(T) = 1 - p = 1/2$$The outcomes of each toss are independent, hence the probability $$P (H H T)$$ is given by the product:$$P (H H T) = P(H) \cdot P(H) \cdot P(T) \\ Your email address will not be published. Each question has five possible answers with one correct answer per question. Here the number of failures is denoted by ‘r’. Hence Example 2 A fair coin is tossed 5 times. Finding the quantity of raw and used materials while making a product. "at least 8 of them have a home insurance with "MyInsurance" means 8 or 9 or 10 have a home insurance with "MyInsurance"The probability that at least 8 out of 10 have have home insurance with the "MyInsurance" is given by\( P( \text{at least 8}) = P( \text{8 or 9 or 10})$$Use the addition rule$$= P(8)+ P(9) + P(10)$$Use binomial probability formula calling "have a home insurance with "MyInsurance" as a "success".$$= P(8 \; \text{successes in 10 trials}) + P(9 \; \text{successes in 10 trials}) + P(10 \; \text{successes in 10 trials})$$$$= \displaystyle{10\choose 8} \cdot 0.8^8 \cdot (1-0.8)^{10-8} + \displaystyle{10\choose 9} \cdot 0.8^9 \cdot (1-0.8)^{10-9} + \displaystyle{10\choose 10} \cdot 0.8^10 \cdot (1-0.8)^{10-10}$$$$= 0.30199 + 0.26843 + 0.10737 = 0.67779$$b)It is a binomial distribution problem with the number of trials is $$n = 500$$.The number of people out of the 500 expected to have a home insurance with "MyInsurance" is given by the mean of the binomial distribution with $$n = 500$$ and $$p = 0.8$$.$$\mu = n p = 500 \cdot 0.8 = 400$$400 people out of the 500 selected at random from that city are expected to have a home insurance with "MyInsurance". The number of trials must be fixed. Samples of 1000 tools are selected at random and tested.a) Find the mean and give it a practical interpretation.b) Find the standard deviation of the number of tools in good working order in these samples.Solution to Example 4When a tool is selected, it is either in good working order with a probability of 0.98 or not in working order with a probability of 1 - 0.98 = 0.02.When selecting a sample of 1000 tools at random, 1000 may be considered as the number of trials in a binomial experiment and therefore we are dealing with a binomial probability problem.a) mean: $$\mu = n p = 1000 \times 0.98 = 980$$In a sample of 1000 tools, we would expect that 980 tools are in good working order .b) standard deviation: $$\sigma = \sqrt{ n \times p \times (1-p)} = \sqrt{ 1000 \times 0.98 \times (1-0.98)} = 4.43$$, Example 5Find the probability that at least 5 heads show up when a fair coin is tossed 7 times.Solution to Example 5The number of trials is $$n = 7$$.The coin being a fair one, the outcome of a head in one toss has a probability $$p = 0.5$$.Obtaining at least 5 heads; is equivalent to showing : 5, 6 or 7 heads and therefore the probability of showing at least 5 heads is given by$$P( \text{at least 5}) = P(\text{5 or 6 or 7})$$Using the addition rule with outcomes mutually exclusive, we have$$P( \text{at least 5 heads}) = P(5) + P(6) + P(7)$$where $$P(5)$$ , $$P(6)$$ and $$P(7)$$ are given by the formula for binomial probabilities with same number of trial $$n$$, same probability $$p$$ but different values of $$k$$.$$\displaystyle P( \text{at least 5 heads} ) = {7\choose 5} (0.5)^5 (1-0.5)^{7-5} + {7\choose 6} (0.5)^6 (1-0.5)^{7-6} + {7\choose 7} (0.5)^7 (1-0.5)^{7-7} \\ = 0.16406 + 0.05469 + 0.00781 = 0.22656$$. Therefore the probability of getting a correct answer in one trial is $$p = 1/5 = 0.2$$It is a binomial experiment with $$n = 20$$ and $$p = 0.2$$.$$P(\text{student answers 15 or more}) = P( \text{student answers 15 or 16 or 17 or 18 or 19 or 20}) \\ = P(15) + P(16) + P(17) + P(18) + P(19) + P(20)$$Using the binomial probability formula$$P(\text{student answers 15 or more}) = \displaystyle{20\choose 15} 0.2^{15} (1-0.2)^{20-15} + {20\choose 16} 0.2^{16} (1-0.2)^{20-16} \\ \quad\quad\quad\quad\quad + \displaystyle {20\choose 17} 0.2^{17} (1-0.2)^{20-17} + {20\choose 18} 0.2^{18} (1-0.2)^{20-18} \\ \quad\quad\quad\quad\quad + \displaystyle {20\choose 19} 0.2^{19} (1-0.2)^{20-19} + {20\choose 20} 0.2^{20} (1-0.2)^{20-20}$$$$\quad\quad\quad\quad\quad \approx 0$$Conclusion: Answering questions randomly by guessing gives no chance at all in passing a test. | 2,950 | 8,656 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-04 | latest | en | 0.87539 |
http://citykidzcdc.com/million-dollar-viih/b9ff4b-quotient-law-limits | 1,638,867,613,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00372.warc.gz | 15,984,394 | 7,845 | Use the Quotient Law to prove that if \lim _{x \rightarrow c} f(x) exists and is nonzero, then \lim _{x \rightarrow c} \frac{1}{f(x)}=\frac{1}{\lim _{x \righta⦠Power law In this section, we establish laws for calculating limits and learn how to apply these laws. In this case there are two ways to do compute this derivative. Following the steps in Examples 1 and 2, it is easily seen that: Because the first two limits exist, the Product Law can be applied to obtain = Now, because this limit exists and because = , the Quotient Law can be applied. When finding the derivative of sine, we have ... Browse other questions tagged limits or ask your own question. Special limit The limit of x is a when x approaches a. Addition law: Subtraction law: Multiplication law: Division law: Power law: The following example makes use of the subtraction, division, and power laws: Use the Quotient Law to prove that if lim x â c f (x) exists and is nonzero, then lim x â c 1 f (x) = 1 lim x â c f (x) solution Since lim x â c f (x) is nonzero, we can apply the Quotient Law: lim x â c 1 f (x) = lim x â c 1 lim x â c f (x) = 1 lim x â c f (x). Quotient Law for Limits. Answer to: Suppose the limits limit x to a f(x) and limit x to a g(x) both exist. The Sum Law basically states that the limit of the sum of two functions is the sum of the limits. If we had a limit as x approaches 0 of 2x/x we can find the value of that limit to be 2 by canceling out the xâs. In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. you can use the limit operations in the following ways. If the limits and both exist, and , then . The law L2 allows us to scale functions by a non-zero scale factor: in order to prove , ... L8 The limit of a quotient is the quotient of the limits (provided the latter is well-defined): By scaling the function , we can take . if . Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share ⦠The quotient limit laws says that the limit of a quotient is equal to the quotient of the limits. Constant Rule for Limits If a , b {\displaystyle a,b} are constants then lim x â a b = b {\displaystyle \lim _{x\to a}b=b} . Also, if c does not depend on x-- if c is a constant -- then Limits of functions at a point are the common and coincidence value of the left and right-hand limits. Formula of subtraction law of limits with introduction and proof to learn how to derive difference property of limits mathematically in calculus. The value of a limit of a function f(x) at a point a i.e., f(a) may vary from the value of f(x) at âaâ. In calculus, the product rule is a formula used to find the derivatives of products of two or more functions.It may be stated as (â
) â² = â² â
+ â
â²or in Leibniz's notation (â
) = â
+ â
.The rule may be extended or generalized to many other situations, including to products of multiple functions, to a rule for higher-order derivatives of a product, and to other contexts. (the limit of a quotient is the quotient of the limits provided that the limit of the denominator is not 0) Example If I am given that lim x!2 f(x) = 2; lim x!2 g(x) = 5; lim x!2 ... More powerful laws of limits can be derived using the above laws 1-5 and our knowledge of some basic functions. The limit of x 2 as xâ2 (using direct substitution) is x 2 = 2 2 = 4 ; The limit of the constant 5 (rule 1 above) is 5 The limit of a quotient is equal to the quotient of numerator and denominator's limits provided that the denominator's limit is not 0. lim xâa [f(x)/g(x)] = lim xâa f(x) / lim xâa g(x) Identity Law for Limits. There is a concise list of the Limit Laws at the bottom of the page. In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Quotient Law states that "The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0)" i.e. Thatâs the point of this example. the product of the limits. The quotient rule follows the definition of the limit of the derivative. 116 C H A P T E R 2 LIMITS 25. 6. Give the ''quotient law'' for limits. Recall from Section 2.5 that the definition of a limit of a function of one variable: Let $$f(x)$$ be defined for all $$xâ a$$ in an open interval containing $$a$$. Then the quotient rule tells us that F prime of X is going to be equal to and this is going to look a little bit complicated but once we apply it, you'll hopefully get a little bit more comfortable with it. 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# What is Equity Multiplier
The ‘Equity Multiplier’ is used to evaluate the financial strength of a company by obtaining a ratio of its total assets to the ratio of the total value of its shareholders.
An equity multiplier is important in determining the percentage of potential risk that a company is going through and its overall leverage status.
## Equity Multiplier Formulae
There are basically two usable formulae of calculating the equity multiplier. One is the simple mathematical division of the company assets by the stockholder’s equity.
The second way is a reciprocal method where the equity ratio of the company is divided by a value of one in the numerator while the ratio itself is in the denominator.
This equity ratio however needs to be obtained first before the application of this method otherwise it will not work. Both of these calculation manners are distinct and each includes the usage of a particular formula.
### Method One
`Equity multiplier = Total assets of company / Stockholders’ equity`
This is a straightforward way of obtaining the ratio by simple division of the company’s total assets by the stockholders’ equity. The stockholders’ equity can be obtained by the following equation:
`Stockholders’ equity = Company’s total assets - Company’s liabilities`
If the stockholders’ equity has even the slightest of error in its calculation of significant figures and decimal points then the multiplier’s exact value will differ.
### Method Two
`Equity multiplier = 1 / Equity Ratio`
This is a slightly advance way yet the easiest way to obtain the desired ratio. This method depends on the fact that a company’s total assets are always equal to the addition of debt with equity which in other words concludes an equation of:
`Company’s total assets = debt + equity`
This method uses equity ratio as the reciprocal (in the denominator of the fraction) and is divided by the value of one.
However for this method to be accurate, the equity ratio itself needs to be obtained first which is quite easily achieved by the following equation:
`Equity ratio = 1 – debt ratio`
This multiplier has an essential role to play in the DuPont Analysis. This analysis splits the ‘Return on Equity’ (ROE) into three parts.
The return on equity is basically the estimation of the efficiency of a company’s usage of investments to promote its earnings growth. The returns on equity divisions are simply financial ratios that include net profit margin, asset turnover and the financial leverage.
These three ratios are combined to calculate the return on equity value. The ROE has a crucial performance in figuring out which of these is underperforming in the business.
## Formula of return on equity
`Return on equity = net profit margin x asset turnover x financial leverage (equity multiplier)`
If the multiplier ratio is not as expected or is going through a pattern of ebb and flow then the ROE value can be severely affected. Conclusively, the equity multiplier can be calculated by two different ways. Each method uses a unique formula
## Equity Multiplier Ratio
This ratio is used to depict the financial strength of a company by showing which part of equity multiplier ratio is dominated.
Since the equity multiplier ratio of a company is obtained by dividing its total assets by the equity of its shareholders, the part of the ratio which is influenced the most shows how overwhelming is the role of that particular part in the company’s overall financial state.
Hence, an equity multiplier ratio of a lower value is considered to be better in terms of financial state in comparison with an equity multiplier ratio of a higher value.
This is because each part of this ratio is showing the command of it over the other and depending on the ratio’s value, a percentage of the company’s financing of its total assets by equity and the financing of its total assets by debt is obtained.
Therefore a ratio of a lower value would indicate that more of the assets are being funded by equity rather than debt and that the company’s financial status is fair enough to be called in an equilibrium.
A simple example used to explain the influence of the equity multiplier ratio on the financial state of a company is by using a scenario of any two companies, Company A and Company B. However, both of these companies must have different equity multiplier ratios.
Suppose the Company A’s total assets are \$100 billion and the equity of its stockholders is \$20 billion. By dividing the total assets by the stockholders’ equity, we get the equity multiplier’s value to be 5.
This represents that Company A is financing 20% of its assets by equity and rest of it by debt. This company therefore is in a larger risk due to its high dependency rate on debt rather than equity.
Similarly, suppose the Company B’s total assets are also \$100 billion but the equity of its stockholders if \$50 billion.
Once again, upon obtaining the equity multiplier ratio we will notice that we have obtained a certain value, which is, of 2. In this case, Company B is financing 50% of its assets by equity and rest by of it by debt.
In contrast, comparing Company A and Company B’s leverage status and debt dependency, we will notice that Company’s B financial strength is evidently and considerably much more than that of Company A since it is only financing its assets by 50% of the debt. On the contrary, Company A is financing 80% of its assets by debt and thus poses a high risk.
Conclusively, if a business or company is profitable enough it would certainly not be dependent on debts since it has enough incoming funds to look after its assets. Thus, in case of a high equity multiplier ratio, a company is more likely to be capitalizing its liabilities with the help of debt rather than the profit obtained since the profit is not enough to fund its payables.
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https://www.hextobinary.com/unit/length/from/arshinru/to/decimeter | 1,603,685,380,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890273.42/warc/CC-MAIN-20201026031408-20201026061408-00004.warc.gz | 744,225,626 | 11,755 | Arshin [Russia] Decimeter
How many Decimeters are in a Arshin [Russia]?
The answer is one Arshin [Russia] is equal to 7.11 Decimeters. Feel free to use our online unit conversion calculator to convert the unit from Arshin [Russia] to Decimeter. Just simply, enter value 1 in Arshin [Russia] and see the result in Decimeter.
How to Convert Arshin [Russia] to Decimeter (arshin to dm)
By using our Arshin [Russia] to Decimeter conversion tool, you know that one Arshin [Russia] is equivalent to 7.11 Decimeter. Hence, to convert Arshin [Russia] to Decimeter, we just need to multiply the number by 7.11. We are going to use very simple Arshin [Russia] to Decimeter conversion formula for that. Pleas see the calculation example given below.
Convert 1 Arshin [Russia] to Decimeter 1 Arshin [Russia] = 1 × 7.11 = 7.11 Decimeter
What is Arshin [Russia] Unit of Measure?
Arshin [Russia] is a unit of measurement for length. It was originated in 18th century in Russia. In Russia, one arshin is equal to 71.12 centimeters.
What is the symbol of Arshin [Russia]?
The symbol of Arshin [Russia] is arshin which means you can also write it as 1 arshin.
What is Decimeter Unit of Measure?
Decimeter is a unit of measurement for length. Decimeter is a decimal fraction of length unit meter and one decimeter is equal to one tenth of a meter.
What is the symbol of Decimeter?
The symbol of Decimeter is dm which means you can also write it as 1 dm.
Arshin [Russia] to Decimeter Conversion Table
Arshin [Russia] [arshin] Decimeter [dm] 1 7.112 2 14.224 3 21.336 4 28.448 5 35.56 6 42.672 7 49.784 8 56.896 9 64.008 10 71.12 100 711.2 1000 7112
Arshin [Russia] to Other Units Conversion Chart
Arshin [Russia] [arshin] Output 1 Arshin [Russia] in Agate Line equals to 392 1 Arshin [Russia] in Alen [Denmark] equals to 1.13 1 Arshin [Russia] in Alen [Scandinavia] equals to 1.19 1 Arshin [Russia] in Alen [Sweden] equals to 1.2 1 Arshin [Russia] in Angstrom equals to 7112000000 1 Arshin [Russia] in Angulam equals to 40.34 1 Arshin [Russia] in Arms Length equals to 1.02 1 Arshin [Russia] in Arpent [Canada] equals to 0.012163502650932 1 Arshin [Russia] in Arpent [France] equals to 0.0099524209347887 1 Arshin [Russia] in Arshin [Iran] equals to 0.68384615384615 1 Arshin [Russia] in Arshin [Iraq] equals to 0.0095463087248322 1 Arshin [Russia] in Astronomical Unit equals to 4.6983687072398e-12 1 Arshin [Russia] in Attometer equals to 711200000000000000 1 Arshin [Russia] in Bamboo equals to 0.22225 1 Arshin [Russia] in Barleycorn equals to 84 1 Arshin [Russia] in Bee Space equals to 109.42 1 Arshin [Russia] in Bicron equals to 711200000000 1 Arshin [Russia] in Bohr equals to 13439661362.86 1 Arshin [Russia] in Braccio equals to 1.02 1 Arshin [Russia] in Braza [Argentina] equals to 0.41109826589595 1 Arshin [Russia] in Braza [Spain] equals to 0.42586826347305 1 Arshin [Russia] in Braza [Texas] equals to 0.42008269344359 1 Arshin [Russia] in Button equals to 1120 1 Arshin [Russia] in Cable equals to 0.0038401727861771 1 Arshin [Russia] in Cable [UK] equals to 0.0038377192982456 1 Arshin [Russia] in Cable [US] equals to 0.0032407407407407 1 Arshin [Russia] in Canna equals to 0.3556 1 Arshin [Russia] in Cape Foot equals to 2.26 1 Arshin [Russia] in Cape Inch equals to 27.11 1 Arshin [Russia] in Cape Rood equals to 0.18823276325696 1 Arshin [Russia] in Centimeter equals to 71.12 1 Arshin [Russia] in Chain equals to 0.035353535353535 1 Arshin [Russia] in Chinese Foot equals to 2.22 1 Arshin [Russia] in Chinese Inch equals to 22.22 1 Arshin [Russia] in Chinese Mile equals to 0.0014224 1 Arshin [Russia] in Chinese Pace equals to 0.44871794871795 1 Arshin [Russia] in Chinese Yard equals to 0.22222222222222 1 Arshin [Russia] in Cuadra equals to 0.0084666666666667 1 Arshin [Russia] in Cuadra [Argentina] equals to 0.0054707692307692 1 Arshin [Russia] in Cubit equals to 1.56 1 Arshin [Russia] in Decimeter equals to 7.11 1 Arshin [Russia] in Dekameter equals to 0.07112 1 Arshin [Russia] in Didot equals to 1886.47 1 Arshin [Russia] in Digit equals to 37.43 1 Arshin [Russia] in Diraa equals to 1.23 1 Arshin [Russia] in Dong equals to 30.26 1 Arshin [Russia] in Douzieme equals to 3780.11 1 Arshin [Russia] in Dra [Iraq] equals to 0.95463087248322 1 Arshin [Russia] in Dra [Russia] equals to 1 1 Arshin [Russia] in Ell equals to 0.62222222222222 1 Arshin [Russia] in Elle [Austria] equals to 0.9126138842551 1 Arshin [Russia] in Elle [Germany] equals to 1.19 1 Arshin [Russia] in Em equals to 168.01 1 Arshin [Russia] in Estadio [Portugal] equals to 0.0027249042145594 1 Arshin [Russia] in Estadio [Spain] equals to 0.0040873563218391 1 Arshin [Russia] in Faden [Austria] equals to 0.37500659108885 1 Arshin [Russia] in Faden [Switzerland] equals to 0.39511111111111 1 Arshin [Russia] in Fathom equals to 0.38888888888889 1 Arshin [Russia] in Faust [Hungary] equals to 6.75 1 Arshin [Russia] in Feet equals to 2.33 1 Arshin [Russia] in Femtometer equals to 711200000000000 1 Arshin [Russia] in Fermi equals to 711200000000000 1 Arshin [Russia] in Finger equals to 6.22 1 Arshin [Russia] in Fingerbreadth equals to 37.33 1 Arshin [Russia] in Fist equals to 7.11 1 Arshin [Russia] in Fod equals to 2.26 1 Arshin [Russia] in Furlong equals to 0.0035353535353535 1 Arshin [Russia] in Fuss [fuß] equals to 2.25 1 Arshin [Russia] in Gaj equals to 0.77777777777778 1 Arshin [Russia] in Gigameter equals to 7.112e-10 1 Arshin [Russia] in Gigaparsec equals to 2.3048422168278e-26 1 Arshin [Russia] in Goad equals to 0.51851851851852 1 Arshin [Russia] in Hairbreadth equals to 7112 1 Arshin [Russia] in Hand equals to 7 1 Arshin [Russia] in Handbreadth equals to 9.36 1 Arshin [Russia] in Hath equals to 1.56 1 Arshin [Russia] in Hectometer equals to 0.007112 1 Arshin [Russia] in Heer equals to 0.0097222222222222 1 Arshin [Russia] in Hvat equals to 0.37500978653211 1 Arshin [Russia] in Inch equals to 28 1 Arshin [Russia] in Jarib [Gantari] equals to 0.017676767676768 1 Arshin [Russia] in Jarib [Shahjahani] equals to 0.014141414141414 1 Arshin [Russia] in Kadi equals to 3.54 1 Arshin [Russia] in Karam equals to 0.42424242424242 1 Arshin [Russia] in Ken equals to 0.39119911991199 1 Arshin [Russia] in Kerat equals to 24.87 1 Arshin [Russia] in Kilofoot equals to 0.0023333333333333 1 Arshin [Russia] in Kilometer equals to 0.0007112 1 Arshin [Russia] in Kiloparsec equals to 2.3048422168278e-20 1 Arshin [Russia] in Kiloyard equals to 0.00077777777777778 1 Arshin [Russia] in Klafter [Austria] equals to 0.37500659108885 1 Arshin [Russia] in Klafter [Switzerland] equals to 0.39511111111111 1 Arshin [Russia] in Klick equals to 0.0007112 1 Arshin [Russia] in Kol equals to 0.98777777777778 1 Arshin [Russia] in Kos equals to 0.00023128455284553 1 Arshin [Russia] in Kyu equals to 2844.8 1 Arshin [Russia] in Lap equals to 0.001778 1 Arshin [Russia] in Lap Pool equals to 0.007112 1 Arshin [Russia] in League equals to 0.00014730610135368 1 Arshin [Russia] in Lieu equals to 0.0001778 1 Arshin [Russia] in Light Year equals to 7.5173899315837e-17 1 Arshin [Russia] in Ligne [France] equals to 335.99 1 Arshin [Russia] in Ligne [Switzerland] equals to 315.25 1 Arshin [Russia] in Line equals to 335.99 1 Arshin [Russia] in Link equals to 3.54 1 Arshin [Russia] in Lug equals to 0.14141414141414 1 Arshin [Russia] in Marathon equals to 0.000016854921694923 1 Arshin [Russia] in Megameter equals to 7.112e-7 1 Arshin [Russia] in Megaparsec equals to 2.3048422168278e-23 1 Arshin [Russia] in Meile [Austria] equals to 0.000093751647772212 1 Arshin [Russia] in Meile [Geographische] equals to 0.000095842081573578 1 Arshin [Russia] in Meile [Germany] equals to 0.000094417524062396 1 Arshin [Russia] in Meter equals to 0.7112 1 Arshin [Russia] in Microinch equals to 28000000 1 Arshin [Russia] in Micrometer equals to 711200 1 Arshin [Russia] in Micron equals to 710659.9 1 Arshin [Russia] in Miglio equals to 0.00047775047775048 1 Arshin [Russia] in Miil [Denmark] equals to 0.000094423791821561 1 Arshin [Russia] in Miil [Sweden] equals to 0.000066548142603163 1 Arshin [Russia] in Mil equals to 28000 1 Arshin [Russia] in Mile equals to 0.00044191919191919 1 Arshin [Russia] in Millimeter equals to 711.2 1 Arshin [Russia] in Millimicron equals to 710659898.48 1 Arshin [Russia] in Mkono equals to 1.56 1 Arshin [Russia] in Muzham equals to 1.52 1 Arshin [Russia] in Myriameter equals to 0.00007112 1 Arshin [Russia] in Nail equals to 12.44 1 Arshin [Russia] in Nanometer equals to 711200000 1 Arshin [Russia] in Nautical League equals to 0.00012800575953924 1 Arshin [Russia] in Nautical Mile equals to 0.00038401727861771 1 Arshin [Russia] in Pace equals to 0.46666666666667 1 Arshin [Russia] in Pace [Roman] equals to 0.48054054054054 1 Arshin [Russia] in Palm equals to 9.33 1 Arshin [Russia] in Palmo [Portugal] equals to 3.23 1 Arshin [Russia] in Palmo [Spain] equals to 3.56 1 Arshin [Russia] in Palmo [Texas] equals to 3.36 1 Arshin [Russia] in Parasang equals to 0.00011853333333333 1 Arshin [Russia] in Parsec equals to 2.3048422168278e-17 1 Arshin [Russia] in Pe equals to 2.13 1 Arshin [Russia] in Perch equals to 0.14141414141414 1 Arshin [Russia] in Perch [Ireland] equals to 0.11111111111111 1 Arshin [Russia] in Pertica equals to 0.24027027027027 1 Arshin [Russia] in Pes equals to 2.4 1 Arshin [Russia] in Petameter equals to 7.1071308058178e-16 1 Arshin [Russia] in Pica equals to 168 1 Arshin [Russia] in Picometer equals to 711200000000 1 Arshin [Russia] in Pie [Argentina] equals to 2.46 1 Arshin [Russia] in Pie [Italy] equals to 2.39 1 Arshin [Russia] in Pie [Spain] equals to 2.55 1 Arshin [Russia] in Pie [Texas] equals to 2.52 1 Arshin [Russia] in Pied Du Roi equals to 2.19 1 Arshin [Russia] in Pik equals to 1 1 Arshin [Russia] in Pike equals to 1 1 Arshin [Russia] in Pixel equals to 2688 1 Arshin [Russia] in Point equals to 2016 1 Arshin [Russia] in Pole equals to 0.1414141417584 1 Arshin [Russia] in Polegada equals to 25.61 1 Arshin [Russia] in Pouce equals to 26.27 1 Arshin [Russia] in Pulgada equals to 30.66 1 Arshin [Russia] in Q equals to 2844.8 1 Arshin [Russia] in Quadrant equals to 7.1110752225303e-8 1 Arshin [Russia] in Quarter equals to 0.0017676767676768 1 Arshin [Russia] in Reed equals to 0.26547219111609 1 Arshin [Russia] in Ri equals to 0.00018111070366296 1 Arshin [Russia] in Ridge equals to 0.11522633744856 1 Arshin [Russia] in Rod equals to 0.14141414141414 1 Arshin [Russia] in Roede equals to 0.07112 1 Arshin [Russia] in Rope equals to 0.1166666704 1 Arshin [Russia] in Royal Foot equals to 2.19 1 Arshin [Russia] in Rute equals to 0.18965333333333 1 Arshin [Russia] in Sadzhen equals to 0.33333333333333 1 Arshin [Russia] in Scandinavian Mile equals to 0.00007112 1 Arshin [Russia] in Scots Foot equals to 2.32 1 Arshin [Russia] in Scots mile equals to 0.00039201852056003 1 Arshin [Russia] in Seemeile equals to 0.00038401727861771 1 Arshin [Russia] in Shackle equals to 0.025925925925926 1 Arshin [Russia] in Shaftment equals to 4.67 1 Arshin [Russia] in Shaku equals to 2.35 1 Arshin [Russia] in Siriometer equals to 4.7548323412926e-18 1 Arshin [Russia] in Smoot equals to 0.41791044776119 1 Arshin [Russia] in Span equals to 3.11 1 Arshin [Russia] in Spat equals to 7.0286460031113e-13 1 Arshin [Russia] in Stadium equals to 0.0038443243243243 1 Arshin [Russia] in Step equals to 0.93333333333333 1 Arshin [Russia] in Story equals to 0.21551515151515 1 Arshin [Russia] in Stride equals to 0.46666666666667 1 Arshin [Russia] in Stride [Roman] equals to 0.48054054054054 1 Arshin [Russia] in Terameter equals to 7.112e-13 1 Arshin [Russia] in Thou equals to 28448 1 Arshin [Russia] in Toise equals to 0.36515388628065 1 Arshin [Russia] in Tsun equals to 19.87 1 Arshin [Russia] in Tu equals to 0.0000044138273443803 1 Arshin [Russia] in Twip equals to 40320 1 Arshin [Russia] in U equals to 16 1 Arshin [Russia] in Vara [Spain] equals to 0.85081947601388 1 Arshin [Russia] in Verge equals to 0.77777777777778 1 Arshin [Russia] in Vershok equals to 16 1 Arshin [Russia] in Verst equals to 0.00066666666666667 1 Arshin [Russia] in Wah equals to 0.3556 1 Arshin [Russia] in Yard equals to 0.77777777777778 1 Arshin [Russia] in Yavam equals to 322.72 1 Arshin [Russia] in Yojan equals to 0.000055239898989899 1 Arshin [Russia] in Zeptometer equals to 711200000000000000000 1 Arshin [Russia] in Zoll [Germany] equals to 27 1 Arshin [Russia] in Zoll [Switzerland] equals to 23.71
Other Units to Arshin [Russia] Conversion Chart
Output Arshin [Russia] [arshin] 1 Agate Line in Arshin [Russia] equals to 0.0025510204081633 1 Alen [Denmark] in Arshin [Russia] equals to 0.88259280089989 1 Alen [Scandinavia] in Arshin [Russia] equals to 0.84364454443195 1 Alen [Sweden] in Arshin [Russia] equals to 0.83494375703037 1 Angstrom in Arshin [Russia] equals to 1.4060742407199e-10 1 Angulam in Arshin [Russia] equals to 0.024789088863892 1 Arms Length in Arshin [Russia] equals to 0.98425196850394 1 Arpent [Canada] in Arshin [Russia] equals to 82.21 1 Arpent [France] in Arshin [Russia] equals to 100.48 1 Arshin [Iran] in Arshin [Russia] equals to 1.46 1 Arshin [Iraq] in Arshin [Russia] equals to 104.75 1 Astronomical Unit in Arshin [Russia] equals to 212839830654.22 1 Attometer in Arshin [Russia] equals to 1.4060742407199e-18 1 Bamboo in Arshin [Russia] equals to 4.5 1 Barleycorn in Arshin [Russia] equals to 0.011905230596175 1 Bee Space in Arshin [Russia] equals to 0.0091394825646794 1 Bicron in Arshin [Russia] equals to 1.4060742407199e-12 1 Bohr in Arshin [Russia] equals to 7.4406636670416e-11 1 Braccio in Arshin [Russia] equals to 0.98425196850394 1 Braza [Argentina] in Arshin [Russia] equals to 2.43 1 Braza [Spain] in Arshin [Russia] equals to 2.35 1 Braza [Texas] in Arshin [Russia] equals to 2.38 1 Button in Arshin [Russia] equals to 0.00089285714285714 1 Cable in Arshin [Russia] equals to 260.4 1 Cable [UK] in Arshin [Russia] equals to 260.57 1 Cable [US] in Arshin [Russia] equals to 308.57 1 Canna in Arshin [Russia] equals to 2.81 1 Cape Foot in Arshin [Russia] equals to 0.44271428571429 1 Cape Inch in Arshin [Russia] equals to 0.036892857142857 1 Cape Rood in Arshin [Russia] equals to 5.31 1 Centimeter in Arshin [Russia] equals to 0.014060742407199 1 Chain in Arshin [Russia] equals to 28.29 1 Chinese Foot in Arshin [Russia] equals to 0.45 1 Chinese Inch in Arshin [Russia] equals to 0.045 1 Chinese Mile in Arshin [Russia] equals to 703.04 1 Chinese Pace in Arshin [Russia] equals to 2.23 1 Chinese Yard in Arshin [Russia] equals to 4.5 1 Cuadra in Arshin [Russia] equals to 118.11 1 Cuadra [Argentina] in Arshin [Russia] equals to 182.79 1 Cubit in Arshin [Russia] equals to 0.64285714285714 1 Decimeter in Arshin [Russia] equals to 0.14060742407199 1 Dekameter in Arshin [Russia] equals to 14.06 1 Didot in Arshin [Russia] equals to 0.00053008998875141 1 Digit in Arshin [Russia] equals to 0.026715410573678 1 Diraa in Arshin [Russia] equals to 0.81552305961755 1 Dong in Arshin [Russia] equals to 0.033042744656918 1 Douzieme in Arshin [Russia] equals to 0.00026454285714286 1 Dra [Iraq] in Arshin [Russia] equals to 1.05 1 Dra [Russia] in Arshin [Russia] equals to 1 1 Ell in Arshin [Russia] equals to 1.61 1 Elle [Austria] in Arshin [Russia] equals to 1.1 1 Elle [Germany] in Arshin [Russia] equals to 0.84364454443195 1 Em in Arshin [Russia] equals to 0.0059519122609674 1 Estadio [Portugal] in Arshin [Russia] equals to 366.99 1 Estadio [Spain] in Arshin [Russia] equals to 244.66 1 Faden [Austria] in Arshin [Russia] equals to 2.67 1 Faden [Switzerland] in Arshin [Russia] equals to 2.53 1 Fathom in Arshin [Russia] equals to 2.57 1 Faust [Hungary] in Arshin [Russia] equals to 0.1481442632171 1 Feet in Arshin [Russia] equals to 0.42857142857143 1 Femtometer in Arshin [Russia] equals to 1.4060742407199e-15 1 Fermi in Arshin [Russia] equals to 1.4060742407199e-15 1 Finger in Arshin [Russia] equals to 0.16071428571429 1 Fingerbreadth in Arshin [Russia] equals to 0.026785714285714 1 Fist in Arshin [Russia] equals to 0.14060742407199 1 Fod in Arshin [Russia] equals to 0.44164791901012 1 Furlong in Arshin [Russia] equals to 282.86 1 Fuss [fuß] in Arshin [Russia] equals to 0.44443194600675 1 Gaj in Arshin [Russia] equals to 1.29 1 Gigameter in Arshin [Russia] equals to 1406074240.72 1 Gigaparsec in Arshin [Russia] equals to 4.3386917885264e+25 1 Goad in Arshin [Russia] equals to 1.93 1 Hairbreadth in Arshin [Russia] equals to 0.00014060742407199 1 Hand in Arshin [Russia] equals to 0.14285714285714 1 Handbreadth in Arshin [Russia] equals to 0.10686164229471 1 Hath in Arshin [Russia] equals to 0.64285714285714 1 Hectometer in Arshin [Russia] equals to 140.61 1 Heer in Arshin [Russia] equals to 102.86 1 Hvat in Arshin [Russia] equals to 2.67 1 Inch in Arshin [Russia] equals to 0.035714285714286 1 Jarib [Gantari] in Arshin [Russia] equals to 56.57 1 Jarib [Shahjahani] in Arshin [Russia] equals to 70.71 1 Kadi in Arshin [Russia] equals to 0.28285714285714 1 Karam in Arshin [Russia] equals to 2.36 1 Ken in Arshin [Russia] equals to 2.56 1 Kerat in Arshin [Russia] equals to 0.040213723284589 1 Kilofoot in Arshin [Russia] equals to 428.57 1 Kilometer in Arshin [Russia] equals to 1406.07 1 Kiloparsec in Arshin [Russia] equals to 43386917885264000000 1 Kiloyard in Arshin [Russia] equals to 1285.71 1 Klafter [Austria] in Arshin [Russia] equals to 2.67 1 Klafter [Switzerland] in Arshin [Russia] equals to 2.53 1 Klick in Arshin [Russia] equals to 1406.07 1 Kol in Arshin [Russia] equals to 1.01 1 Kos in Arshin [Russia] equals to 4323.68 1 Kyu in Arshin [Russia] equals to 0.00035151856017998 1 Lap in Arshin [Russia] equals to 562.43 1 Lap Pool in Arshin [Russia] equals to 140.61 1 League in Arshin [Russia] equals to 6788.59 1 Lieu in Arshin [Russia] equals to 5624.3 1 Light Year in Arshin [Russia] equals to 13302489415889000 1 Ligne [France] in Arshin [Russia] equals to 0.0029762373453318 1 Ligne [Switzerland] in Arshin [Russia] equals to 0.0031721034870641 1 Line in Arshin [Russia] equals to 0.0029762373453318 1 Link in Arshin [Russia] equals to 0.28285714285714 1 Lug in Arshin [Russia] equals to 7.07 1 Marathon in Arshin [Russia] equals to 59329.85 1 Megameter in Arshin [Russia] equals to 1406074.24 1 Megaparsec in Arshin [Russia] equals to 4.3386917885264e+22 1 Meile [Austria] in Arshin [Russia] equals to 10666.48 1 Meile [Geographische] in Arshin [Russia] equals to 10433.83 1 Meile [Germany] in Arshin [Russia] equals to 10591.25 1 Meter in Arshin [Russia] equals to 1.41 1 Microinch in Arshin [Russia] equals to 3.5714285714286e-8 1 Micrometer in Arshin [Russia] equals to 0.0000014060742407199 1 Micron in Arshin [Russia] equals to 0.0000014071428571429 1 Miglio in Arshin [Russia] equals to 2093.14 1 Miil [Denmark] in Arshin [Russia] equals to 10590.55 1 Miil [Sweden] in Arshin [Russia] equals to 15026.72 1 Mil in Arshin [Russia] equals to 0.000035714285714286 1 Mile in Arshin [Russia] equals to 2262.86 1 Millimeter in Arshin [Russia] equals to 0.0014060742407199 1 Millimicron in Arshin [Russia] equals to 1.4071428571429e-9 1 Mkono in Arshin [Russia] equals to 0.64285714285714 1 Muzham in Arshin [Russia] equals to 0.6561670416198 1 Myriameter in Arshin [Russia] equals to 14060.74 1 Nail in Arshin [Russia] equals to 0.080357142857143 1 Nanometer in Arshin [Russia] equals to 1.4060742407199e-9 1 Nautical League in Arshin [Russia] equals to 7812.15 1 Nautical Mile in Arshin [Russia] equals to 2604.05 1 Pace in Arshin [Russia] equals to 2.14 1 Pace [Roman] in Arshin [Russia] equals to 2.08 1 Palm in Arshin [Russia] equals to 0.10714285714286 1 Palmo [Portugal] in Arshin [Russia] equals to 0.30933633295838 1 Palmo [Spain] in Arshin [Russia] equals to 0.28121484814398 1 Palmo [Texas] in Arshin [Russia] equals to 0.29766591676041 1 Parasang in Arshin [Russia] equals to 8436.45 1 Parsec in Arshin [Russia] equals to 43386917885264000 1 Pe in Arshin [Russia] equals to 0.4685601799775 1 Perch in Arshin [Russia] equals to 7.07 1 Perch [Ireland] in Arshin [Russia] equals to 9 1 Pertica in Arshin [Russia] equals to 4.16 1 Pes in Arshin [Russia] equals to 0.4171822272216 1 Petameter in Arshin [Russia] equals to 1407037561742100 1 Pica in Arshin [Russia] equals to 0.005952380952381 1 Picometer in Arshin [Russia] equals to 1.4060742407199e-12 1 Pie [Argentina] in Arshin [Russia] equals to 0.40607142857143 1 Pie [Italy] in Arshin [Russia] equals to 0.41892857142857 1 Pie [Spain] in Arshin [Russia] equals to 0.39142857142857 1 Pie [Texas] in Arshin [Russia] equals to 0.39678571428571 1 Pied Du Roi in Arshin [Russia] equals to 0.45674915635546 1 Pik in Arshin [Russia] equals to 0.99831271091114 1 Pike in Arshin [Russia] equals to 0.99831271091114 1 Pixel in Arshin [Russia] equals to 0.00037202383295838 1 Point in Arshin [Russia] equals to 0.00049603177727784 1 Pole in Arshin [Russia] equals to 7.07 1 Polegada in Arshin [Russia] equals to 0.039046681664792 1 Pouce in Arshin [Russia] equals to 0.038062429696288 1 Pulgada in Arshin [Russia] equals to 0.032619047619048 1 Q in Arshin [Russia] equals to 0.00035151856017998 1 Quadrant in Arshin [Russia] equals to 14062570.97 1 Quarter in Arshin [Russia] equals to 565.71 1 Reed in Arshin [Russia] equals to 3.77 1 Ri in Arshin [Russia] equals to 5521.48 1 Ridge in Arshin [Russia] equals to 8.68 1 Rod in Arshin [Russia] equals to 7.07 1 Roede in Arshin [Russia] equals to 14.06 1 Rope in Arshin [Russia] equals to 8.57 1 Royal Foot in Arshin [Russia] equals to 0.45674915635546 1 Rute in Arshin [Russia] equals to 5.27 1 Sadzhen in Arshin [Russia] equals to 3 1 Scandinavian Mile in Arshin [Russia] equals to 14060.74 1 Scots Foot in Arshin [Russia] equals to 0.43089145106862 1 Scots mile in Arshin [Russia] equals to 2550.9 1 Seemeile in Arshin [Russia] equals to 2604.05 1 Shackle in Arshin [Russia] equals to 38.57 1 Shaftment in Arshin [Russia] equals to 0.21428571428571 1 Shaku in Arshin [Russia] equals to 0.42604049493813 1 Siriometer in Arshin [Russia] equals to 210312357665200000 1 Smoot in Arshin [Russia] equals to 2.39 1 Span in Arshin [Russia] equals to 0.32142857142857 1 Spat in Arshin [Russia] equals to 1422749131991.2 1 Stadium in Arshin [Russia] equals to 260.12 1 Step in Arshin [Russia] equals to 1.07 1 Story in Arshin [Russia] equals to 4.64 1 Stride in Arshin [Russia] equals to 2.14 1 Stride [Roman] in Arshin [Russia] equals to 2.08 1 Terameter in Arshin [Russia] equals to 1406074240719.9 1 Thou in Arshin [Russia] equals to 0.000035151856017998 1 Toise in Arshin [Russia] equals to 2.74 1 Tsun in Arshin [Russia] equals to 0.050337457817773 1 Tu in Arshin [Russia] equals to 226560.74 1 Twip in Arshin [Russia] equals to 0.000024801587301587 1 U in Arshin [Russia] equals to 0.0625 1 Vara [Spain] in Arshin [Russia] equals to 1.18 1 Verge in Arshin [Russia] equals to 1.29 1 Vershok in Arshin [Russia] equals to 0.0625 1 Verst in Arshin [Russia] equals to 1500 1 Wah in Arshin [Russia] equals to 2.81 1 Yard in Arshin [Russia] equals to 1.29 1 Yavam in Arshin [Russia] equals to 0.0030986361079865 1 Yojan in Arshin [Russia] equals to 18102.86 1 Zeptometer in Arshin [Russia] equals to 1.4060742407199e-21 1 Zoll [Germany] in Arshin [Russia] equals to 0.037035995500562 1 Zoll [Switzerland] in Arshin [Russia] equals to 0.042182227221597 | 8,763 | 23,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, 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http://www.loopandbreak.com/hsl/ | 1,618,118,344,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00374.warc.gz | 160,045,760 | 13,028 | ### hsl()
Colors also can be defined the Hue-saturation-lightness model (HSL) using the hsl() functional notation. The advantage of HSL over RGB is that it is far more intuitive: you can guess at the colors you want, and then tweak. It is also easier to create sets of matching colors (by keeping the hue the same and varying the lightness/darkness, and saturation).
Hue is represented as an angle of the color circle (i.e. the rainbow represented in a circle). This angle is given as a unitless <number>. By definition red=0=360, and the other colors are spread around the circle, so green=120, blue=240, etc. As an angle, it implicitly wraps around such that -120=240 and 480=120.
Saturation and lightness are represented as percentages.
100% is full saturation, and0% is a shade of grey.
100% lightness is white, 0% lightness is black, and 50% lightness is “normal”.
```hsl(0, 100%,50%) /* red */
hsl(30, 100%,50%)
hsl(60, 100%,50%)
hsl(90, 100%,50%)
hsl(120,100%,50%) /* green */
hsl(150,100%,50%)
hsl(180,100%,50%)
hsl(210,100%,50%)
hsl(240,100%,50%) /* blue */
hsl(270,100%,50%)
hsl(300,100%,50%)
hsl(330,100%,50%)
hsl(360,100%,50%) /* red */
hsl(120,100%,25%) /* dark green */
hsl(120,100%,50%) /* green */
hsl(120,100%,75%) /* light green */
hsl(120,100%,50%) /* green */
hsl(120, 67%,50%)
hsl(120, 33%,50%)
hsl(120, 0%,50%)
hsl(120, 60%,70%) /* pastel green */
``` | 460 | 1,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-17 | latest | en | 0.824419 |
http://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-5-section-5-5-the-substitution-rule-5-5-exercises-page-418/3 | 1,524,270,938,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944848.33/warc/CC-MAIN-20180420233255-20180421013255-00569.warc.gz | 435,946,551 | 12,573 | ## Calculus: Early Transcendentals 8th Edition
$$\int x^2\sqrt{x^3+1}dx=\frac{2}{9}\sqrt{(x^3+1)^3}+C$$
$$A=\int x^2\sqrt{x^3+1}dx$$ Let $u=x^3+1$ Then $du=(x^3+1)'dx=3x^2dx$. So $x^2dx=\frac{1}{3}du$. Substitute into $A$, we have $$A=\int \sqrt u(\frac{1}{3})du$$ $$A=\frac{1}{3}\int u^{1/2}du$$ $$A=\frac{1}{3}\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{u^{3/2}}{\frac{9}{2}}+C$$ $$A=\frac{2\sqrt{u^3}}{9}+C$$ $$A=\frac{2}{9}\sqrt{(x^3+1)^3}+C$$ | 240 | 446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-17 | latest | en | 0.351768 |
https://www.reddit.com/r/askscience/comments/3q25sx/how_good_of_a_conductor_is_our_brain/ | 1,539,731,170,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510893.26/warc/CC-MAIN-20181016221847-20181017003347-00312.warc.gz | 1,035,926,269 | 66,698 | Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts
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Posted by2 years ago
Archived
## How good of a conductor is our brain?
no idea if this should be tagged physics or human body,sorry in advance.
I was just wondering,since our brain sends electric signals,how much of them is lost along the way?
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level 1
11 points · 2 years ago · edited 2 years ago
tl;dr--Any kind of tissue is a rather poor conductor when compared to physical conductors like copper. However, neural tissue is designed to conduct signals through more active, complicated means. Signal velocities can be quite good in the body, and neurons are fantastic signal conductors compared to other tissues, but nowhere near as high as physical conductors.
First, some background;
The question itself is a little difficult to address because the way we define conductivity for metals and conductivity for neural tissue is very different. For physical, rather than biological, systems, we employ what's known as Ohm's Law to determine conductivity. Ohm's Law states that the Current density that results from an applied electric field (resulting from a battery, say), is proportional the electrical field multiplied by a constant called the conductivity. In mathematical terms, it's expressed by:
J = pE, where J is the current density, E is the applied electric field, and p is the conductivity. Clearly, the greater the conductivity, the greater the resulting current given a certain applied field.
But the way a neuron conducts an electrical signal is NOT through the generation of an electric field that spans the length of the whole wire (the axon). Rather, as /u/vasavasorum says, it's an active process that involves continually regenerating a potential difference called an action potential. Therefore, Ohm's Law doesn't really apply over the length of a neuron's projection to another neuron, because the way signals are conducted over physical wires verses neurons are VERY different.
Nevertheless, I'll still try to answer this question using some type of analogy. I guess we could make a comparison between wires and neurons by the speed at which they conduct signals.
Membranes leak ions, and ion concentration differences are what are ultimately responsible for signal propagation in a neuron. The fact that membranes are leaky are what dramatically slows signal propagation in neurons. We make membranes less leaky by an insulating material called myelin in our bodies, and this dramatically increases signal propagation velocities in our brains. Further still, you can increase nerve diameter to increase conduction velocities. Despite this, the highest conduction velocities we get in neurons is a measly 120 m/s. Compare this to wires which communicate at 1/100 the speed of light (that sounds small, but this translates to very nearly 3 million meters per second), and you'll think to yourselves that neurons are actually quite poor conductors.
But they're the best that our bodies have, and they're great for what we need.
level 2
2 points · 2 years ago
This was a brilliant answer and addresses OP's question way more objectively than what I said.
Thank you!
level 1
8 points · 2 years ago
That's not how neural signaling works.
As a signal travels down a neuron, none of it is lost due to resistance, because at every step along the way the neuron uses energy (derived from the ion gradient across the membrane, which is derived from ATP) to boost the signal to full strength. So it doesn't matter how long the neuron is, as long as you have enough ion gradient, the strength of the signal doesn't degrade.
(And at interconnections between neurons a completely different thing happens where multiple neurons can feed into a single neuron, at which point the question about signal loss doesn't really make sense.)
level 2
Population Genetics | Landscape Ecology | Landscape Genetics2 points · 2 years ago
I'm not sure who downvoted this, but it's correct. Axons are like chains of repeaters, preventing signal degradation.
level 1
2 points · 2 years ago · edited 2 years ago
I don't know enough physics to tell you with certainty about loss of energy, but maybe this will help:
The electric signals travelling along neurons aren't the same as the electric currents that travel through wires. The latter occurs by electrons moving through the wires, while in the brain what happens is that the normal ionic gradients between the inside of a neuron and the outside of a neuron changes. The resting gradient makes the outside of the cell stay positively charged and the inside stay negatively charged, which means it's polarized. When a neurotransmitter binds to a neuronal receptor, it may cause depolarization, which means that positive ions will go inside the cell (usually sodium and calcium). The ions go inside through voltage-dependant ion channels that open up when they detect changes in voltage (i.e. first neurotransmitter causes ions to go inside the cell, changes in voltage caused by these ions opens up voltage-dependant ion channels downstream), further changing the ion gradient onwards (or downstream), until it reaches the end of the neuron's axon and causes local physiological changes that translate to a certain function.
All in all, although it is an electric current, it's one caused by ion movement, there are no actual electrons moving through the neuronal cell.
TL;DR: Ion gradient changes sequentially throughout the axon and causes a particular biological function in the end of the axon. No actual electrons moving, so it doesn't behave exactly like electric currents in wires.
PS.: I'd really like a physicist to correct me or add extra info on electric currents caused by ionic movement. Sorry I couldn't fully aswer you, OP.
PPS.: Some neurotransmitters cause hyperpolarization and are deemed to be inhibitors.
level 2
3 points · 2 years ago
I have a degree in physics and a background in biomedical sciences. Maybe my response fills in what you're looking for?
Though your response is definitely great!
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Problem Type: Divide a Whole Number by Unit Fractions.
This worksheet supports the following Common Core State Standard:
CCSS.Math.Content.5.NF.B.7.b
Interpret division of a whole number by a unit fraction, and compute such quotients. For example, create a story context for 4 ÷ (1/5), and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that 4 ÷ (1/5) = 20 because 20 x (1/5) = 4.
Publisher: National Governors Association Center for Best Practices, Council of Chief State School Officers, Washington D.C.
Internet4classrooms is a collaborative effort by Susan Brooks and Bill Byles. | 320 | 1,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-35 | longest | en | 0.828409 |
https://socratic.org/questions/how-do-you-factor-x-2-3x-18-0 | 1,575,720,000,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540499389.15/warc/CC-MAIN-20191207105754-20191207133754-00468.warc.gz | 561,909,002 | 6,056 | # How do you factor x^2 +3x - 18 = 0?
May 23, 2015
${x}^{2} + 3 x - 18$
We can Split the Middle Term of this expression to factorise it.
In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:
${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot \left(- 18\right) = - 18$
and,
${N}_{1} + {N}_{2} = b = 3$
After trying out a few numbers we get ${N}_{1} = 6$ and ${N}_{2} = - 3$
$6 \cdot \left(- 3\right) = - 18$ and $6 + \left(- 3\right) = 3$
${x}^{2} + 3 x - 18 = {x}^{2} + 6 x - 3 x - 18$
$x \left(x + 6\right) - 3 \left(x + 6\right)$
color(green)((x-3)color(red)((x + 6) is the factorised form. | 285 | 667 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2019-51 | latest | en | 0.763297 |
https://www.gamedev.net/forums/topic/613997-hulls-and-cob-calculation/ | 1,547,907,424,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583668324.55/warc/CC-MAIN-20190119135934-20190119161934-00207.warc.gz | 793,859,036 | 23,890 | # Hulls and COB calculation
This topic is 2638 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
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I have some questions on calculating COB for a mesh. Assume that the hull density is non uniform. Knowing the total mass of the hull, how can I calculate COB?
My instincts tell me that I will need to integrate the volume starting from the lowest part of the keel up, along the +y axis, until I arrive at d for a plane p(<0,1,0.>,d) where the volume of the hull below p * density of water = mass of hull. If so than the centroid of that mesh should be the COB. I'm also aware that it should be possible to use some sort of binary subdivision to speed this up.
Am I on the right track.
@Mods Sorry for the double post, firefox glitched, feel free to remove
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# Graded Problem Set #2 - Done.docx - MTH 243 – PCC(Trude...
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MTH 243 – PCC (Trude) Name:___________Jin Situ__ _________ GRADED PROBLEM SET #2 Answer each of the following questions completely. When possible to answer using a complete sentence and offering explanation, please do so. There are a total of 20 points possible in the assignment. This data along the right is from the U.S. Department of Health and Human Services, National Center for Health Statistics, Third National Health and Nutrition Exam Survey. Recorded are the pulse rates from people who identify as female. a. [4 pts] Fill in the table below outlining the summary statistics Sample Size 40 Mean 76.3 Standard Deviation 12.4986 Minimum 60 First Quartile 68 Median 74 Third Quartile 80 Maximum 124 b. [3 pts] Construct a histogram for the data with your number line starting at 60 and using a bin width of 10. (Either by hand or using technology) c. [2 pts] Interpret the standard deviation in a complete sentence using the numerical value and the context. The pulse rates from the women got survey from the U.S. Department of Health and Human Services, National Center for Health Statistics. From the highest pulse rate 124
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Unformatted text preview: down to 60. The standard deviation of the pulse rates for this group of women are 12.4986. d. [2 pts] Are there any outliers in this data set? (Show work) IQR=80-68=12 LF=68-1.5(12)=50 UF=80+1.5(12)=98 So 124 is the outliner. e. [3 pts] Construct a boxplot for the data. (Either by hand or using technology) 60 60 60 64 64 64 64 68 68 68 68 68 72 72 72 72 72 72 72 72 76 76 76 76 76 76 80 80 80 80 80 80 88 88 88 88 88 96 104 124 MTH 243 – PCC (Trude) Name:___________Jin Situ__ _________ f. [2 pts] Is the shape symmetric, left-skewed, or right skewed? Right skewed. g. [2 pts] Would you expect the mean or median to be larger based on the shape of the distribution? Explain. Yes, I would expect on mean. Because it is a right shewed, and the mean is higher than the median. h. [2 pts] Summarize your findings by describing the shape, center, spread, and unusual features of the data. The data is right skewed with a median pulse rate of 74 and IQR of 12. There was an outlier of 124....
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http://www.yougonumbers.com/hexadecimal/0x78 | 1,502,888,830,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886101966.48/warc/CC-MAIN-20170816125013-20170816145013-00558.warc.gz | 714,879,971 | 5,273 | Miscellaneous & Conversions For 120
Written English One Hundred Twenty Roman Numeral 120 CXX RGB Colour (# 078 or #000078)
Decimal / Number 120 Binary 1111000 Ternary 11110 Hexadecimal (Hex) 0x78 Quaternary 1320 Quinary 440 Senary (Heximal) 320 Octal 170 Duodecimal (Dozenal) a0
Mathematics
Prime Factors: 2 x 2 x 2 x 3 x 5 = 120 Factorial: 120! 6.68950291345E+198 Harshad Number: 120 True Triangular Number Of 120 7260 Square Root: √120 10.9544511501 Cube Root (Root 3): 3√120 4.93242414866 Tesseract Root (Root 4): 4√120 3.30975091965 Quintal / nth Root. (Root 5) 5√120 2.6051710847 ¾ Root 36.2565047683 Fibonnaci # 120 5.35835925499E+24 Recipricol 1/ 120 0.00833333333333 Quarter 1/4 30 or 30/120 Half 1/2 60 or 60/120 Third 1/3 40 or 40/120
2% Fibonacci
38% Numbers
20% Binary | 308 | 788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-34 | longest | en | 0.340905 |
https://estebantorreshighschool.com/useful-about-equations/simple-equation.html | 1,632,737,205,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00196.warc.gz | 273,250,228 | 10,843 | ## What are the types of equation?
Types of equationslinear equation for degree one.quadratic equation for degree two.cubic equation for degree three.quartic equation for degree four.quintic equation for degree five.sextic equation for degree six.septic equation for degree seven.octic equation for degree eight.
## What is the golden rule for solving equations?
Do unto one side of the equation, what you do to the other! When solving math equations, we must always keep the ‘scale’ (or equation) balanced so that both sides are ALWAYS equal.
## What are the 4 steps to solving an equation?
We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal.
## What are the 3 types of equations?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form.
## How do you explain an equation?
What is an equation? In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. The most basic and common algebraic equations in math consist of one or more variables.
## What is the formula for algebra?
n! = n(n − 1)! = n(n − 1)(n − 2)! = ….Solution:
More topics in Algebra Formulas
Factoring Formulas Percentage Formula
Direct Variation Formula Inverse Variation Formula
Equation Formula Series Formula
Function Notation Formula Foil Formula
## What is simple algebra?
Basic algebra is the field of mathematics that it one step more abstract than arithmetic. Remember that arithmetic is the manipulation of numbers through basic math functions. Algebra introduces a variable, which stands for an unknown number or can be substituted for an entire group of numbers.
## What are the four basic rules of algebra?
Basic Rules and Properties of AlgebraCommutative Property of Addition. a + b = b + a. Examples: real numbers. Commutative Property of Multiplication. a × b = b × a. Examples: real numbers. Associative Property of Addition. (a + b) + c = a + (b + c) Examples: real numbers. Associative Property of Multiplication. (a × b) × c = a × (b × c) Examples: real numbers.
You might be interested: Pkb equation
## What are the 5 steps to solving an equation?
The 5 Steps of Problem SolvingA “Real World” Math Drama. Step #1: Stop and Think Before Doing Anything. Step #2: English-to-Equation Translation. Step #3: Solve for Whatever You’re Interested In. Step #4: Make Sure You Understand the Result. Step #5: Use Your Result to Solve Other Problems. Wrap Up.
## What are the rules of equations?
A General Rule for Solving EquationsSimplify each side of the equation by removing parentheses and combining like terms.Use addition or subtraction to isolate the variable term on one side of the equation.Use multiplication or division to solve for the variable.
### Releated
#### How to write a regression equation
What is a regression equation example? A regression equation is used in stats to find out what relationship, if any, exists between sets of data. For example, if you measure a child’s height every year you might find that they grow about 3 inches a year. That trend (growing three inches a year) can be […]
#### Solving an absolute value equation
How do you find the absolute value? Absolute Value means and “−6” is also 6 away from zero. More Examples: The absolute value of −9 is 9. The absolute value of 3 is 3. Can you solve problems using absolute value? Solving absolute value equations is as easy as working with regular linear equations. The […] | 805 | 3,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-39 | latest | en | 0.908966 |
http://www.math.utexas.edu/pipermail/maxima/2006/001352.html | 1,369,335,082,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703682988/warc/CC-MAIN-20130516112802-00028-ip-10-60-113-184.ec2.internal.warc.gz | 585,574,125 | 1,768 | # [Maxima] How to compose multi-variate polynomials?
Robert Dodier robert.dodier at gmail.com
Sun May 7 11:06:14 CDT 2006
```Hi Jonathan,
> H maps 2N-dimensional space to N-dim'al space, and
> f,g1,g2 each map N-dim'al space to N-dim'al space.
>
> We'll want to test if
>
> H(g1(xvec), g2(xvec))
>
> equals f(xvec) .
I'm afraid I don't see what's going on here.
Maybe this point can be clarified with an example.
> define(funmake(f, makelist(concat( 'x,j) , j , 1,N)) , ''expression_in_xvec)
Maxima allows subscripted variables x[1], x[2], x[3], ...
I don't completely understand what you need to do, but
it seems likely that working with subscripted variables
is going to be easier than constructing symbols x1, x2, x3, ...
For example, when expression_in_xvec is an expression
in the variables x[1], x[2], x[3], ... you can just write
f (x) := ''expression_in_xvec;
and then f ([a, b, c, ...]) causes a, b, c, ... to be substituted for
x[1], x[2], x[3], ....
Hope this helps,
Robert Dodier
``` | 321 | 1,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2013-20 | latest | en | 0.834809 |
https://tbc-python.fossee.in/convert-notebook/Analog_Integrated_Circuits_by_R.S._Tomar/Chapter5_1.ipynb | 1,718,215,530,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861183.54/warc/CC-MAIN-20240612171727-20240612201727-00875.warc.gz | 523,733,324 | 51,628 | # Chapter5 - Linear applications of op-amps¶
## Ex 5.1 - page : 121¶
In [1]:
from __future__ import division
V1=2 #V
V2=3 #V
V3=4 #V
V4=5 #V
R1=10 #kohm
R2=15 #kohm
R3=22 #kohm
R4=50 #kohm
Rf=10 #kohm
Vout=-Rf/R1*V1-Rf/R2*V2-Rf/R3*V3-Rf/R4*V4 #V
print "Output voltage of the circuit is %0.2f V " %Vout
Output voltage of the circuit is -6.82 V
## Ex 5.2 - page : 129¶
In [3]:
from __future__ import division
Rf=240 #kohm
#Vout=-4*Vx+3*Vy
#case 1st
Vy=0 #V(But Vx is not=0)
#Vox=-Rf/R1*Vx=-4*Vx
R1=Rf/4 #kohm
#case 2nd
Vx=0 #V(But Vy is not=0)
#Voy=(1+Rf/R1)*R2*Vy/(R1+R2)=3*Vy
R2=3/(1+Rf/R1)*R1/((1-3/(1+Rf/R1)))
print "Resistance R1 is %0.f kohm " %R1
print "Resistance R2 is %0.f kohm " %R2
Resistance R1 is 60 kohm
Resistance R2 is 90 kohm
## Ex 5.3 - page : 130¶
In [5]:
from __future__ import division
V1=-2 #V
V2=3 #V
R1=50 #kohm
R2=100 #kohm
Rf=250 #kohm
#I1+I2=If with IB=0 & Vx=0
Vout=-(V1/R1+V2/R2)*Rf #V
print "Output Voltage is %0.1f V " %Vout
Output Voltage is 2.5 V
## Ex 5.4 - page : 130¶
In [7]:
from __future__ import division
V1=-2 #V
V2=3 #V
R1=12 #kohm
R2=12 #kohm
R3=10 #kohm
Rf=12 #kohm
Ri=12 #kohm
Rt=2 #kohm
Vyx=200*10**-6 #V
Vout=Rf/Ri*(1+2*R3/Rt)*Vyx #V
Vout*=1000 #mV
print "Output Voltage is %0.1f mV " %Vout
Output Voltage is 2.2 mV
## Ex 5.5 - page : 131¶
In [9]:
from __future__ import division
R1max=50 #kohm(Potentiometer)
R4=10 #kohm
R3=10 #kohm
R1=R1max #kohm
print "Resistance R2 is %0.f kohm " %R2
print "Minimum value of resistance R1 is %0.f kohm " %R1min
Resistance R2 is 100 kohm
Minimum value of resistance R1 is 1 kohm
## Ex 5.6 - page : 132¶
In [14]:
from __future__ import division
R3=1 #kohm
Rt=5 #kohm
Ri=1.8 #kohm
R1=1.8 #kohm
Rf=18 #kohm
R2=18 #kohm
Vs=15 #V
AoL=2*10**5 #Gain(for 741C)
Rio=2#Mohm
Ro=75#Mohm
fo=5 #Hz
fBW=1 #MHz
print "Differential gain is %0.2f " %Ad
Beta=(R3+Rt)/(2*R3+Rt) #unitless
Rix=Rio*10**6*(1+AoL*Beta) #ohm
print "Input impedence, Rix is %0.2e ohm " %Rix
print "Output impedence is %0.1e Rof ohm " %Rof
#Answer in the book is wrong for Rix.
Differential gain is 14.00
Input impedence, Rix is 3.43e+11 ohm
Output impedence is 5.2e-03 Rof ohm
## Ex 5.8 - page : 134¶
In [15]:
from __future__ import division
Ri=10 #kohm
Rf=15 #kohm
Vs=9 #V
#Part (a)
Ra=120 #ohm
Rb=120 #ohm
Rc=120 #ohm
Rd=120 #ohm
Vx=0 #V
Vy=0 #V (as Bridge is balanced)
Vout=(Vy-Vx)*Rf/Ri #V
print "(a) Output Voltage is %0.2f V " %Vout
#Part (b)
Ra=120 #ohm
Rb=120 #ohm
Rc=120 #ohm
Rd=150 #ohm
Vx=Rb*Vs/(Ra+Rb) #V
Vy=Rc*Vs/(Rc+Rd)#V
Vyx=Vy-Vx #V
Vout=(Vy-Vx)*Rf/Ri #V
print "(b) Output Voltage is %0.2f V " %Vout
(a) Output Voltage is 0.00 V
(b) Output Voltage is -0.75 V
## Ex 5.9 - page : 135¶
In [17]:
from __future__ import division
Vin=2 #V
Rf=2*2/(2+2)+2 #kohm
R1=1 #kohm
Vout=-Rf/R1*Vin #V
print "Output Voltage is %0.f V " %Vout
Output Voltage is -6 V
## Ex 5.11 - page : 144¶
In [21]:
from __future__ import division
import math
G=20 #dB(Gain)
f3dB=2 #kHz
Cf=0.05 #micro F
Rf=1/(f3dB*1000*2*math.pi*Cf/1000000)/1000 #kohm
G=10**(G/20) #Gain(unitless)
Ri=Rf*1000/G #ohm
print "Resistance Rf is %0.1f kohm " %Rf
print "Resistance Ri is %0.f ohm " %Ri
# Answer in wrong in thetextbook.
Resistance Rf is 1.6 kohm
Resistance Ri is 159 ohm
## Ex 5.13 - page : 146¶
In [19]:
%matplotlib inline
import matplotlib.pylab as plt
import numpy as np
from __future__ import division
t2=50 #ms(After open the switch)
R=40 #kohm
C=0.2 #micro F
V2=3 #V
Vin=5 #V
#For Ideal op-amp V1=V2
t1=0 #s
Vout1=V2 #V
V1=V2 #V
t2=t2*10**-3 #s
f=lambda T:(Vin-V1)
def integrate(a,b,f):
# def function before using this
# f=lambda t:200**2*t**2
#a=lower limit;b=upper limit;f is a function
import numpy
N=1000 # points for iteration
t=numpy.linspace(a,b,N)
ft=f(t)
ans=numpy.sum(ft)*(b-a)/N
ans/=3
ans**=1.0/2
return ans
Vout2=-1/(R*10**3*C*10**-6)*integrate(0,t2,f)+Vout1 #V
#Here we have t=0 switch closed Vout=3V
t=np.array([t1*1000,t2*1000]) #ms
Vout=np.array([Vout1,Vout2]) #V
plt.plot(t, Vout)
plt.title('Vout Vs time after switch is opened')
plt.xlabel('t(ms)')
plt.ylabel('Vout(V)')
plt.show()
#Plot in the textbook is not accurate.
## Ex 5.14 : page - 147¶
In [4]:
from __future__ import division
R1=1 #kohm
R2=1 #kohm
R3=1 #kohm
Rf=R2+R3 #kohm
Vin=1 #V
#Capacitor remains open circuited for steady state in both cases.
Vout=-Rf/R1*Vin #V
print "Output Voltage is %0.2f V " %Vout
Output Voltage is -2.00 V
## Ex 5.16 - page : 148¶
In [6]:
from __future__ import division
#From the given equationVout=-integrate('5*Vx+2*Vy+4*Vz','t',0,t) :
R1Cf=1.0/5 #ratio
R2Cf=1.0/2 #ratio
R3Cf=1.0/4 #ratio
print "Various design parameters are : "
Cf=10 #micro F##Chosen for the design
print "Capacitance is %0.2f micro F " %Cf
R1=R1Cf/(Cf*10**-6)/1000 #kohm
R2=R2Cf/(Cf*10**-6)/1000 #kohm
R3=R3Cf/(Cf*10**-6)/1000 #kohm
print "Resistance R1 is %0.2f kohm " %R1
print "Resistance R2 is %0.2f kohm " %R2
print "Resistance R3 is %0.2f kohm " %R3
Various design parameters are :
Capacitance is 10.00 micro F
Resistance R1 is 20.00 kohm
Resistance R2 is 50.00 kohm
Resistance R3 is 25.00 kohm
## Ex 5.17 - page : 153¶
In [7]:
from __future__ import division
f=10 #kHz
Rf=3.2 #kohm
Ci=0.001 #micro F
dt=5 #micro seconds
dVin=5-(-5) #V(When voltage changes from -5V to +5V)
Vout=-Rf*1000*Ci*10**-6*dVin/(dt*10**-6) #V
print "When voltage changes from -5V to +5V, The output Voltage is %0.2f V " %Vout
dVin=-5-(+5) #V(When voltage changes from +5V to -5V)
Vout=-Rf*1000*Ci*10**-6*dVin/(dt*10**-6) #V
print "When voltage changes from +5V to -5V, The output Voltage is %0.2f V " %Vout
When voltage changes from -5V to +5V, The output Voltage is -6.40 V
When voltage changes from +5V to -5V, The output Voltage is 6.40 V
## Ex 5.18 page : 154¶
In [12]:
from __future__ import division
import math
fmin=200 #Hz
fmax=1 #kHz
fa=fmax #kHz
print "Various design parameters are : "
Ci=0.05 #micro F##Chosen for the design
print "Capacitance Ci is %0.2f micro F " %Ci
fb=10*fa #kHz
Rf=1/(2*math.pi*fa*10**3*Ci*10**-6)/1000 #kohm
print "Resistance Rf is %0.1f kohm " %Rf
Ri=1/(2*math.pi*fb*10**3*Ci*10**-6) #ohm
print "Resistance Ri is %0.f ohm " %Ri
Cf=Ri*Ci/(Rf*10**3) #micro F
print "Capacitance Cf is %0.3f micro F " %Cf
# Answer in the textbook is not accurate.
Various design parameters are :
Capacitance Ci is 0.05 micro F
Resistance Rf is 3.2 kohm
Resistance Ri is 318 ohm
Capacitance Cf is 0.005 micro F
## Ex 5.19 - page : 156¶
In [15]:
from __future__ import division
import math
fmax=100 #Hz
fa=fmax #Hz
print "Various design parameters are : "
Ci=0.1 #micro F##Chosen for the design
print "Capacitance Ci is %0.2f micro F " %Ci
Rf=1/(2*math.pi*fa*Ci*10**-6)/1000 #kohm
print "Resistance Rf is %0.1f kohm " %Rf
print "Use f=15 kohm"
fb=15*fa #kHz
Ri=1/(2*math.pi*fb*Ci*10**-6)/1000 #kohm
print "Resistance Ri is %0.2f ohm " %Ri
print "Use Ri=1 kohm"
Cf=Ri*Ci/Rf #micro F
print "Capacitance Cf is %0.3f micro F " %Cf
#Answer in the book is not accurate for Cf.
Various design parameters are :
Capacitance Ci is 0.10 micro F
Resistance Rf is 15.9 kohm
Use f=15 kohm
Resistance Ri is 1.06 ohm
Use Ri=1 kohm
Capacitance Cf is 0.007 micro F
## Ex 5.20 -page : 157¶
In [17]:
from __future__ import division
f=50 #Hz
T=1/f #s(Period)
Ci=0.05 #micro F
RiCi=0.01*T #Given
Ri=RiCi/(Ci*10**-6)/1000 #kohm
print "Resistance Ri is %0.2f kohm " %Ri
#Vout=-.002*dVin/dt given
#On comparing with Vout=-Rf*Ci*dVin/dt
RfCi=0.002 #on comparing
Rf=RfCi/(Ci*10**-6)/1000 #kohm
print "Resistance Rf is %0.2f kohm " %Rf
Cf=Ri*Ci/Rf #micro F
print "Capacitance Cf is %0.3f micro F " %Cf
Resistance Ri is 4.00 kohm
Resistance Rf is 40.00 kohm
Capacitance Cf is 0.005 micro F | 3,505 | 7,620 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.549586 |
http://www.phy.ntnu.edu.tw/ntnujava/index.php?action=printpage;topic=173.0 | 1,611,388,595,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703536556.58/warc/CC-MAIN-20210123063713-20210123093713-00694.warc.gz | 182,539,020 | 2,813 | # NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
## Easy Java Simulations (2001- ) => Dynamics => Topic started by: Fu-Kwun Hwang on May 16, 2005, 07:59:39 am
Title: Condition for stable equilibrium Post by: Fu-Kwun Hwang on May 16, 2005, 07:59:39 am Under the influence of gravity: the condition for stable equilibrium is the center of gravity [b:22e0df7a63](c.g.)[/b:22e0df7a63] must be lower than the supporting point. (potential energy is minimum) Title: Re: Condition for stable equilibrium Post by: lookang on February 11, 2010, 05:38:31 pm great applet! i am trying to re-customize it to allow for 2 mass variables, 2 distances.i defined(omega)/dt = -(m1*g*(xl-x) +m2*g*(xr-x))/(inertia1+inertia2)but it didn't evolution as i thought it would in real life.i will work on it again.btw, the time is off by one hour, it is 6.39 pm instead of the time shown 5.39pm. for ur info ;D Title: Re: Condition for stable equilibrium Post by: Fu-Kwun Hwang on February 11, 2010, 05:49:54 pm If the center of gravity for those two objects is (xc,yc) and the supporting point is (x,y)The conditions for stable equilibrium is (x=xc and y>yc) when supporting bar is horizontal.The who system can be model like a pendulum, the supporting point is the tip of the pendulum, and assume all the mass is located at the center of the gravity. For the time offset: Please click the PROFILE (http://www.phy.ntnu.edu.tw/ntnujava/index.php?action=profile) link,then click Look and Layout Preferences from the left menu.You will find "Time Offset:" in the table, change the value from 0 to 1 (or click auto detect link)then click Change Profile button , then everything should be fine for your time format. Title: Re: Condition for stable equilibrium Post by: lookang on February 12, 2010, 07:53:20 am Quote from: Fu-Kwun Hwang on February 11, 2010, 05:49:54 pmIf the center of gravity for those two objects is (xc,yc) and the supporting point is (x,y)The conditions for stable equilibrium is (x=xc and y>yc) when supporting bar is horizontal.The who system can be model like a pendulum, the supporting point is the tip of the pendulum, and assume all the mass is located at the center of the gravity. This is amazing insight, thanks for the tips, looks like i need to change quite a bit of evolution equation....LOLwill remember your tips! thx! Title: Re: Condition for stable equilibrium Post by: Fu-Kwun Hwang on February 12, 2010, 10:18:43 am The first applet was model in the same way. Use the center of gravity relative to the supporting point as a pendulum to model the system. You can check out how it was done. | 740 | 2,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-04 | latest | en | 0.854038 |
http://en.opasnet.org/w/New_house_stock_production_in_Finland | 1,553,157,523,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202506.45/warc/CC-MAIN-20190321072128-20190321094128-00240.warc.gz | 67,575,325 | 22,846 | # Building stock in Helsinki
During Decision analysis and risk management 2015 course, this page was used to collect student contributions. To see them, look at an archived version. The page has since been updated for its main use. Data in the archived tables was moved: Table 2. Energy parametres of buildings, Table 4. Energy sinks, Table 5. Changes in energy efficiency, Table 6. Important energy parametres. Tables 3, 7, and 8 did not contain data and were removed.
## Question
What is the building stock in Helsinki and its projected future?
```## This code is Op_en7115/ on page [[Building stock in Helsinki]]. library(OpasnetUtils) library(ggplot2) objects.latest("Op_en7115", code_name = "stockBuildings") stockBuildings <- EvalOutput(stockBuildings) ggplot(stockBuildings@output, aes(x = Time, weight = stockBuildingsResult, fill = Heating)) + geom_bar(binwidth = 5) + theme_gray(base_size = 24) + labs( title = "Current building stock (floor area) by heating type \n and year of construction", x = "Construction year", y = expression("Floor area ( "*m^2*" )")) ggplot(stockBuildings@output, aes(x = Time, weight = stockBuildingsResult, fill = Building)) + geom_bar(binwidth = 5) + theme_gray(base_size = 24) + labs( title = "Current building stock (floor area) by building type \n and year of construction", x = "Construction year", y = expression("Floor area ( "*m^2*" )")) ```
## Rationale
This part contains the data needed for calculations about the building stock in Helsinki. It shows the different building and heating types in Helsinki, and how much and what kind of renovations are done for the existing building stock in a year, including how much and how old building stock is demolished. This data is used in further calculations in the model.
There is also some other important data that wasn't used in the model's calculations. These include more accurate renovation statistics for residential buildings, U-value changes for renovations and thermal transmittance of different parts of residential buildings. This data is found under Data not used.
### Carbon neutral Helsinki 2035
```# This is code Op_en/7115 on page [[Building stock in Helsinki]] library(OpasnetUtils) library(plotly) objects.latest("Op_en6007", code_name = "miscellaneous") # [[OpasnetUtils/Drafts]] truncateIndex objects.latest("Op_en6007", code_name = "hnh2035") # [[OpasnetUtils/Drafts]] pushIndicatorGraph objects.latest("Op_en7237", code_name = "intermediates") # [[Helsinki energy decision 2015]] buildings etc. buildings <- EvalOutput(buildings) tmp <- truncateIndex(buildings,"Building",6) colnames(tmp@output)[colnames(tmp@output)=="EnergySavingPolicy"] <- "Scenario" #> unique(tmp\$Scenario) #[1] BAU Energy saving moderate Energy saving total #[4] WWF energy saving levels(tmp\$Scenario) <- c("BAU","NA","tavoite","NA") tmp <- tmp[tmp\$Scenario!="NA",] p_buildings_b <- plot_ly( oapply(tmp[tmp\$Scenario=="BAU",], c("Building","Time"),sum)@output, x = ~Time, y = ~buildingsResult, color = ~Building, type = 'scatter', mode = 'lines') %>% layout( title="Rakennusala talotyypeittäin", xaxis=list(title="Vuosi"), yaxis=list(title="Rakennusala (m2)") ) p_buildings_h <- plot_ly( oapply(tmp[tmp\$Scenario=="BAU",], c("Heating","Time"),sum)@output, x = ~Time, y = ~buildingsResult, color = ~Heating, type = 'scatter', mode = 'lines') %>% layout( title="Rakennusala lämmitysmuodoittain", xaxis=list(title="Vuosi"), yaxis=list(title="Rakennusala (m2)") ) pushIndicatorGraph(p_buildings_b, "https://hnh.teamy.fi/v1/indicator/12/") pushIndicatorGraph(p_buildings_h, "https://hnh.teamy.fi/v1/indicator/13/") ```
### Building stock
These tables are based on FACTA database classifications and their interpretation for assessments. This data is used for modelling. The data is large and can be seen from the Opasnet Base. Technical parts on this page are hidden for readability. Building types should match Energy use of buildings#Baseline energy consumption.
```library(OpasnetUtils) library(ggplot2) # [[Building stock in Helsinki]], building stock, locations by city area (in A Finnish coordinate system) #stockBuildings <- Ovariable("stockBuildings", ddata = "Op_en7115.stock_details") #colnames(stockBuildings@data)[colnames(stockBuildings@data) == "Built"] <- "Time" #colnames(stockBuildings@data)[colnames(stockBuildings@data) == "Postal code"] <- "City_area" # [[Building stock in Helsinki]] dat <- opbase.data("Op_en7115.stock_details")[ , c( # "Rakennus ID", "Sijainti", "Valmistumisaika", # "Julkisivumateriaali", "Käyttötarkoitus", # "Lämmitystapa", "Polttoaine", # "Rakennusaine", # "Varusteena koneellinen ilmanvaihto", # "Perusparannus", # "Kunta rakennuttajana", # "Energiatehokkuusluokka", # "Varusteena aurinkopaneeli", "Tilavuus", "Kokonaisala", "Result" # Kerrosala m2 )] colnames(dat) <- c("City_area", "Time", "Building types in Facta", "Heating types in Facta", "Tilavuus", "Kokonaisala", "Kerrosala") dat\$Time <- as.numeric(substring(dat\$Time, nchar(as.character(dat\$Time)) - 3)) #dat <- dat[dat\$Time != 2015 , ] # This is used to compare numbers to 2014 statistics. dat\$Time <- as.numeric(as.character((cut(dat\$Time, breaks = c(0, 1885 + 0:26*5), labels = as.character(1885 + 0:26*5))))) dat\$Tilavuus <- as.numeric(as.character(dat\$Tilavuus)) dat\$Kokonaisala <- as.numeric(as.character(dat\$Kokonaisala)) dat\$Kerrosala <- as.numeric(as.character(dat\$Kerrosala)) build <- tidy(opbase.data("Op_en7115.building_types")) colnames(build)[colnames(build) == "Result"] <- "Building" heat <- tidy(opbase.data("Op_en7115.heating_types")) colnames(heat)[colnames(heat) == "Result"] <- "Heating" ###################### # Korjaus ######################## temp <- as.character(heat\$Heating) temp[temp == "District heating"] <- "District" temp[temp == "Light oil"] <- "Oil" temp[temp == "Fuel oil"] <- "Oil" heat\$Heating <- temp ######################################## dat <- merge(merge(dat, build), heat)#[c("City_area", "Time", "Building", "Heating", "stockBuildingsResult")] dat\$Kerrosala[is.na(dat\$Kerrosala)] <- dat\$Kokonaisala[is.na(dat\$Kerrosala)] * 0.8 # If floor area is missing, estimate from total area. cat("Kerrosala ilman 2015 (m^2)\n") oprint(aggregate(dat["Kerrosala"], by = dat["Building"], FUN = sum, na.rm = TRUE)) cat("Kokonaisala ilman 2015 (m^2)\n") oprint(aggregate(dat["Kokonaisala"], by = dat["Building"], FUN = sum, na.rm = TRUE)) cat("Tilavuus ilman 2015 (m^3)\n") oprint(aggregate(dat["Tilavuus"], by = dat["Building"], FUN = sum, na.rm = TRUE)) temp <- aggregate(dat["Kerrosala"], by = dat[c("Time", "Building", "Heating")], FUN =sum, na.rm = TRUE) colnames(temp)[colnames(temp) == "Kerrosala"] <- "stockBuildingsResult" stockBuildings <- Ovariable("stockBuildings", data = temp) objects.store(stockBuildings) cat("Ovariable stockBuildings stored.\n") ```
### Construction and demolition
It is assumed that construction occurs at a constant rate so that there is an increase of 42% in 2050 compared to 2013. Energy efficiency comes from Energy use of buildings.
```# This code is Op_en7115/changeBuildings on page [[Building stock in Helsinki]] library(OpasnetUtils) changeBuildings <- Ovariable("changeBuildings", dependencies = data.frame( Name = c( "stockBuildings", "efficiencyShares" ), Ident = c( "Op_en7115/stockBuildings", # [[Building stock in Helsinki]] "Op_en5488/efficiencyShares" # [[Energy use of buildings]] ) ), formula = function(...) { out <- oapply(stockBuildings, cols = c("Time", "Constructed"), FUN = sum) out <- out * 0.013125 * 5 * efficiencyShares # linear increase 42% from 2013 to 2050 out@output <- out@output[as.numeric(as.character(out@output\$Time)) >= 2015 , ] return(out) } ) objects.store(changeBuildings) cat("Ovariable changeBuildings stored.\n") ```
Fraction of houses demolished per year.
Demolition rate(% /a)
ObsAgeRate
100
2501
310001
```# This code is Op_en7115/demolitionRate on page [[Building stock in Helsinki]] library(OpasnetUtils) demolitionRate <- Ovariable('demolitionRate', dependencies = data.frame(Name = "dummy"), formula = function(...) { temp <- tidy(opbase.data('Op_en7115', subset = 'Demolition rate')) temp\$Age <- round(as.numeric(as.character(temp\$Age))) out <- as.data.frame(approx( temp\$Age, temp\$Result, n = (max(temp\$Age) - min(temp\$Age) + 1), method = "constant" )) colnames(out) <- c("Age", "demolitionRateResult") out\$demolitionRateResult <- out\$demolitionRateResult / 100 * 10 # For ten-year intervals out <- Ovariable("demolitionRate", output = out, marginal = c(TRUE, FALSE)) return(out) } ) objects.store(demolitionRate) cat("Object demolitionRate stored.\n") ```
### Heating type conversion
The fraction of heating types in the building stock reflects the situation at the moment of construction and not currently. The heating type conversion corrects this by changing a fraction of heating methods to a different one at different timepoints. Cumulative fraction, other timepoints will be interpolated.
Yearly_heating_converted_factor(m2/m2)
ObsHeating_fromHeating_toTimeResult
1OilGeothermal20050
2OilGeothermal20150.5
3OilGeothermal20251
```library(OpasnetUtils) heatTypeConversion <- Ovariable("heatTypeConversion", dependencies = data.frame( Name = c( "buil", # stock at different timepoints "obstime" ) ), formula = function(...) { dat <- opbase.data("Op_en7115", subset = "Yearly_heating_converted_factor") colnames(dat)[colnames(dat) == "Time"] <- "Obsyear" dat\$Obs <- NULL out <- data.frame() temp <- unique(dat[c("Heating_from", "Heating_to")]) for (i in 1:nrow(temp)) { onetype <- merge(temp[i,], dat) tempout <- merge(obstime@output, onetype, all.x = TRUE)[c("Obsyear","Result")] tempout <- merge(tempout, temp[i,]) for (j in (1:nrow(tempout))[is.na(tempout\$Result)]) { a <- onetype\$Obsyear[which.min(abs(as.numeric(as.character(onetype\$Obsyear)) - as.numeric(as.character(obstime\$Obsyear[j]))))] tempout\$Result[j] <- onetype\$Result[a] } out <- rbind(out, tempout) } out <- Ovariable(output = out, marginal = colnames(out) != "Result") colnames(out@output)[colnames(out@output) == "Heating_from"] <- "Heating" out <- buil * out out1 <- out out1\$Result <- - out1\$Result out1\$Heating_to <- NULL out\$Heating <- out\$Heating_to out\$Heating_to <- NULL out@output <- rbind(out1@output, out@output) #sum(out\$Result) #nrow(out1@output)*2 - nrow(out@output) return(out) } ) objects.store(heatTypeConversion) cat("Ovariable heatTypeConversionstored.\n") ```
### Renovations
Estimates from Laura Perez and Stephan Trüeb, unibas.ch N:\YMAL\Projects\Urgenche\WP9 Basel\Energy_scenarios_Basel_update.docx
Fraction of houses renovated per year(%)
ObsAgeResultDescription
100Estimates from Laura Perez and Stephan Trüeb
2200Assumption Result applies to buildings older than the value in the Age column.
3251
4301
5501
61001
710001
```library(OpasnetUtils) renovationRate <- Ovariable('renovationRate', dependencies = data.frame(Name = "dummy"), formula = function(...) { temp <- tidy(opbase.data('Op_en7115', subset = 'Fraction of houses renovated per year')) temp\$Age <- round(as.numeric(as.character(temp\$Age))) out <- as.data.frame(approx( temp\$Age, temp\$Result, n = (max(temp\$Age) - min(temp\$Age) + 1), method = "constant" )) colnames(out) <- c("Age", "renovationRateResult") out\$renovationRateResult <- out\$renovationRateResult / 100 out <- Ovariable("renovationRate", output = out, marginal = c(TRUE, FALSE)) return(out) } ) objects.store(renovationRate) cat("Object renovationRate stored.\n") ```
Popularity of renovation types(%)
ObsRenovationFractionDescription
1None0
2Windows65
3Technical systems30
4Sheath reform5
5General0
```library(OpasnetUtils) renovationShares <- Ovariable("renovationShares", dependencies = data.frame(Name = "dummy"), formula = function(...) { out <- Ovariable("raw", ddata = 'Op_en7115', subset = 'Popularity of renovation types') out <- findrest((out), cols = "Renovation", total = 100) / 100 renovationyear <- Ovariable("renovationyear", data = data.frame( Obsyear = factor(c(2015, 2025, 2035, 2045, 2055, 2065)), Result = 1 )) out <- out * renovationyear # renovation shares repeated for every potential renovation year. out@output\$Renovation <- factor(out@output\$Renovation, levels = c( "None", "General", "Windows", "Technical systems", "Sheath reform" ), ordered = TRUE) return(out) } ) objects.store( renovationShares # Fraction of renovation type when renovation is done. ) cat("Objects renovationShares stored.\n") ```
### Locations of city areas
Locations of city areas (hidden for readability).
### Data not used
This contains data that was not used in the model's calculations. This includes renovation rates, the rates of heat flowing out of buildings and total floor areas of multiple types of buildings in Helsinki. The floor area data is also found in the background data of this page, which was used in the model.
Building Baseline 2020 2025 2050 Year of baseline Description Residential 27884795 32472388 34890241 44069914 2014 Building stock of Helsinki area, 2014 Public 4537025 4764475 4945952 5855546 2014 Building stock of Helsinki area, 2014 Industrial 3277271 3306063 3360467 3640854 2014 Building stock of Helsinki area, 2014 Other 10861972 11406505 11840973 13806423 2014 Building stock of Helsinki area, 2014
Notes
• Estimates were based on Siemens City Performance toolin seuraava kokous 2.2 and some derived calculations on BUILDING STOCK CALCULATION 2015.
• How to get the numbers for the baseline floor area for residential, public, industrial and other: Residential floor area was named as residential together, public by summing the floor area of health care, education and common buildings, industrial buildings were as such and other buildings comprise of business, traffic, office and storage buildings.
• Ref. Helsinki master plan for 2050: there are 860 000 citizens living in Helsinki (ref. www.yleiskaava.fi, visio2050); Residental buildings => fast growth
• Prediction of citizen number in Helsinki in 2020, 2030, 2040 and 2050 was used for calculations (ref. Helsingin 30% päästövähennysselvitys).
• Helsinki’s climate policy: 30% reduction in emissions: In 2010 the proportion of jobs in services and public sectors was 94%, and in industry 6%. In 2020 the proportion of jobs in services and public sectors is estimated to be 96%, and in industry 4%. Public and other buildings => between fast growth option and basic option, Industry=> Basic option
• Prediction of job number in Helsinki in 2020, 2030, 2040 and 2050 was used for calculations (ref. Helsingin 30% päästövähennysselvitys).
• Tables one and two The presentation of Tables 1 and 2
Technical notes:
Sheet 4_Input Buildings (Area Demand). Priority 1. Auxiliaries PPT. Absolute increase/decrease rate will be based on the inhabitants projected in time.
This is another list building types that was considered but rejected as too complex: Residential buildings, Government & public administration buildings, Commercial offices buildings, Data centers buildings, Education and K12 and universitiy buildings, Hospitals and healthcare buildings, Hotels and hospitality and leisure buildings, Exhibitions and fairs and halls buildings, Retail and stores and shops buildings, Warehouses & shopping mall buildings, Industrial buildings, Non residential buildings unspecified.
• There was a problem with missing data. There is more than 400000 m^2 floor area that is missing; this is estimated from total area that is available for these buildings. For other buildings, there is more than 400000 m^2 total area missing from buildings where floor area is given. See statistical analysis [1]. This was corrected by inputation so that is floor area was missing, 0.8*total_area was used instead [2].
Renovations per year made in residental buildings owned by Helsinki city, by construction year of the buildings.[1]
Construction year Balcony glasses Windows Julkisivujen peruskorjaus Vesikattojen peruskojaus Lämmönvaihtimen uusiminen Patteriverkoston säätö Kylpyhuonekalusteiden vaihto Patteriventtiilien vaihto New balcony doors LTO-laitteen asennus Water consumption measurements
-20 0,0 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 % 1,1 %
21-25 0,0 % 10,3 % 1,2 % 11,1 % 10,3 % 10,3 % 1,2 % 10,3 % 1,2 % 10,3 % 10,3 %
26-30 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 9,5 % 0,0 %
31-35 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 % 0,0 %
36-40 0,0 % 0,0 % 0,0 % 0,0 % 4,2 % 4,2 % 0,0 % 0,0 % 0,0 % 4,2 % 0,0 %
41-45 0,0 % 16,7 % 0,0 % 16,7 % 16,7 % 16,7 % 0,0 % 16,7 % 0,0 % 16,7 % 16,7 %
46-50 0,0 % 5,2 % 0,0 % 7,4 % 7,4 % 7,4 % 0,0 % 7,4 % 0,0 % 5,2 % 5,2 %
51-55 0,0 % 11,3 % 0,0 % 8,8 % 8,8 % 8,8 % 0,0 % 8,8 % 0,0 % 8,8 % 16,2 %
56-60 0,0 % 5,4 % 0,0 % 4,9 % 6,2 % 7,1 % 0,0 % 6,2 % 4,5 % 4,5 % 5,4 %
61-65 0,0 % 1,5 % 1,3 % 0,8 % 2,9 % 2,4 % 1,0 % 2,4 % 0,9 % 0,9 % 2,9 %
66-70 0,6 % 2,9 % 1,2 % 2,8 % 1,4 % 2,3 % 1,1 % 1,1 % 0,1 % 1,1 % 1,1 %
71-75 3,2 % 3,1 % 3,4 % 2,9 % 3,1 % 2,6 % 0,2 % 1,1 % 0,2 % 0,2 % 0,2 %
76-80 0,1 % 2,7 % 0,1 % 0,7 % 2,0 % 1,7 % 1,1 % 1,2 % 0,2 % 0,4 % 0,2 %
81-85 1,0 % 2,8 % 0,7 % 2,3 % 3,3 % 4,8 % 3,5 % 0,0 % 0,0 % 0,0 % 0,8 %
86-90 0,0 % 1,3 % 0,0 % 2,1 % 6,1 % 1,6 % 0,7 % 1,8 % 0,3 % 0,3 % 1,0 %
91-95 0,6 % 0,3 % 0,0 % 3,9 % 8,6 % 1,9 % 5,1 % 0,8 % 0,2 % 0,0 % 1,3 %
96-00 0,1 % 0,0 % 0,0 % 0,6 % 1,2 % 1,0 % 1,5 % 1,0 % 0,0 % 0,0 % 4,2 %
01-05 2,9 % 0,0 % 0,0 % 0,0 % 1,2 % 1,0 % 0,0 % 1,0 % 0,0 % 0,0 % 0,7 %
06-10 1,7 % 0,0 % 0,0 % 0,0 % 0,5 % 0,5 % 0,0 % 0,5 % 0,0 % 0,0 % 0,0 %
#: . In the document there are similar tables for total renovations from 2010 onwards to years 2016, 2020 and 2050. --Heta (talk) 09:28, 16 June 2015 (UTC) (type: truth; paradigms: science: relevant defense)
Toimenpiteiden vaikutukset yksittäisessä kohteessa ja toimenpiteisiin liittyviä huomautuksia.[1]
Action The feature in question Difference to before Unit Notes
Glass for balconies U-value for windows -0,3 W/m2,K Säästö 1-4% rakennustasolla
Changing the windows U-value for windows -1 W/m2,K Vanhoista osa kaksilasisia ja osa kolmilasisia. Uudes 1,0 W/m2,K tai alle
Julkisivun peruskorjaus U-value of walls -0,2 W/m2,K U-arvo puolitetaan eli n. 100 mm lisäeristys
Vesikattojen peruskorjaus Yläpohjan U-arvo -0,15 W/m2,K Oletetaan 50% lisäeristys U-arvo puoleen eli n. 100 mm lisäerstys
Balcony door change U-value of doors -0,5 W/m2,K Tiivistyminen tuo lisäsäästöä
Construction decade Thermal transmittance factors for building components (W/m2K) Ventilation and leakage air rates (1/h) Floor Roof Walls Windows Outdoors Supply air through the heat recovery unit Supply air bypassing the heat recovery unit Leakage air Before 1980 Single family house 0.52 0.32 0.54 2.14 1.18 0.30 0.05 0.20 Row house 0.52 0.36 0.56 2.15 1.00 0.3 0.05 0.20 Apartment building 0.59 0.37 0.61 2.18 1.40 0.37 0.00 0.10 1980's Single family house 0.30 0.21 0.28 1.70 1.00 0.30 0.05 0.15 Row house 0.32 0.22 0.30 1.70 1.00 0.30 0.05 0.15 Apartment building 0.34 0.23 0.29 1.80 1.40 0.35 0.00 0.10 1990's Single family house 0.25 0.20 0.25 1.70 1.00 0.30 0.05 0.15 Row house 0.32 0.22 0.28 1.70 1.00 0.30 0.05 0.15 Apartment building 0.332 0.22 0.28 1.75 1.40 0.38 0.00 0.10 2000's Single family house 0.24 0.17 0.24 1.40 1.00 0.30 0.05 0.13 Row house 0.28 0.18 0.26 1.50 1.00 0.45 0.05 0.15 Apartment building 0.28 0.18 0.26 1.50 1.40 0.55 0.00 0.10 2010's Single family house 0.16 0.09 0.17 1.00 1.00 0.30 0.05 0.10 Row house 0.16 0.09 0.17 1.00 1.00 0.50 0.05 0.15 Apartment building 0.16 0.09 0.17 1.00 1.00 0.60 0.00 0.10
Helsinki energy decision 2015
In English
Assessment Main page | Helsinki energy decision options 2015
Helsinki data Building stock in Helsinki | Helsinki energy production | Helsinki energy consumption | Energy use of buildings | Emission factors for burning processes | Prices of fuels in heat production | External cost
Models Building model | Energy balance | Health impact assessment | Economic impacts
Related assessments Climate change policies in Helsinki | Climate change policies and health in Kuopio | Climate change policies in Basel
In Finnish
Yhteenveto Helsingin energiapäätös 2015 | Helsingin energiapäätöksen vaihtoehdot 2015 | Helsingin energiapäätökseen liittyviä arvoja | Helsingin energiapäätös 2015.pptx
## References
1. HAESS Final report, Tampere University of Technology, 2010
2. MK Mattinen, J Heljo, J Vihola, A Kurvinen, S Lehtoranta, A Nissinen: Modeling and visualisation of residential sector energy consumption and greenhouse gas emissions | 6,763 | 20,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-13 | longest | en | 0.905815 |
https://askfilo.com/math-question-answers/tangents-are-drawn-from-the-origin-to-the-curve-ysin-x-prove-that-their-points | 1,719,006,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00422.warc.gz | 93,973,273 | 34,757 | World's only instant tutoring platform
Question
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Solving time: 4 mins
# Tangents are drawn from the origin to the curve . Prove that their points of contact lie on the curve
## Text solutionVerified
Let the point of contact be . Then, Slope of the tangent to the curve at is . Since the tangent passes through the origin, its equation We have and lies on the curve
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Question Text Tangents are drawn from the origin to the curve . Prove that their points of contact lie on the curve Updated On Aug 3, 2023 Topic Application of Derivatives Subject Mathematics Class Class 12 Answer Type Text solution:1 Video solution: 2 Upvotes 283 Avg. Video Duration 13 min | 329 | 1,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-26 | latest | en | 0.868643 |
https://en.jiansujichilun.com/article-692/ | 1,675,739,072,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00608.warc.gz | 227,382,833 | 12,481 | # Analysis of motion non-uniformity of chain drives
chain entrySprocketAfter forming a broken line, the motion of the chain drive is very similar to that of the belt drive wound on a regular polygon wheel, see Figure 9.The side length corresponds to the chain pitch p, and the number of sides corresponds to the number of sprocket teeth z.For each revolution of the sprocket, the distance the chain moves is zp. Let z1 and z2 be the number of teeth of the two sprockets, p is the pitch (mm), and n1 and n2 are the rotational speeds of the two sprockets (r/min). The average speed v (m/s) is
v=z1pn1/60*1000=z2pn2/60*1000 (4)
In fact, both the instantaneous chain speed and the instantaneous ratio of a chain drive vary.The analysis is as follows: The tight side of the chain is in a horizontal position during transmission, see Figure 6.9.Assuming that the driving wheel rotates at an equal angular velocity ω1, its indexed peripheral speed is R1ω1.When the chain link enters the drive wheel, its pin always changes its position as the sprocket rotates.When at the instant of angle β, the instantaneous speed of the horizontal movement of the chain is equal to the horizontal component of the peripheral speed of the pin.i.e. chain speed v
v=cosβR1ω1 (6)
image
The variation range of the angle is between ±φ1/2, φ1=360. /z1.When β=0, the chain speed is large, vmax=R1ω1; when β=±φ1/2, the chain speed is small, vmin=R1ω1cos(φ1/2).Therefore, even when the driving sprocket rotates at a constant speed, the chain speed v changes.The cycle changes every time a chain pitch is rotated, see Figure 10.In the same way, the instantaneous speed v`=R1ω1sinβ of the vertical movement of the chain also changes periodically, so that the chain shakes up and down.
image
slaveSprocketSince the chain speed v≠constant and the constant change of γ angle (Fig. 9), its angular speed ω2=v/R2cosγ also changes.
Obviously, the instantaneous transmission ratio cannot get a constant value.Therefore, the chain drive works unstable. | 519 | 2,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-06 | latest | en | 0.872251 |
http://www.chegg.com/homework-help/mathematical-reasoning-writing-and-proof-2nd-edition-chapter-3.2pa2-solutions-9780131877184 | 1,469,805,082,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830091.67/warc/CC-MAIN-20160723071030-00104-ip-10-185-27-174.ec2.internal.warc.gz | 355,955,226 | 16,086 | View more editions
# TEXTBOOK SOLUTIONS FOR Mathematical Reasoning Writing and Proof 2nd Edition
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Chapter: Problem:
Complete the following truth table. What does this truth table prove?
P Q Q P P ⟶ Q Q ⟶ P T T T F F T F F
Consider the following proposition again:
For each integer n, if n2 is an odd integer, then n is an odd integer.
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Chapter: Problem:
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Corresponding Textbook
Mathematical Reasoning Writing and Proof | 2nd Edition
9780131877184ISBN-13: 0131877186ISBN: Ted SundstromAuthors: | 202 | 738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-30 | latest | en | 0.70357 |
https://leetcode.weippig.com/golang/988.-smallest-string-starting-from-leaf | 1,685,504,536,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00205.warc.gz | 385,051,006 | 145,105 | # 988. Smallest String Starting From Leaf
Meidum
You are given the `root` of a binary tree where each node has a value in the range `[0, 25]` representing the letters `'a'` to `'z'`.
Return the lexicographically smallest string that starts at a leaf of this tree and ends at the root.
As a reminder, any shorter prefix of a string is lexicographically smaller.
• For example, `"ab"` is lexicographically smaller than `"aba"`.
A leaf of a node is a node that has no children.
Example 1:
Input: root = [0,1,2,3,4,3,4]
Output:
"dba"
Example 2:
Input: root = [25,1,3,1,3,0,2]
Output:
Example 3:
Input: root = [2,2,1,null,1,0,null,0]
Output:
"abc"
Constraints:
• The number of nodes in the tree is in the range `[1, 8500]`.
• `0 <= Node.val <= 25`
### 解題
Runtime: 3 ms, faster than 100%
Memory Usage: 4.5 MB, less than 92%
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func smallestFromLeaf(root *TreeNode) string {
return helper(root, "")
}
func helper(root *TreeNode, str string) string {
if root == nil {
return str
}
newStr := string(byte(root.Val + 97)) + str
if root.Left == nil && root.Right == nil {
return newStr
}
if root.Left == nil {
return helper(root.Right, newStr)
}
if root.Right == nil {
return helper(root.Left, newStr)
}
return min(helper(root.Left, newStr), helper(root.Right, newStr))
}
func min(a, b string) string{
if a < b {
return a
}
return b
} | 430 | 1,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-23 | longest | en | 0.480353 |
https://rdrr.io/cran/Irescale/f/README.md | 1,660,414,445,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00579.warc.gz | 445,569,184 | 8,702 | README.md In Irescale: Calculate and Rectify Moran's I
Many Geographical Analysis utilizes spatial autocorrelation, that allows us to study the geographical evolution from different points of view. One measurement for spatial autocorrelation is Moran's I, that is based on Pearson’s correlation coefficient in general statistics
Performing the Analysis
This package offers a straight fordward to perform the whole analisys by using the function `rescaleI` which requires an input file with a specific format you can see it at [Loading data] section
```{r whole_analysis} library(Irescale) fileInput<-system.file("testdata", "chen.csv", package="Irescale") data<-loadFile(fileInput) scaledI<-rescaleI(data,samples=1000, scalingUpTo="MaxMin") fn = file.path(tempdir(),"output.csv",fsep = .Platform\$file.sep) saveFile(fn,scaledI) if (file.exists(fn)) #Delete file if it exists file.remove(fn)
``````
## Analysis Step by Step
The analysis can be done following the steps
The input file^[The data used in this example is taken from [@chen2009].] should have the following format.
- The first column represents an unique id for the record.
- The second and third column represent the latitute and longitud of where the sample was taken
- The fourth and beyond represents the different measured variables
```{r}
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
``````
To load data to performe the analysis is quite simple. The function `loadFile` provides the interface to make it. loadFile returns a list with two variables, `data` and `varOfInterest`, the first one represents a vector with latitude and longitude; `varOfInterest` is a matrix with all the measurements from the field.
``````library(Irescale)
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
``````
If the data has a chessboard shape,the file is organized in rows and columns, where the rows represent latitute and columns longitude, the measurements are in the cell. The function `loadChessBoard` can be used to load into the analysis.
``````library(Irescale)
fileInput<-"../inst/testdata/chessboard.csv"
``````
Calculate Distance
Once the data is loaded, The distance matrix, the distance between all the points might be calcualted. The distance can be calculated using `calculateEuclideanDistance' if the points are taken in a geospatial location.
``````library(Irescale)
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
distM<-calculateEuclideanDistance(input\$data)
distM[1:5,1:5]
``````
If the data is taken from a chessboard a like field, the Manhattan distance can be used.
``````library(Irescale)
fileInput<-"../inst/testdata/chessboard.csv"
distM<-calculateManhattanDistance(input\$data)
distM[1:5,1:5]
``````
Calculate Weighted Distance Matrix
The weighted distance matrix can be calculated it using the function `calculateWeightedDistMatrix`, however it is not required to do it, because 'calculateMoranI' does it.
``````library(Irescale)
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
distM<-calculateEuclideanDistance(input\$data)
distW<-calculateWeightedDistMatrix(distM)
distW[1:5,1:5]
``````
Moran's I
It is time to calculate the spatial autocorrelation statistic Morans' I. The function `calcualteMoranI`, which requires the distance matrix, and the variable you want are interested on.
``````library(Irescale)
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
distM<-calculateEuclideanDistance(input\$data)
I<-calculateMoranI(distM = distM,varOfInterest = input\$varOfInterest)
I
``````
Resampling Method for I
The scaling process is made using Monte Carlo resampling method. The idea is to shuffle the values and recalculate I for at least 1000 times. In the code below, after resampling the value of I, a set of statistics are calculated for that generated vector.
``````library(Irescale)
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
distM<-calculateEuclideanDistance(input\$data)
I<-calculateMoranI(distM = distM,varOfInterest = input\$varOfInterest)
vI<-resamplingI(1000,distM, input\$varOfInterest) # This is the permutation
statsVI<-summaryVector(vI)
statsVI
``````
Plotting Distribution (Optional)
To see how the value of I is distribuited, the method `plotHistogramOverlayNormal` provides the functionality to get a histogram of the vector generated by resampling with a theorical normal distribution overlay.
``````library(Irescale)
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
distM<-calculateEuclideanDistance(input\$data)
I<-calculateMoranI(distM = distM,varOfInterest = input\$varOfInterest)
vI<-resamplingI(1000,distM, input\$varOfInterest) # This is the permutation
statsVI<-summaryVector(vI)
plotHistogramOverlayNormal(vI,statsVI, main=colnames(input\$varOfInterest))
``````
Rescaling I
Once we have calculated the null distribution via resampling, you need to scale by centering and streching. The method `iCorrection`, return an object with the resampling vector rescaled, and all the summary for this vector, the new value of I is returned in a variable named `newI`
``````library(Irescale)
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
distM<-calculateEuclideanDistance(input\$data)
I<-calculateMoranI(distM = distM,varOfInterest = input\$varOfInterest)
vI<-resamplingI(1000,distM, input\$varOfInterest) # This is the permutation
statsVI<-summaryVector(vI)
corrections<-iCorrection(I,vI)
corrections\$newI
``````
Calculate P-value
In order to provide a significance to this new value, you can calculate the pvalue using the method `calculatePvalue`. This method requires the scaled vector, you get this vector,`scaledData`, the scaled I, `newI` and the mean of the `scaledData`.
``````library(Irescale)
fileInput<-system.file("testdata", "chen.csv", package="Irescale")
distM<-calculateEuclideanDistance(input\$data)
I<-calculateMoranI(distM = distM,varOfInterest = input\$varOfInterest)
vI<-resamplingI(1000,distM, input\$varOfInterest) # This is the permutation
statsVI<-summaryVector(vI)
corrections<-iCorrection(I,vI)
pvalueIscaled<-calculatePvalue(corrections\$scaledData,corrections\$newI,corrections\$summaryScaledD\$mean)
pvalueIscaled
``````
Stability Analysis
In order to determine how many iterations it is necessary to run the resampling method, it is possible to run a stability analysis. This function draw a chart in log scale (10^x) of the number of interations needed to achieve the stability in the Monte Carlo simulation.
``````fileInput<-system.file("testdata", "chen.csv", package="Irescale") | 1,664 | 6,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | longest | en | 0.736941 |
https://coderegimetech.com/rubiks-nxnxn-cube-algorithms/ | 1,685,547,098,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00325.warc.gz | 220,255,678 | 13,378 | # nxnxn Cube Algorithms
In this article, we have uploaded the nxnxn Cube Algorithms PDF to help you. The Professor’s Cube is an extension of the Rubik’s Cube and the Rubik’s Revenge. It is made of five rotating slices, from which it follows that the Professor’s Cube is composed by 98 cubies: 8 corner cubies (possessing 3 stickers each), 36 edge cubies (2 stickers), and 54 remaining center cubies (one sticker only). At an first glance the Professor’s Cube turns out to share a remarkable feature with the Rubik’s Cube: one cubie in each face, namely the most central one, is fixed. Below we have provided the download link for Cube Algorithms PDF.
### nxnxn Cube Algorithms PDF – Details
This represents a big difference with respect to Rubik’s Revenge and any cube with an even number of slices, where each center cube can be moved. Furthermore, the number of edge cubies is exactly the sum of the 24 edges(twelve pairs, in particular) of the Rubik’s Revenge and the 12 edges of the Rubik’s Cube: we will refer to the formers as coupled edges (indicated by black spots in Figure 1), while to the letters as single edgesThe Professor’s Cube is an extension of the Rubik’s Cube and the Rubik’s Revenge. It is made of five rotating slices, from which it follows that the Professor’s Cube is composed by 98 cubies: 8 corner cubies (possessing 3 stickers each), 36 edge cubies (2 stickers), and 54 remaining center cubies (one sticker only).
### nxnxn Cube Algorithms PDF – Method
• The nxnxn professor’s cube was first invented by Udo Krell in 1981.
• It became very famous in all of Rubik’s Cube.
• nxnxn cube has 8 corners, 36 edges, and 54 centres.
• Even now it is very famous among children.
• And they keep on searching how to solve nxnxn cube and it’s algorithms.
Rubik cube notation: The rotations of the six sides of the puzzle are marked with the letters:
F (Front) – Forward
R (Right) – Right
U (Up) – up
D (Down) – Down
L (Left) – Left
B (Back) – Back | 507 | 1,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-23 | latest | en | 0.909756 |
https://www.bartleby.com/solution-answer/chapter-41-problem-24e-elementary-geometry-for-college-students-7e-7th-edition/9781337614085/in-quadrilateral-abcd-the-midpoints-of-opposite-sides-are-joined-to-form-two-intersecting-segments/e9a60f54-757b-11e9-8385-02ee952b546e | 1,571,857,323,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00292.warc.gz | 805,768,805 | 55,556 | Chapter 4.1, Problem 24E
### Elementary Geometry For College St...
7th Edition
Alexander + 2 others
ISBN: 9781337614085
Chapter
Section
### Elementary Geometry For College St...
7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem
# In quadrilateral ABCD, the midpoints of opposite sides are joined to form two intersecting segments. Try drawing other quadrilaterals and joining opposite midpoints. What can you conclude about these segments in each case?
To determine
To conclude:
When the midpoints of the consecutive sides are joined in order.
Explanation
Given:
In quadrilateral ABCD, the midpoints of opposite sides are joined to form two intersecting segments.
Corollary:
A quadrilateral is a four sided closed figure.
From the given quadrilateral ABCD, the midpoints of opposite sides are joined to form two intersecting segments...
### Still sussing out bartleby?
Check out a sample textbook solution.
See a sample solution
#### The Solution to Your Study Problems
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Get Started | 251 | 1,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-43 | latest | en | 0.897219 |
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Forum Rules: Read This Before Posting
### Topic: electrolysis question (Read 7636 times)
0 Members and 1 Guest are viewing this topic.
#### funboy
• Regular Member
• Posts: 51
• Mole Snacks: +2/-4
##### electrolysis question
« on: December 19, 2006, 07:44:21 PM »
By the electrolysis of water, 11.2 L of oxygen at STP was prepared
What charge was required
1 mol of a gas at STP = 22.4 L
Therefore 11.2 L of a gas at STP = 1/2 mole
H2O --> 2H+ + 1/2O2 + 2e-
So 2H+ and 1/2 O2 accompanied by 2 X (9.64 X 10^4) electrons yeilds 1 mol of H2O
(1/2 mole of O would require 2 moles of electrons or 2 X (9.64 X 10^4)
Given that I = Q/T
All I know is Q
Question says nothing about T and I need to calculate I
What am I doing wrong
#### Borek
• Mr. pH
• Deity Member
• Posts: 25518
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• Gender:
• I am known to be occasionally wrong.
##### Re: electrolysis question
« Reply #1 on: December 19, 2006, 07:48:55 PM »
What charge was required
(...)
Question says nothing about T and I need to calculate I
What do you need I for if the question - as you have entered it - asks for charge?
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#### funboy
• Regular Member
• Posts: 51
• Mole Snacks: +2/-4
##### Re: electrolysis question
« Reply #2 on: December 19, 2006, 11:18:04 PM »
By the electrolysis of water, 11.2 L of oxygen at STP was prepared
What charge was required
1 mol of a gas at STP = 22.4 L
Therefore 11.2 L of a gas at STP = 1/2 mole
H2O --> 2H+ + 1/2O2 + 2e-
So 2H+ and 1/2 O2 accompanied by 2 X (9.64 X 10^4) electrons yeilds 1 mol of H2O
(1/2 mole of O would require 2 moles of electrons or 2 X (9.64 X 10^4)
Charge required is 2 moles which equals 1.93 X 10^5
Question B
If a current of 0.5A was used, how long did it take
I = Q/T
.5 = 1.93 X 10^5 / X
.5x = 1.93 X 10^5
x = 9.65 X 10 ^4
x / 3.6 X 10^3 = 26.8
Am I close??
Thanks
Chris
#### Borek
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##### Re: electrolysis question
« Reply #3 on: December 20, 2006, 04:17:44 AM »
Not without units. 26.8 of what? Otherwise OK.
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#### Jabus
• Regular Member
• Posts: 16
• Mole Snacks: +0/-0
##### Re: electrolysis question
« Reply #4 on: January 03, 2007, 03:06:58 PM »
with I = Q/t
I don't see why the x is used...couldn't it just be:
t = 1.93 x 10^5 / 0.5A
which is about 386000 seconds and divide by 3600 to get hours so it took roughly 107 hours?
#### pepe02ar
• New Member
• Posts: 5
• Mole Snacks: +0/-0
##### Re: electrolysis question
« Reply #5 on: January 04, 2007, 06:14:22 PM »
Hello. Those results are wrong. Remeber that the Water cannot conduce elecricity without the precense of a compound wich could transmit electricity. You need (for an electrolisys) a ionic salt (for example) disolved in the Water (an electrolite solution). The Faraday's Laws apllies to the quantity of electrolite descomposed in the electrolisis. Bye.
#### Borek
• Mr. pH
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• Gender:
• I am known to be occasionally wrong.
##### Re: electrolysis question
« Reply #6 on: January 04, 2007, 06:36:48 PM »
Hello. Those results are wrong. Remeber that the Water cannot conduce elecricity without the precense of a compound wich could transmit electricity. You need (for an electrolisys) a ionic salt (for example) disolved in the Water (an electrolite solution). The Faraday's Laws apllies to the quantity of electrolite descomposed in the electrolisis. Bye.
You are right that there is an ionic substance needed. But the rest of your post is wrong - put some NaOH into water and do the calculations to see that you will get exactly the same result, and that NaOH will not get consumed - so in fact you will be electrolysing water.
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#### jennielynn_1980
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##### Re: electrolysis question
« Reply #7 on: January 11, 2007, 03:26:46 PM »
with I = Q/t
I don't see why the x is used...couldn't it just be:
t = 1.93 x 10^5 / 0.5A
which is about 386000 seconds and divide by 3600 to get hours so it took roughly 107 hours?
I did the question the same way as Jabus. Is this a correct way of doing it?
#### Jabus
• Regular Member
• Posts: 16
• Mole Snacks: +0/-0
##### Re: electrolysis question
« Reply #8 on: January 11, 2007, 06:13:09 PM »
After looking into it, I'm pretty sure our result is the correct one Jennie as funboys result would be in C/s. Since the question asks for 'how long did it take' I'm fairly sure our answer is correct. But I suppose someone double checking wouldn't hurt
#### Borek
• Mr. pH
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• Gender:
• I am known to be occasionally wrong.
##### Re: electrolysis question
« Reply #9 on: January 11, 2007, 06:50:36 PM »
Fact the funboy solved the question using x instead of t is not important - replace x with t and it will work OK.
However, he did a mistake in his math which I have not catched earlier. Note how his result is exactly 4 times too low (4*26.8=107.2). That's not a coincidence.
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#### jennielynn_1980
• Full Member
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##### Re: electrolysis question
« Reply #10 on: January 11, 2007, 08:04:35 PM »
Thanks for clearing that up Borek I knew he was using x instead of t but I wasn't sure how exactly he got his answer of 26.8 that's all.
#### pepe02ar
• New Member
• Posts: 5
• Mole Snacks: +0/-0
##### Re: electrolysis question
« Reply #11 on: February 02, 2007, 04:24:11 PM »
Hello. Sorry by the late. Ok. I was wrong, I doesn't take care that details . Ok. We got this redox:
2 H2O ---- 2 H+ + O2 + 4e-
The Faraday's definition consist that a Faraday (unit of charge) es equal to the charge of ONE MOLE of electrons. In this reaction we can see that four electrons by 2 molecules of Water there are in the change, or there are four electrons by one molecule of Oxygen. So, for one mole of Oxygen, we got four moles of electrons, or, wich is the same, 4 F (four Faraday's). The problem says that 1/2 mole of Oxygen was obtained (in STP), and ask for the charge requeride. Make your own conclusions.
P.D.: Isn't beautifull, the chemistry ?
P.P.D.: Sorry by my bad english. | 2,137 | 6,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-10 | latest | en | 0.908761 |
https://uicheritagegarden.org/and-pdf/1703-fundamental-theorem-of-calculus-part-1-and-2-examples-pdf-423-854.php | 1,638,057,898,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00352.warc.gz | 641,405,306 | 9,070 | # Fundamental Theorem Of Calculus Part 1 And 2 Examples Pdf
File Name: fundamental theorem of calculus part 1 and 2 examples .zip
Size: 1116Kb
Published: 31.01.2021
Let's recast the first example from the previous section. What about the second approach to this problem, in the new form? We summarize this in a theorem. First, we introduce some new notation and terms. That is, the left hand side means, or is an abbreviation for, the right hand side.
## 5.3: The Fundamental Theorem of Calculus Basics
The second part of the fundamental theorem tells us how we can calculate a definite integral. Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes. Determine the new limits of integration. Refine the limits of integration. Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website.
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## 5.3: The Fundamental Theorem of Calculus Basics
Exercises 3. Problems Each chapter ends with a list of the solutions to all the odd-numbered exercises. The examples in this section can all be done with a basic knowledge of indefinite integrals and will not require the use of the substitution rule. Functions defined by integrals: challenge problem Opens a modal Practice. Second Fundamental Theorem of Calculus. Functions defined by definite integrals accumulation functions 4 questions.
## fundamental theorem of calculus problems and solutions pdf
The examples in this section can all be done with a basic knowledge of indefinite integrals and will not require the use of the substitution rule. The authors are thankful to students Aparna Agarwal, Nazli Jelveh, and Michael Wong for their help with checking some of the solutions. Let Fbe an antiderivative of f, as in the statement of the theorem. Functions defined by definite integrals accumulation functions 4 questions.
In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function. Unfortunately, so far, the only tools we have available to calculate the value of a definite integral are geometric area formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section we look at some more powerful and useful techniques for evaluating definite integrals. These new techniques rely on the relationship between differentiation and integration.
#### Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives
The Fundamental Theorem of Calculus is often claimed as the central theorem of elementary calculus. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. Just select one of the options below to start upgrading. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Proof: Suppose that. Fundamental Theorem of Calculus in Descent Lemma. The Fundamental Theorem of Calculus Part 2 i.
In this section we are going to concentrate on how we actually evaluate definite integrals in practice. Recall that when we talk about an anti-derivative for a function we are really talking about the indefinite integral for the function. This should explain the similarity in the notations for the indefinite and definite integrals. Also notice that we require the function to be continuous in the interval of integration. This was also a requirement in the definition of the definite integral.
Here's how to figure them out. Shed the societal and cultural narratives holding you back and let step-by-step Stewart Calculus textbook solutions reorient your old paradigms. The Fundamental theorem of calculus links these two branches.
Мидж. Ты меня слышишь. От ее слов повеяло ледяным холодом: - Джабба, я выполняю свои должностные обязанности.
Затаив дыхание, Сьюзан дважды щелкнула по конверту. - Северная Дакота, - прошептала она еле слышно. - Посмотрим, кто ты. Сьюзан прочитала открывшееся сообщение, которое состояло из одной строчки, потом прочитала его еще .
Пора.
Беккер толкнул двойную дверь и оказался в некотором подобии кабинета. Там было темно, но он разглядел дорогие восточные ковры и полированное красное дерево. На противоположной стене висело распятие в натуральную величину. Беккер остановился. Тупик.
Мне нужен только ключ. - Какой ключ. Стратмор снова вздохнул. - Тот, который тебе передал Танкадо.
Мне нужно передохнуть хотя бы несколько минут, - подумал. В нескольких милях от этого места человек в очках в железной оправе сидел на заднем сиденье фиата, мчавшегося по проселочной дороге. - Клуб Колдун, - повторил он, напомнив таксисту место назначения. Водитель кивнул, с любопытством разглядывая пассажира в зеркало заднего вида. - Колдун, - пробурчал он себе под нос.
Определенно .
0 Response | 1,382 | 6,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-49 | latest | en | 0.887656 |
http://math.stackexchange.com/questions/39165/continuous-function-on-the-circle-and-real-line | 1,469,421,864,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824204.27/warc/CC-MAIN-20160723071024-00051-ip-10-185-27-174.ec2.internal.warc.gz | 160,222,307 | 17,815 | # Continuous function on the circle and real line
If $\varphi$ is a continuous function defined on the unit circle $T =\{z \in \mathbb{C} : |z| = 1 \}$, then $f (t) = \varphi(e^{it})$ is a continuous periodic function with period $2\pi$. Conversely, if $f$ is a continuous function on the real line of period $2\pi$, there exists a unique continuous function $\varphi$ defined on the unit circle $T$ such that $f (t) = \varphi(e^{it})$. Is my guess correct? Can anyone lead me a proof for this?
-
How continuous is "continuous"? – J. M. May 15 '11 at 6:40
To detail a little bit J.M.'s question, because I wonder too : when you say $\varphi$ is a continuous function defined on the circle, you don't specify its image... what are the domains and codomains of your functions? After that it'll be easier to tell. – Patrick Da Silva May 15 '11 at 7:16
Well in Artin's Algebra there is a Lemma: The continuous homomorphisms $\varphi: \mathbb{R}^{+} \to U_{1}$ are of the form $\varphi(x)=e^{icx}$ for some $c \in \mathbb{R}$. – user9413 May 15 '11 at 7:17
@user10805: you are right. Surely there is a unique function $\phi$ such that $f(t)=\phi(e^{it})$, so you want to know why $\phi$ is continuous. The map $\mathbb{R}\to T$, $t\mapsto e^{it}$, has locally a homeomorphism, so $\phi$ is continuous iff $f$ is. – user8268 May 15 '11 at 7:49
Let $X$ be a topological space and $f\colon \mathbb R \to X$ be continuous and $2\pi$-periodic. Since $f$ is $2\pi$-periodic $\phi\colon T \to X$ is well-defined by $$\phi(e^{it}) = f(t).$$ On the other hand with the condition $$\forall t\in \mathbb R\colon f(t) = \phi(e^{it}),$$ there is only this possible definition, showing the uniqueness. Finally let us prove the continuity of $\phi$. Let $\tau\colon T \to [0,2\pi)$ be defined by $$\tau(e^{it}) = t$$ for an adequate $t\in [0,2\pi)$. Then $\tau$ is certainly continuous (but not homeomorphic). Continuity now follows from $\phi = f \circ \tau$. | 616 | 1,943 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2016-30 | latest | en | 0.844013 |
https://www.nftartrank.com/what-are-the-results-of-wave-action/ | 1,701,946,721,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00476.warc.gz | 1,014,374,748 | 23,973 | # What Are The Results Of Wave Action?
The results of wave action are the patterns that are created as waves of people move through a space.
## Which type of coastal erosion involves destructive waves picking up which material hurling them at the base of a cliff?
Coastal erosion is a type of erosion that involves destructive waves picking up materials hurling them at the base of a cliff.
## How does wave action affect rocky shores?
Wave action affects rocky shores in a variety of ways. The water’s surface tension causes waves to break against the shore, while the height of the waves affects the speed and direction of the waves. This can cause the shore to move, and can also cause the waves to form and break more quickly than if the shore was flat.
## What is hazard and risk?
Hazard is a term used to describe the potential risk of something happening. Risk is the potential probability of something happening.
## What causes rocky shores?
A rocky shore is caused by theroundedness of the earth’s surface. This causes waves to break on the shore, and this in turn makes the shore harder and more resistant to erosion.
## Is when certain types of cliff erode as a result of weak acids in the sea?
The ocean has a large amount of acids and bases, which create cliffs when they react.
## How can a cliff be damaged by the action of waves Brainly?
Cliffs can be damaged by the action of waves because the weight of the waves can break through the cliff’s surface.
## What causes waves in physics?
The cause of waves in physics is the interaction of particles in a system.
## How do waves affect landforms?
Waves can cause landforms to change by breaking the surface of the water, creating waves that can push or pull objects along the shoreline.
## How does hydraulic action occur?
Hydraulic action occurs when fluid is forced through a small opening by the action of a hydraulic pump. The fluid is forced up the tube and into the pump. The fluid pressure inside the pump is greater than the atmospheric pressure outside the pump. This difference in pressure causes the fluid to be forced up the tube and into the pump.
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## Why waves are formed?
A wave is created when the pressure of the water droplets meets the surface of the water. This causes the water droplets to collide, and the pressure of the water droplets causes them to push and pull each other along the surface of the water.
## What are the importance of waves?
Waves are important because they create the sound that we hear. They are also important because they help us understand the ocean.
## What are 4 features formed by wave erosion?
1. Wave erosion creates valleys and ridges on the ocean floor.2. Wave erosion can cause the ocean floor to be covered in sand, shellfish, and other marine life.3. Wave erosion can create channels and canals in the ocean floor.4. Wave erosion can create cliffs and peaks in the ocean floor.
## What type of hazard is wave action?
Wave action is the movement of water in a channel or river.
## What are the effects of waves?
The effects of waves are that they cause a displacement of water or air, an increase in the speed of sound, and a change in the direction of sound.
## What are the impacts of wave erosion?
The impacts of wave erosion depend on the type of wave, the severity of the erosion, and the location of the erosion. Wave erosion can cause the displacement of sand and other material, the creation of dune systems, and the loss of coastline.
## What are the 3 causes of waves?
The three causes of waves are:-The displacement of a particle (vibrations)-The change in the amplitude (strength) of a wave-The change in the frequency (speed) of a wave
## Which of the following are produced by wave erosion?
A. WavesB. SandsC. Sands and waterC. Sands and water
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## What is wave action in geography?
Wave action is the movement of water droplets and molecules in response to the wind.
## What are the four main types of wave action?
A wave is a type of energy that travels in a straight line. The speed of a wave is the measure of its energy. Wave action is the process by which waves interact with each other.
## How do waves affect marine life?
Waves can impact marine life in a number of ways. For example, a wave can cause a surge in water pressure, which can in turn cause marine life to become stranded or stranded animals can be pulled out of the water. Additionally, waves can also cause the release of pollutants from seafood, which can have negative consequences for marine life.
## How does wave refraction affect the coastline?
The waves that travel along the coast are refracted in different ways. Some waves are reflected back and sent towards the shore, while others are sent towards the water. The angle at which the wave is reflected is determined by the height of the wave and the distance from the shore.
## What are the types of hazards?
There are many types of hazards, but some of the most common are: fires, floods, earthquakes, and falls.
## How do organisms deal with wave action?
There is no one answer to this question as organisms vary in their responses to wave action. Some organisms may be more resistant to wave action than others, and some may be more sensitive to wave action. Additionally, organisms may be more or less affected by wave action depending on their location and size.
## What are the results of different coastal processes?
The coastlines of the world are changing. The ocean is warming, and as the water warms, it starts to produce more sediment. The sediments build up in the ocean, and as the water gets more and more sediment, it becomes harder and harder to find food and water. The ocean starts to make more and more noise, and it starts to create more and more pollution.
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## How would you minimize the effect of wave action to the beach shoreline?
There are a few ways to minimize the effect of wave action to the beach shoreline. One way is to create a wave-proof barrier by filling in a large area of the beach with sand and then filling in the space with more sand when the waves reach the barrier. Another way is to place large rocks or trees in the beach area to create a buffer between the shoreline and the waves.
## How does wave action change a coastline?
A coastline is a long, thin strip of land that runs along the coast. The shorelines of oceans and seas are usually made up of several waves that have crashed together. The energy of the waves creates a friction that pulls the water towards the shore.
## What are examples of hazards?
There are many examples of hazards in the world, but some of the most common are falls, burns, accidents, and being in the wrong place at the wrong time.
## Which of the following wave actions is a process of erosion?
The process of erosion is the movement of water over a surface.
## What is wave action in biology?
In biology, wave action is the action of waves in a medium. It is the result of the displacement of particles in a medium by the action of waves.
## What are the causes and effects of waves?
There is no one answer to this question as waves can be caused by both natural and man-made phenomena. Some of the most common causes of waves include wind, water, and seismic activity. Additionally, waves can be affected by temperature, humidity, and other factors. | 1,565 | 7,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-50 | latest | en | 0.947013 |
https://greprepclub.com/forum/many-researchers-have-chipped-away-at-the-edges-of-vera-s-ar-15953.html | 1,590,693,152,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399830.24/warc/CC-MAIN-20200528170840-20200528200840-00333.warc.gz | 375,855,986 | 27,568 | It is currently 28 May 2020, 11:12
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# Many researchers have chipped away at the edges of Vera‘s ar
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Many researchers have chipped away at the edges of Vera‘s ar [#permalink] 02 Dec 2019, 10:25
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Many researchers have chipped away at the edges of Vera‘s argument, (i) ______ his take on the data or his interpretation of historical documents. But no one has directly (ii) ______ the heart of his argument: that we have wildly underestimated the impact that animals, especially large ones, had on the environment. Even Vera‘s critics say they appreciate the debate he has stirred up, if only because it has made them (iii) ______ their convictions.
Blank (i) Blank (ii) Blank (iii)
embracing confronted reaffirm
quibbling with supported reexamine
deliberately overlooking apprehended revivify
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Re: Many researchers have chipped away at the edges of Vera‘s ar [#permalink] 03 Dec 2019, 16:43
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Many researchers have chipped away at the edges of Vera‘s argument, (i) ______ his take on the data or his interpretation of historical documents. But no one has directly (ii) ______ the heart of his argument: that we have wildly underestimated the impact that animals, especially large ones, had on the environment. Even Vera‘s critics say they appreciate the debate he has stirred up, if only because it has made them (iii) ______ their convictions.
Blank (i):
Clues are - "many researchers chipped away" at Vera's argument -> indicating some kind of "attack"
->"quibbling with"
Blank (ii):
Clue - "but" indicating opposite direction. Earlier they were attacking, it now they seem to accept some part of his argument, which is backed by the remainder of the sentence -"that we..."
Also be careful of "no one" -> basically no one argued with him on this front. So though there was a change in the flow, due to the appearance of "no one" we are looking for a word with negative connotation
indicating "confronted" ("supported" - opposite to what we need, "apprehended" meaning understood -> again opposite ...these would have made sense had there been "everyone" instead of "no one")
Blank (iii)
These arguments have initiated some kind of trigger that is making them revisit, analyze their convictions. All options are positive and very close to each other. "revivify" and "reaffirm" are quite strong words, but we dont have any indication that Vera's argument was so strong that these critics would be proselytized.
"reexamine" is the best fist - where the argument is just a trigger for them to revisit their convictions
Re: Many researchers have chipped away at the edges of Vera‘s ar [#permalink] 03 Dec 2019, 16:43
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# Many researchers have chipped away at the edges of Vera‘s ar
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Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®. | 1,001 | 4,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-24 | latest | en | 0.907463 |
https://blogs.mathworks.com/pick/2011/12/02/will-my-flow-be-turbulent/?s_tid=blogs_rc_1 | 1,726,531,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00752.warc.gz | 115,119,687 | 26,124 | # Will my flow be turbulent?
Brett's Pick this week is Moody, by Tom Davis.
I am not a mechanical engineer, though I did study fluid mechanics in my biomedical engineering past. In fluid flows, the Reynolds number is a dimensionless parameter that describes the ratio of inertial to viscous forces. This relationship is significant because, among other things, it can be used to predict the nature of the flow--whether laminar or turbulent. In biomechanics, fully developed fluid flows are typically of low Reynolds numbers, certainly below the critical threshold at which flow becomes turbulent.
However, in classical mechanics and hydraulics, flows are typically through long, often rigid, pipes. Flows become "fully developed," and can be of much higher Reynolds numbers. In these regimes, assuming one knows the characteristics of the pipe, one can calculate the Darcy friction factor to determine the (pressure) head loss during flow.
Okay, geeking out a bit. The relevance here is that the interplay of Reynolds numbers and Darcy factors can be difficult to deal with; Moody diagrams allow us to relate the two graphically, and to see at a glance whether, given a Darcy Factor and a Reynolds number, flow will be laminar or turbulent for pipes of a specified roughness.
Creating a Moody diagram is no trivial task, though. Or at least, it wasn't until Tom shared his MATLAB code for creating one. Function 'Moody' allows you to specify units (SI or Imperial) and paper size (A4 or Letter), and the name of an output file, and it will create a beautiful Moody diagram for you!
Thanks for sharing that, Tom! As always, comments to this blog post are welcome. Or leave a comment for Tom here.
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To leave a comment, please click here to sign in to your MathWorks Account or create a new one. | 394 | 1,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.926227 |
https://aboutcivil.org/strength-of-materials.html | 1,521,392,981,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645830.10/warc/CC-MAIN-20180318165408-20180318185408-00090.warc.gz | 513,950,042 | 6,139 | Strength of Materials
Analysis of internal forces in bodies
This procedure consists of finding out the effect of:
2. Twisting
3. Bending
4. Combinations of axial, twisting and bending loadings.
Definition
It is the combination of physical, mathematical, and computer laws and techniques to predict the behavior of a solid materials that are subjected to mechanical or thermal loadings. It is the branch of mechanics that deals with the behavior of solid matter under external actions. The external actions may be:
• External Force
• Temperature Change
• Displacement
Applications of Solid Mechanics
This field has a wide range of applications, laws and concepts of solid mechanics are used:
• In Civil Engineering to design foundations and structures
• In Geo-Mechanics to model shape of planets, tectonics and predict earthquakes
• In Mechanical Engineering to design load bearing components for vehicles, power generation and transmission
SOME IMPORTANT TERMINOLOGIES
Stress
When an external force is applied on a body, it undergoes deformation which is resisted by the body. The magnitude of the resisting force is numerically equal to the applied force. This internal resisting force per unit area of the body is known as stress.
Stress = Resistive Force/Area
In equation form: s = P/A,
Units are: N/m2, kN/m2 MPa (Mega Pascal), Psi (lb/in2), psf (lb/ft2), Ksi (kips/in2), ksf (kips/ft2)
Strain
When a body is subjected to an external force, there is some change of dimension in the body. Numerically the strain is equal to the ratio of change in length to the original length of the body.
Strain = Change in length/Original length
In equation form: e = dL/L
Units: m/m, mm/m, In/in, in/ft
Shear Stress(t) and Shear Strain
The two equal and opposite forces act tangentially on any cross sectional plane of the body tending to slide one part of the body over the other part.The stress induced is called shear stress and the corresponding strain is known as shear strain.
Elastic Limit:
The maximum stress that can be applied to a metal without producing permanent deformation is known as Elastic Limit. When stress is applied on a body its dimensions change, these changes can be reversed if the stress applied do not cross a certain limit. This certain limit within which the material when unloaded will re-gain its original dimensions is known as Elastic Limit. Beyond the elastic limit the changes will be permanent and cannot be reversed without an external force. Brittle materials tend to break at or shortly past their elastic limit, while ductile materials deform at stress levels beyond their elastic limit.
Hooke’s law
This law states that when a material is loaded, within its elastic limit, the stress is directly proportional to the strain.
To view animation of Stress Strain Relationship Click Here
s = Ee
E = s/e
Its unit is same as that of Stress
Where,
E is Young’s modulus
s is Stress
e is Strain
Yield Point or Yield Stress
It is the lowest stress in a material at which the material begins to exhibit plastic properties. Beyond this point an increase in strain occurs without an increase in stress which is called Yielding.
Ultimate Strength
It is the maximum stress that a material can withstand while being stretched or pulled before necking
Strain Hardening
It is the strengthening of a metal by plastic deformation because of dislocation(irregular) movements within the crystal structure of the material. Any material with a reasonably high melting point such as metals and alloys can be strengthened by this method. | 762 | 3,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-13 | latest | en | 0.901742 |
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# C3 graph with a constant Watch
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1. can anyone help with part c please? I am not sure if a is positive or negative so do I just assume a is 0? Thanks Attachment 584256584258
Attached Images
2. expand it first
you should be able to solve for the intersection and roots once you do so
3. (Original post by coconut64)
can anyone help with part c please? I am not sure if a is positive or negative so do I just assume a is 0? Thanks
You can't assume it's zero.
In this case "a" cannot be negative, otherwise you'll be raising a negative number to a real power - what's - which can't be sketched in a standard graph as it involves complex numbers.
Note that if a >1, the exponential will be increasing.
If a < 1 then it's decreasing.
And if a=1 ?
Can you take it from there.
4. (Original post by anonmouse221)
expand it first
you should be able to solve for the intersection and roots once you do so
wHAT DO i EXPAND?
5. (Original post by ghostwalker)
You can't assume it's zero.
In this case "a" cannot be negative, otherwise you'll be raising a negative number to a real power - what's - which can't be sketched in a standard graph as it involves complex numbers.
Note that if a >1, the exponential will be increasing.
If a < 1 then it's decreasing.
And if a=1 ?
Can you take it from there.
if a is 1 then I will get 1/x +1 and 1+1 for a^x+a...
6. (Original post by coconut64)
if a is 1 then I will get 1/x +1 and 1+1 for a^x+a...
Confusing. How can you get two different expressions when you set a=1, into part c?
However, if a=1, then part c reduces to 1+1, and so, f(x)=2, i.e. constant.
7. (Original post by ghostwalker)
Confusing. How can you get two different expressions when you set a=1, into part c?
However, if a=1, then part c reduces to 1+1, and so, f(x)=2, i.e. constant.
I am slightly lost too. There are 3 parts in part c and each part requires you to draw a graph. I don't get what you mean by f(x)=2 , so it is just a straight line?? Thanks
8. (Original post by coconut64)
I am slightly lost too. There are 3 parts in part c and each part requires you to draw a graph. I don't get what you mean by f(x)=2 , so it is just a straight line?? Thanks
Sorry, I thought you were just looking at the last part of part C, which is also called (c).
This is a slightly unfair question, IMO, as the shape of the graph depends on the value of "a", If a >1, then its increasing, if a < 1 it's decreasing, and if a=1, it's a constant - i.e. just a straight line at y=2.
9. (Original post by ghostwalker)
Sorry, I thought you were just looking at the last part of part C, which is also called (c).
This is a slightly unfair question, IMO, as the shape of the graph depends on the value of "a", If a >1, then its increasing, if a < 1 it's decreasing, and if a=1, it's a constant - i.e. just a straight line at y=2.
Sorry about the confusion! So what do I set a as ? It says a is just an arbitrary constant....
10. (Original post by coconut64)
Sorry about the confusion! So what do I set a as ? It says a is just an arbitrary constant....
It's arbitrary so, you can't set it as anything. I'd suggest drawing three graphs, one for each of the different scenarios.
11. (Original post by ghostwalker)
It's arbitrary so, you can't set it as anything. I'd suggest drawing three graphs, one for each of the different scenarios.
thanks
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https://www.cnclathing.com/guide/guide-to-cutting-speeds-and-feeds-definition-selection-calculation-and-more-cnclathing | 1,716,192,629,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00224.warc.gz | 615,992,486 | 21,693 | > > Guide to Cutting Speeds and Feeds: Definition, Selection, Calculation and More | CNCLATHING
> > Guide to Cutting Speeds and Feeds: Definition, Selection, Calculation and More | CNCLATHING
Guide to Cutting Speeds and Feeds: Definition, Selection, Calculation and More | CNCLATHING
2020.6.17
Proper cutting speeds and feeds setting are the basis of efficient operation on CNC machines. In this article, let’s learn about speeds and feeds definition, selection, calculation and more. Check out how to determine cutting speed and feed rate.
What are Speeds and Feeds? - Difference Between Cutting Speed and Feed
Speeds and feeds refer to two separate velocities – cutting speed and feed rate in machining, but they have a combined effect on the cutting process, determine the rate of material removal, surface finish, and power requirements.
Cutting speed, also called surface speed, is the relative velocity between the cutting tool and the surface of the workpiece is being cut. The feed rate is the relative velocity at which the cutter is advanced along the workpiece, or defined as the distance of tool travels during one spindle revolution.
Cutting Speed Factors & How to Determine Cutting Speed?
Cutting speed may also be defined as the rate at the part surface, that is, how fast the material moves past the cutting edge of the tool. For different operations, the definition of surface may vary. In drilling and milling, the surface is the outside diameter of the tool. In turning and boring, the surface can be defined on either side of the depth of cut (the starting surface or end surface). Cutting speeds are expressed in feet per minute (imperial) or meters per minute (metric).
The material to be machined and cutting tool materials will have a large impact on the calculation of cutting speed. The depth of the cut and the feed rate will also affect the speed, but not to as great as the component hardness. The optimum cutting speed will be different if the types of material and machining conditions are changed. The softer the material of workpiece, the higher the cutting speed, when the cutting tool material is stronger, the cutting speed increases.
On a rotary tool like a drill bit or milling cutter, cutting speed is equal to how fast its periphery spins relative to the stock clamped to the worktable, it’s the same with the tools used on lathes, except their cutting speeds are measured by how fast the rotary material moves past the edge of the turning tool. What’s the recommended cutting speed for different materials? Check out Cutting Speed Chart for turning, drilling and more CNC machining operations when processing a wide range of metals.
Feed Rate Selection - How to Determine Feed Rate on Lathe
The feed of a lathe is determined by the speed of the lead screw or feed rod. The speed is controlled by the change gears in the quick change gearbox. For rough cutting, the goal is only to remove excess materials from the stock, a coarse feed should be applied. For finishing cut, a good surface finish is important and the process should complete the size of the machining parts, a fine feed should be adopted. For general CNC turning using an HSS cutting tool, a feed rate of .005 – .020 inches per revolution for roughing and a .002 – .004 inches per revolution for finishing are recommended.
How Do You Calculate Cutting Speed and Feed?
When calculating the cutting speed in meters per minute, the spindle speed of the machine (n) and diameter of the part must be known. When calculating the feed rate, spindle speed, feed per tooth and the number of flutes should be known.
Cutting Speed and Feed Rate Formula
Cutting Speed: Vc = (π* D * n)/1000
Feed: Vf = n * fz * Z
Vc: cutting speed (m/min)
π: 3.14
D: diameter (mm)
n: spindle speed (min-1)
Vf = feed (mm/min)
fz = feed per tooth (mm/tooth)
Z = number of flutes
Setting Speeds on a Lathe Machine
Too high cutting speed will make the tool more easily damaged, while too low speed may be lost lots of time and reduce production efficiency. How to set the cutting speed on a lathe?
– When speeds measured in revolutions per minute: changed by stepped pulleys or gear levers
– For belt-driven lathe: speeds obtained by changing flat belt and back gear drive.
– For geared-head lathe: speeds changed by moving speed levers into proper positions according to r/min chart fastened to headstock. But don’t change the speeds when the machine is running. | 963 | 4,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.932506 |
http://www.newton.dep.anl.gov/askasci/env99/env235.htm | 1,419,168,435,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802771253.22/warc/CC-MAIN-20141217075251-00124-ip-10-231-17-201.ec2.internal.warc.gz | 715,128,673 | 3,853 | Reading a Seismograph ```Name: Jason C. Status: Student Age: 13 Location: N/A Country: N/A Date: January 27, 2004 ``` Question: How do you interpret data collected from a seismograph into the Richter scale? Replies: The concept of the Richter scale is relatively straightforward, but the actual application is complicated and depends on the specific characteristics of the seismograph. The original equation used by Professor Richter in 1935 said that the magnitude of an earthquake was equal to the base 10 logarithm of the ground motion in millimeters measured on a certain type of seismograph plus a correction factor related to the distance of the earthquake. The distance is calculated from the difference in arrival time for different types of waves that travel at different speeds. So, for a constant distance between an earthquake's hypocenter and the seismograph, the ground motion has to increase by a factor of 10 to cause an increase of 1 on the Richter scale. Professor Richter was trying to characterize the energy of a seismic event, not damage. So, going from a 2.0 to a 4.0 magnitude event does not imply twice the damage. In fact, a rule of thumb is that the energy released increases by a factor of 30 for each 1.0 increase in magnitude. So, compared to a magnitude 3 event, a magnitude 8 earthquake releases 30*30*30*30*30 or 24.3 million times more energy and causes the amplitude (size) of the ground motions to be 100,000 times greater. The energy in a magnitude 8 earthquake is equivalent to 1 billion tons of TNT, or 30 jumbo thermonuclear weapons. The original Richter scale was based on just a few instruments of a certain type and for Southern California geologic conditions. In the seven decades since it was created, many adjustments have been made for different regions, wave types, and instruments. Also, newer measurements, referred to as the "seismic moment" and "moment magnitude", have been developed to address some of the Richter scale's shortcomings. Andy Johnson Click here to return to the Environmental and Earth Science Archives
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Update: June 2012 | 570 | 2,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2014-52 | latest | en | 0.93983 |
https://dcodesnippet.com/what-was-the-date-45-days-ago/ | 1,713,890,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00542.warc.gz | 177,326,582 | 18,523 | How To Calculate The Date 45 Days Ago – Easy Methods And Tips
//
Thomas
Learn how to the date 45 days ago using manual methods or a tool. Consider and with different lengths for accurate results.
Determining the Date 45 Days Ago
Calculating the Date
Determining the date 45 days ago can be done using various methods. One way is to manually the date by subtracting 45 days from the current date. Another method is to use a or tool, which can provide an accurate and quick result. Additionally, it is important to take into account and with different lengths to ensure the accuracy of the calculation.
Using a Calendar or Date Calculator
A convenient way to determine the date 45 days ago is by using a or . These tools are designed to simplify date calculations and provide accurate results. By inputting the current date and subtracting 45 days, the calculator will automatically generate the date 45 days prior. This method eliminates the need for manual calculations and reduces the chances of errors.
Subtracting 45 Days from the Current Date
If you prefer a manual approach, you can subtract 45 days from the current date to determine the date 45 days ago. Start by identifying the current date and then subtracting 45 days from it. For example, if the current date is June 1st, subtracting 45 days would result in April 17th. This method requires a basic understanding of arithmetic and can be used without the need for external tools.
Considering Leap Years and Months with Different Lengths
When determining the date 45 days ago, it is essential to consider and with different lengths. Leap years occur every four years and add an extra day to the month of February. This means that if the date 45 days ago falls within a leap year, the calculation needs to account for the additional day.
Months also vary in length, with some having 30 days and others having 31. To accurately determine the date 45 days ago, it is important to consider the specific month in question. For example, if the current date is February 15th, subtracting 45 days would take you back to January 1st, considering the shorter length of February.
By taking into account and with different lengths, you can ensure the accuracy of your calculation when determining the date 45 days ago. Whether using a or , or manually subtracting days, considering these factors will result in an accurate and reliable date.
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+1 803-820-9654 | 523 | 2,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-18 | latest | en | 0.910808 |
https://www.shvoong.com/how-to-optimize-qr-codes-for-display-size-and-scanning-distance/ | 1,653,224,724,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00034.warc.gz | 1,129,651,796 | 27,415 | # How to Optimize QR Codes for Display Size and Scanning Distance
Img Source - WeareTeachers
QR codes are a quick and error-free way to market your brand or products. You can use them in multiple ways to share content with your target audience. If you are planning to use QR codes for marketing purposes, this article will tell you how you can optimise your codes with respect to scanning distance and display size.
## The Connection Between QR Codes and Scanning Distance
When designing a QR code, you must consider the distance between the QR code and the consumer, says Create QR Code. People try to minimise the size of QR codes, so that they do not take a lot of space on a banner, or magazine page. However, it should be done without compromising the scannability of the code.
Often marketers are asked to fit a QR code in a small space. So they end up designing a code that is too small to be scanned, according to SmallBizTrends. If you are planning to design a QR code, you can avoid this issue by considering a few factors, including the colour, placement, and density of the QR code.
If you design a QR code for a magazine, and then display the same code on the TV as well, people will have to leave their couch to scan a small QR code, which makes the code inefficient. A QR code for magazines is scanned from a short distance, so it should be small in size. On the other hand, a QR code that is to be displayed on a TV screen should be large, so that people don’t have to leave their couch to reach the TV and scan the code.
## Variables to Consider When Designing a QR Code
QR code density, scanning distance, and display size are the important variables you need to remember when generating a QR code. The scanning distance has a direct relationship with code density and size.
• Low-density codes are small in size, because their modules do not merge together. If you reduce the QR code density while keeping its area the same, you can increase the functional distance.
• High-density codes need more area, because they have several modules which can merge together if they are not provided with proper spacing. If you increase the density of a QR code while keeping the display area constant, you can reduce the functional distance.
• The density of a QR code depends on the length of a URL. The longer the URL, the higher the density of the code will be. It is calculated by multiplying the characters of a URL with the error correction value.
## Optimization of a QR Code
Some QR code generating tools automatically optimise the density of the code, while others require manual optimization. As mentioned earlier, your URL’s number of characters determines the code density. The number of URL characters depends on different elements, including tracking parameters, keywords, and domain.
There are total four error correction values: 7%, 15%, 25%, and 30% (source). Only one of these values is used for one QR code. When the percentage of error correction is increased, the value of the density also increases. If you want to decrease the size of a code, reduce the number of URL characters to get better results. | 655 | 3,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | longest | en | 0.925247 |
https://www.bartleby.com/solution-answer/chapter-13-problem-54p-structural-analysis-6th-edition/9781337630931/solve-problem-1339-for-the-loading-shown-in-fig-p1339-and-the-support-settlements-of-10-mm-at-a/2707c406-ac76-11e9-8385-02ee952b546e | 1,586,352,146,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371813538.73/warc/CC-MAIN-20200408104113-20200408134613-00228.warc.gz | 806,401,088 | 53,672 | # Solve Problem 13.39 for the loading shown in Fig. P13.39 and the support settlements of 10 mm at A , 65 mm at C , 40 mm at E , and 25 mm at G . 13.37 through 13.45 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.37–P13.45 using the method of consistent deformations. FIG.P13.39, P13.54
### Structural Analysis
6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter
Section
### Structural Analysis
6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 13, Problem 54P
Textbook Problem
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## Solve Problem 13.39 for the loading shown in Fig. P13.39 and the support settlements of 10 mm at A, 65 mm at C, 40 mm at E, and 25 mm at G.13.37 through 13.45 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.37–P13.45 using the method of consistent deformations.FIG.P13.39, P13.54
To determine
Calculate the support reactions for the given beam using method of consistent deformation.
Sketch the shear and bending moment diagrams for the given beam.
### Explanation of Solution
Given information:
The settlement at A (ΔA) is 10mm=0.01m.
The settlement at C (ΔC) is 65mm=0.065m.
The settlement at E (ΔE) is 40mm=0.04m.
The settlement at G (ΔG) is 25mm=0.025m.
The moment of inertia (I) is 500×106mm4.
The modulus of elasticity (E) is 200 GPa.
The structure is given in the Figure.
Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.
• For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
• For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
• For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as negative and the counter clockwise moment as positive.
Calculation:
Sketch the free body diagram of the structure as shown in Figure 1.
Calculate the degree of indeterminacy of the structure:
Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.
The beam is supported by 5 support reactions and the number of equilibrium equations is 3.
Therefore, the degree of indeterminacy of the beam is i=2.
Select the bending moments at the supports E and C as redundant.
Let θCL and θCR are the slopes at the ends C of the left and right spans of the primary beam due to external loading.
Let θEL and θER are the slopes at the ends E of the left and right spans of the primary beam due to external loading.
Let fCCL and fCCR are the flexibility coefficient representing the slopes at C of the left and right spans due to unit value of redundant MC.
Let fEEL and fEER are the flexibility coefficient representing the slopes at E of the left and right spans due to unit value of redundant ME.
Let fCE be the flexibility coefficient representing the slope at point C of the primary beam due to unit value at point E.
Use beam deflection formulas:
Calculate the value of θCL using the relation:
θCL=Pa(L2a2)6EIL
Substitute 120kN for P, 6m for a, and 10m for L.
θCL=120×6(10262)6EI(10)=768kNm2EI
Substitute 500×106mm4 for E and 200 GPa for E.
Calculate the value of θCR using the relation:
θCR=Pb(L2b2)6EIL
Substitute 120kN for P, 4m for b, 2I for I, and 10m for L.
θCR=120×4(10242)6E(2I)(10)=336kNm2EI
Substitute 500×106mm4 for E and 200 GPa for E.
Calculate the value of θEL using the relation:
θEL=Pa(L2a2)6EIL
Substitute 120kN for P, 6m for a, 2I for I, and 10m for L.
θEL=120×6(10262)6E(2I)(10)=384kNm2EI
Substitute 500×106mm4 for E and 200 GPa for E.
Calculate the value of θER using the relation:
θER=PL216EI
Substitute 150kN for P and 8m for L.
θER=150(8)216EI=600kNm2EI
Substitute 500×106mm4 for E and 200 GPa for E.
Calculate the value of fCCL using the relation:
fCCL=ML3EI
Substitute 1kNm for M and 10 m for L.
fCCL=1(10)3EI=10kNm2/kNm3EI
Substitute 500×106mm4 for E and 200 GPa for E.
Calculate the value of fCCR using the relation:
fCCR=ML6EI
Substitute 1kNm for M and 10 m for L.
fCCL=1(10)6EI=10kNm2/kNm6EI
Substitute 500×106mm4 for E and 200 GPa for E.
Calculate the value of fCE and fEC using the relation:
fCE=fEC=ML6EI
Substitute 1kNm for M, 2I for I, and 10 m for L.
fCE=fEC=1(10)6E(2I)=10kNm2/kNm12EI
Substitute 500×106mm4 for E and 200 GPa for E.
Calculate the value of fEEL using the relation:
fEEL=ML6EI
Substitute 1kNm for M and 10 m for L.
fEEL=1(10)6EI=10kNm2/kNm6EI
Substitute 500×106mm4 for E and 200 GPa for E.
fEEL=10kNm2/kNm6×200GPa×106kN/m21GPa×500×106mm4×(1m1,000mm)4=0
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A+ Guide to Hardware (Standalone Book) (MindTap Course List) | 1,813 | 6,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-16 | longest | en | 0.838211 |
https://sage-answer.com/how-many-atoms-of-chlorine-are-in-ccl4/ | 1,713,600,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00600.warc.gz | 451,078,729 | 42,986 | # How many atoms of chlorine are in CCl4?
## How many atoms of chlorine are in CCl4?
four chlorine atoms
A molecule of carbon tetrachloride contains one carbon and four chlorine atoms.
## How many chlorine atoms would be in a compound with the name tetrachloride?
Give the formula for each molecule. The name carbon tetrachloride implies one carbon atom and four chlorine atoms, so the formula is CCl 4.
How many atoms are in the molecule CCl4?
Properties. In the carbon tetrachloride molecule, four chlorine atoms are positioned symmetrically as corners in a tetrahedral configuration joined to a central carbon atom by single covalent bonds.
How many atoms are in silicon tetrachloride?
3.1Computed Properties
Property Name Property Value Reference
Heavy Atom Count 5 Computed by PubChem
Formal Charge 0 Computed by PubChem
Complexity 19.1 Computed by Cactvs 3.4.8.18 (PubChem release 2021.05.07)
Isotope Atom Count 0 Computed by PubChem
### How many moles of chlorine atoms are in one mole of CCl4?
Answer: You can see that one mole of carbon tetrachloride contains one mole of carbon (and 4 moles of chlorine). So 2.98 moles will contain 2.98 moles of carbon.
### How many chlorine atoms are in PCl5?
Phosphorus Pentachloride or PCl5 is a compound formed by chemical elements Phosphorus (Atomic number: 15, symbol: P) and Chlorine (Atomic number: 17, symbol: Cl). A Phosphorus Pentachloride molecule consists of 1 atom of phosphorus for 5 atoms of chlorine.
How many chlorine atoms are in each set?
Chemical formula is the representation of the number of atoms present in the molecule or molecular compound . Therefore, three carbon tetrachloride has = 12 chlorine atoms.
What is the molecular formula for chlorine?
Cl-
Chloride/Formula
#### What is the formula of silicon tetrachloride?
SiCl4
Silicon tetrachloride/Formula
#### How many chlorine atoms are there in 2 moles of chlorine?
How many moles of carbon atoms in chlorine atoms does it take to make 1 mole of carbon tetrachloride molecules?
You can see that one mole of carbon tetrachloride contains one mole of carbon (and 4 moles of chlorine).
How many atoms are in n2o5?
2 Nitrogen atoms
N₂O₅ is named Dinitrogen Pentoxide because it has 2 Nitrogen atoms and 5 Oxygen atoms.
## How many atoms of chlorine are in a tetrachloride molecule?
In each molecule of carbon tetrachloride there are four atoms of chlorine. The prefix tetra- means ‘four’. Q: How many atoms of chlorine are in the carbon tetrachloride molecule?
## How many atoms of chlorine are in a carbon molecule?
How many atoms of chlorine are in the carbon tetrachloride molecule? In each molecule of carbon tetrachloride there are four atoms of chlorine. The prefix tetra- means ‘four’.
What kind of chemical is silicon tetrachloride?
Silicon tetrachloride is a colorless, fuming liquid with a pungent odor. It is decomposed by water to hydrochloric acid with evolution of heat. It is corrosive to metals and tissue in the presence of moisture. It is used in smoke screens, to make various silicon containing chemicals, and in chemical analysis. | 778 | 3,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.89377 |
http://english.turkcebilgi.com/circumradius | 1,394,452,283,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010776091/warc/CC-MAIN-20140305091256-00045-ip-10-183-142-35.ec2.internal.warc.gz | 63,397,044 | 10,759 | In geometry, the circumscribed circle or circumcircle of a polygon is a circle which passes through all the vertices of the polygon. The centre of this circle is called the circumcenter.
A polygon which has a circumscribed circle is called a cyclic polygon. All regular simple polygons, all triangles and all rectangles are cyclic.
A related notion is the one of a minimum bounding circle, which is the smallest circle that completely contains the polygon within it. Not every polygon has a circumscribed circle, as the vertices of a polygon do not need to all lie on a circle. Yet any polygon has a unique minimum bounding circle, which may be constructed by a linear time algorithm.[1] Even if a polygon has a circumscribed circle, it may not coincide with its minimum bounding circle; for example, for an obtuse triangle, the minimum bounding circle has the hypotenuse as diameter and does not pass through the opposite vertex.
## Circumcircles of triangles
All triangles are cyclic, i.e. every triangle has a circumscribed circle.
The circumcenter of a triangle can be found as the intersection of the three perpendicular bisectors. (A perpendicular bisector is a line that forms a right angle with one of the triangle's sides and intersects that side at its midpoint.) This is because the circumcenter is equidistant from any pair of the triangle's points, and all points on the perpendicular bisectors are equidistant from those points of the triangle.
In coastal navigation, a triangle's circumcircle is sometimes used as a way of obtaining a position line using a sextant when no compass is available. The horizontal angle between two landmarks defines the circumcircle upon which the observer lies.
The circumcentre's position depends on the type of triangle:
• If and only if a triangle is acute (all angles smaller than a right angle), the circumcenter lies inside the triangle
• If and only if it is obtuse (has one angle bigger than a right angle), the circumcenter lies outside
• If and only if it is a right triangle, the circumcenter lies on one of its sides (namely, the hypotenuse). This is one form of Thales' theorem.
The circumcentre of an acute triangle is inside the triangle The circumcentre of a right triangle is on the hypotenuse The circumcentre of an obtuse triangle is outside the triangle
The diameter of the circumcircle can be computed as the length of any side of the triangle, divided by the sine of the opposite angle. (As a consequence of the law of sines, it doesn't matter which side is taken: the result will be the same.) The triangle's nine-point circle has half the diameter of the circumcircle.
In any given triangle, the circumcenter is always collinear with the centroid and orthocenter. The line that passes through all of them is known as the Euler line.
The isogonal conjugate of the circumcenter is the orthocenter.
The useful minimum bounding circle of three points is defined either by the circumcircle (where three points are on the minimum bounding circle) or by the two points of the longest side of the triangle (where the two points define a diameter of the circle.). It is common to confuse the minimum bounding circle with the circumcircle.
The circumcircle of three collinear points is an infinitely large circle. Nearly collinear points often cause problems and errors in computation of the circumcircle.
Circumcircles of triangles have an intimate relationship with the Delaunay triangulation of a set of points.
### Circumcircle equations
The circumcircle is given in Cartesian coordinates by the equation
where A, B and C are the vertices of the triangle, and the solution for v is the circumcircle. (Note A2 = Ax2 + Ay2.)
Given
, , ,
we then have av2 − 2Svb = 0 and, assuming the three points were not in a line (otherwise the circumcircle is that line that can also be seen as a generalized circle with S at infinity), (vS/a)2 = b/a + S2/a2, giving the circumcenter S/a and the circumradius √ (b/a + S2/a2). This approach should also work for the circumsphere of a tetrahedron.
An equation for the circumcircle in trilinear coordinates x : y : z is a/x + b/y + c/z = 0. An equation for the circumcircle in barycentric coordinates x : y : z is 1/x + 1/y + 1/z = 0.
The isogonal conjugate of the circumcircle is the line at infinity, given in trilinear coordinates by ax + by + cz = 0 and in barycentric coordinates by x + y + z = 0.
### Coordinates of circumcenter
The circumcenter has trilinear coordinates (cos , cos , cos ) where are the angles of the triangle. The circumcenter has barycentric coordinates
where are edge lengths ( respectively) of the triangle.
### The angles at which the circle meets the sides
The angles at which the circumscribed circle meet the sides of the triangle coincide with angles at which sides meet each other. The side opposite angle α meets the circle twice: once at each end; in each case at angle α (similarly for the other two angles).
## Triangle centers on the circumcircle of triangle ABC
In this section, the vertex angles are labeled A, B, C and all coordinates are trilinear coordinates:
• Steiner point = bc/ (b2c2) : ca/ (c2a2) : ab/(a2b2) = the nonvertex point of intersection of the circumcircle with the Steiner ellipse. (The Steiner ellipse, with center = centroid(ABC), is the ellipse of least area that passes through A, B, and C. An equation for this ellipse is 1/(ax) + 1/(by) + 1/(cz) = 0.)
• Tarry point = sec (A + ω) : sec (B + ω) : sec (C + ω) = antipode of the Steiner point
• Focus of the Kiepert parabola = csc (BC) : csc (CA) : csc (AB)
Quadrilaterals that can be circumscribed have particular properties including the fact that opposite angles are supplementary angles (adding up to 180° or π radians).
## References
• ^ Megiddo, N. (1983). "Linear-time algorithms for linear programming in R3 and related problems". SIAM Journal on Computing 12: 759–776.
• Kimberling, Clark (1998). "Triangle centers and central triangles". Congressus Numerantium 129: i–xxv, 1–295.
Geometry (Greek γεωμετρία; geo = earth, metria = measure) is a part of mathematics concerned with questions of size, shape, and relative position of figures and with properties of space. Geometry is one of the oldest sciences.
POLYGONE is an Electronic Warfare Tactics Range located on the border between France and Germany. It is one of only two in Europe, the other being RAF Spadeadam.
The range, also referred to as the Multi-national Aircrew Electronic Warfare Tactics Facility (MAEWTF), is
In geometry, the centre (or center, in American English) of an object is a point in some sense in the middle of the object. If geometry is regarded as the study of isometry groups then the centre is a fixed point of the isometries.
simple polygon is a polygon whose sides do not intersect. They are also called Jordan polygons, because the Jordan curve theorem can be used to prove that such a polygon divides the plane into two regions, the region inside it and the region outside it.
A triangle is one of the basic shapes of geometry: a polygon with three corners or and three sides or edges which are straight line segments.
In Euclidean geometry any three non-collinear points determine a triangle and a unique plane, i.e.
rectangle is defined as a quadrilateral where all four of its angles are right angles.
From this definition, it follows that a rectangle has two pairs of parallel sides; that is, a rectangle is a parallelogram.
In computational complexity theory, an algorithm is said to take linear time, or O(n) time, if the asymptotic upper bound for the time it requires is proportional to the size of the input, which is usually denoted n.
A triangle is one of the basic shapes of geometry: a polygon with three corners or and three sides or edges which are straight line segments.
In Euclidean geometry any three non-collinear points determine a triangle and a unique plane, i.e.
perpendicular (or orthogonal) to each other if they form congruent adjacent angles. The term may be used as a noun or adjective. Thus, referring to Figure 1, the line AB is the perpendicular to CD through the point B.
Bisection is the general activity of dividing something into two parts.
Bisect may mean:
• bisection, in geometry, dividing something into two equal parts
• bisect (philately), the use of postage stamp halves
Pilotage is the use of fixed visual references on the ground or sea by means of sight or radar to guide oneself to a destination, sometimes with the help of a map or nautical chart. People use pilotage for activities such as guiding vessels and aircraft, hiking and Scuba diving.
A position line is a line that can be identified both on a nautical chart or aeronautical chart and by observation out on the surface of the earth. The intersection of two position lines is a fix that used in position fixing to identify the navigator's location.
For the history and development of the sextant see Reflecting instruments
A sextant
COMPASS is an acronym for COMPrehensive ASSembler. COMPASS is a macro assembly language on Control Data Corporation's 3000 series, and on the 60-bit CDC 6000 series, 7600 and
If and only if, in logic and fields that rely on it such as mathematics and philosophy, is a logical connective between statements which means that the truth of either one of the statements
hypotenuse of a right triangle is the triangle's longest side; the side opposite the right angle. The length of the hypotenuse of a right triangle can be found using the Pythagorean theorem, which states that the square of the length of the hypotenuse equals the sum of the squares
Thales' theorem (named after Thales of Miletus) states that if A, B and C are points on a circle where the line AC is a diameter of the circle, then the angle ABC is a right angle.
diameter (Greek words diairo = divide and metro = measure) of a circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circle. The diameters are the longest chords of the circle.
trigonometric functions (also called circular functions) are functions of an angle. They are important in the study of triangles and modeling periodic phenomena, among many other applications.
angle (in full, plane angle) is the figure formed by two rays sharing a common endpoint, called the vertex of the angle. The magnitude of the angle is the "amount of rotation" that separates the two rays, and can be measured by considering the length of circular arc swept
law of sines (or sine law, sine formula) is a statement about arbitrary triangles in the plane. If the sides of the triangle are a, b and c and the angles opposite to those sides are A, B and C
In geometry, the nine-point circle is a circle that can be constructed for any given triangle. It is so named because it passes through nine significant points, six lying on the triangle itself (unless the triangle is obtuse).
In geometry, the centroid or barycenter of an object in -dimensional space is the intersection of all hyperplanes that divide into two parts of equal moment about the hyperplane. Informally, it is the "average" of all points of .
In geometry, an altitude of a triangle is a straight line through a vertex and perpendicular to (i.e. forming a right angle with) the opposite side or an extension of the opposite side.
Euler line, named after Leonhard Euler, is a line determined from any triangle that is not equilateral; it passes through several important points determined from the triangle. In the image, the Euler line is shown in red.
In geometry, the isogonal conjugate of a point P with respect to a triangle ABC is constructed by reflecting the lines PA, PB, and PC about the angle bisectors of A, B, and C. | 2,646 | 11,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2014-10 | longest | en | 0.923603 |
http://en.wikipedia.org/wiki/Johnson-Nyquist_noise | 1,432,329,895,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207926736.56/warc/CC-MAIN-20150521113206-00249-ip-10-180-206-219.ec2.internal.warc.gz | 81,839,311 | 19,856 | Johnson–Nyquist noise
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These three circuits are all equivalent: (A) A resistor at nonzero temperature, which has Johnson noise; (B) A noiseless resistor in series with a noise-creating voltage source (i.e. the Thévenin equivalent circuit); (C) A noiseless resistance in parallel with a noise-creating current source (i.e. the Norton equivalent circuit).
Johnson–Nyquist noise (thermal noise, Johnson noise, or Nyquist noise) is the electronic noise generated by the thermal agitation of the charge carriers (usually the electrons) inside an electrical conductor at equilibrium, which happens regardless of any applied voltage. The generic, statistical physical derivation of this noise is called the fluctuation-dissipation theorem, where generalized impedance or generalized susceptibility is used to characterize the medium.
Thermal noise in an ideal resistor is approximately white, meaning that the power spectral density is nearly constant throughout the frequency spectrum (however see the section below on extremely high frequencies). When limited to a finite bandwidth, thermal noise has a nearly Gaussian amplitude distribution.[1]
History
This type of noise was first measured by John B. Johnson at Bell Labs in 1926.[2][3] He described his findings to Harry Nyquist, also at Bell Labs, who was able to explain the results.[4]
Noise voltage and power
Thermal noise is distinct from shot noise, which consists of additional current fluctuations that occur when a voltage is applied and a macroscopic current starts to flow. For the general case, the above definition applies to charge carriers in any type of conducting medium (e.g. ions in an electrolyte), not just resistors. It can be modeled by a voltage source representing the noise of the non-ideal resistor in series with an ideal noise free resistor.
The one-sided power spectral density, or voltage variance (mean square) per hertz of bandwidth, is given by
$\overline {v_{n}^2} = 4 k_\text{B} T R$
where kB is Boltzmann's constant in joules per kelvin, T is the resistor's absolute temperature in kelvin, and R is the resistor value in ohms (Ω). Use this equation for quick calculation, at room temperature:
$\sqrt{\overline {v_{n}^2}} = 0.13 \sqrt{R} ~\mathrm{nV}/\sqrt{\mathrm{Hz}}.$
For example, a 1 kΩ resistor at a temperature of 300 K has
$\sqrt{\overline {v_{n}^2}} = \sqrt{4 \cdot 1.38 \cdot 10^{-23}~\mathrm{J}/\mathrm{K} \cdot 300~\mathrm{K} \cdot 1~\mathrm{k}\Omega} = 4.07 ~\mathrm{nV}/\sqrt{\mathrm{Hz}}.$
For a given bandwidth, the root mean square (RMS) of the voltage, $v_{n}$, is given by
$v_{n} = \sqrt{\overline {v_{n}^2}}\sqrt{\Delta f } = \sqrt{ 4 k_\text{B} T R \Delta f }$
where Δf is the bandwidth in hertz over which the noise is measured. For a 1 kΩ resistor at room temperature and a 10 kHz bandwidth, the RMS noise voltage is 400 nV.[5] A useful rule of thumb to remember is that 50 Ω at 1 Hz bandwidth correspond to 1 nV noise at room temperature.
A resistor in a short circuit dissipates a noise power of
$P = {v_{n}^2}/R = 4 k_\text{B} \,T \Delta f.$
The noise generated at the resistor can transfer to the remaining circuit; the maximum noise power transfer happens with impedance matching when the Thévenin equivalent resistance of the remaining circuit is equal to the noise generating resistance. In this case each one of the two participating resistors dissipates noise in both itself and in the other resistor. Since only half of the source voltage drops across any one of these resistors, the resulting noise power is given by
$P = k_\text{B} \,T \Delta f$
where P is the thermal noise power in watts. Notice that this is independent of the noise generating resistance.
Noise current
The noise source can also be modeled by a current source in parallel with the resistor by taking the Norton equivalent that corresponds simply to divide by R. This gives the root mean square value of the current source as:
$i_n = \sqrt {{4 k_\text{B} T \Delta f } \over R}.$
Thermal noise is intrinsic to all resistors and is not a sign of poor design or manufacture, although resistors may also have excess noise.
Noise power in decibels
Signal power is often measured in dBm (decibels relative to 1 milliwatt). From the equation above, noise power in a resistor at room temperature, in dBm, is then:
$P_\mathrm{dBm} = 10\ \log_{10}(k_\text{B} T \Delta f \times 1000)$
where the factor of 1000 is present because the power is given in milliwatts, rather than watts. This equation can be simplified by separating the constant parts from the bandwidth:
$P_\mathrm{dBm} = 10\ \log_{10}(k_\text{B} T \times 1000) + 10\ \log_{10}(\Delta f)$
which is more commonly seen approximated for room temperature (T = 300 K) as:
$P_\mathrm{dBm} = -174 + 10\ \log_{10}(\Delta f)$
where $\Delta f$ is given in Hz; e.g., for a noise bandwidth of 40 MHz, $\Delta f$ is 40,000,000.
Using this equation, noise power for different bandwidths is simple to calculate:
Bandwidth $(\Delta f )$ Thermal noise power Notes
1 Hz −174 dBm
10 Hz −164 dBm
100 Hz −154 dBm
1 kHz −144 dBm
10 kHz −134 dBm FM channel of 2-way radio
15 kHz −132.24 dBm One LTE subcarrier
100 kHz −124 dBm
180 kHz −121.45 dBm One LTE resource block
200 kHz −121 dBm GSM channel
1 MHz −114 dBm Bluetooth channel
2 MHz −111 dBm Commercial GPS channel
3.84 MHz −108 dBm UMTS channel
6 MHz −106 dBm Analog television channel
20 MHz −101 dBm WLAN 802.11 channel
40 MHz −98 dBm WLAN 802.11n 40 MHz channel
80 MHz −95 dBm WLAN 802.11ac 80 MHz channel
160 MHz −92 dBm WLAN 802.11ac 160 MHz channel
1 GHz −84 dBm UWB channel
Thermal noise on capacitors
Thermal noise on capacitors is referred to as kTC noise. Thermal noise in an RC circuit has an unusually simple expression, as the value of the resistance (R) drops out of the equation. This is because higher R contributes to more filtering as well as to more noise. The noise bandwidth of the RC circuit is 1/(4RC),[6] which can substituted into the above formula to eliminate R. The mean-square and RMS noise voltage generated in such a filter are:[7]
$\overline {v_{n}^2} = k_\text{B} T / C$
$v_{n} = \sqrt{ k_\text{B} T / C }.$
Thermal noise accounts for 100% of kTC noise, whether it is attributed to the resistance or to the capacitance.
In the extreme case of the reset noise left on a capacitor by opening an ideal switch, the resistance is infinite, yet the formula still applies; however, now the RMS must be interpreted not as a time average, but as an average over many such reset events, since the voltage is constant when the bandwidth is zero. In this sense, the Johnson noise of an RC circuit can be seen to be inherent, an effect of the thermodynamic distribution of the number of electrons on the capacitor, even without the involvement of a resistor.
The noise is not caused by the capacitor itself, but by the thermodynamic equilibrium of the amount of charge on the capacitor. Once the capacitor is disconnected from a conducting circuit, the thermodynamic fluctuation is frozen at a random value with standard deviation as given above.
The reset noise of capacitive sensors is often a limiting noise source, for example in image sensors. As an alternative to the voltage noise, the reset noise on the capacitor can also be quantified as the electrical charge standard deviation, as
$Q_{n} = \sqrt{ k_\text{B} T C }.$
Since the charge variance is $k_\text{B} T C$, this noise is often called kTC noise.
Any system in thermal equilibrium has state variables with a mean energy of kT/2 per degree of freedom. Using the formula for energy on a capacitor (E = ½CV2), mean noise energy on a capacitor can be seen to also be ½C(kT/C), or also kT/2. Thermal noise on a capacitor can be derived from this relationship, without consideration of resistance.
The kTC noise is the dominant noise source at small capacitors.
Noise of capacitors at 300 K
Capacitance $\sqrt{ k_\text{B} T / C }$ Electrons
1 fF 2 mV 12.5 e
10 fF 640 µV 40 e
100 fF 200 µV 125 e
1 pF 64 µV 400 e
10 pF 20 µV 1250 e
100 pF 6.4 µV 4000 e
1 nF 2 µV 12500 e
Noise at very high frequencies
Main article: Planck's law
The above equations are good approximations at frequencies below about 80 gigahertz (EHF). In the most general case, which includes up to optical frequencies, the power spectral density of the voltage across the resistor R, in V2/Hz is given by:[8]
$\Phi (f) = \frac{2 R h f}{e^{\frac{h f}{k_\text{B} T}} - 1}$
where f is the frequency, h Planck's constant, kB Boltzmann constant and T the temperature in kelvin. If the frequency is low enough, that means:
$f \ll \frac{k_\text{B} T}{h}$
(this assumption is valid until few terahertz at room temperature) then the exponential can be approximated by the constant and linear terms of its Taylor series. The relationship then becomes:
$\Phi (f) \approx 2 R k_\text{B} T.$
In general, both R and T depend on frequency. In order to know the total noise it is enough to integrate over all the bandwidth. Since the signal is real, it is possible to integrate over only the positive frequencies, then multiply by 2. Assuming that R and T are constants over all the bandwidth $\Delta f$, then the root mean square (RMS) value of the voltage across a resistor due to thermal noise is given by
$v_n = \sqrt { 4 k_\text{B} T R \Delta f },$
that is, the same formula as above.
References
1. ^ John R. Barry, Edward A. Lee, and David G. Messerschmitt (2004). Digital Communications. Sprinter. p. 69. ISBN 9780792375487.
2. ^ "Proceedings of the American Physical Society: Minutes of the Philadelphia Meeting December 28, 29, 30, 1926", Phys. Rev. 29, pp. 367-368 (1927) – a February 1927 publication of an abstract for a paper - entitled "Thermal agitation of electricity in conductors" - presented by Johnson during the December 1926 APS Annual Meeting
3. ^ J. Johnson, "Thermal Agitation of Electricity in Conductors", Phys. Rev. 32, 97 (1928) – details of the experiment
4. ^ H. Nyquist, "Thermal Agitation of Electric Charge in Conductors", Phys. Rev. 32, 110 (1928) – the theory
5. ^ Google Calculator result for 1 kΩ room temperature 10 kHz bandwidth
6. ^ Kent H. Lundberg, See pdf, page 10: http://web.mit.edu/klund/www/papers/UNP_noise.pdf
7. ^ R. Sarpeshkar, T. Delbruck, and C. A. Mead, "White noise in MOS transistors and resistors", IEEE Circuits Devices Mag., pp. 23–29, Nov. 1993. Also here
8. ^ L.B. Kish, "Stealth communication: Zero-power classical communication, zero-quantum quantum communication and environmental-noise communication", Applied Physics Lett. 87 (2005), Art. No. 234109; http://arxiv.org/abs/physics/0508135
This article incorporates public domain material from the General Services Administration document "Federal Standard 1037C" (in support of MIL-STD-188). | 2,896 | 10,841 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 24, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2015-22 | latest | en | 0.917046 |
https://www.scientificlib.com/en/Mathematics/DynamicalSystem/DynamicalSystem.html | 1,685,722,515,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00750.warc.gz | 1,093,344,063 | 13,870 | # .
The dynamical system concept is a mathematical formalization for any fixed "rule" which describes the time dependence of a point's position in its ambient space. Examples include the mathematical models that describe the swinging of a clock pendulum, the flow of water in a pipe, and the number of fish each spring in a lake.
At any given time a dynamical system has a state given by a set of real numbers (a vector) which can be represented by a point in an appropriate state space (a geometrical manifold). Small changes in the state of the system correspond to small changes in the numbers. The evolution rule of the dynamical system is a fixed rule that describes what future states follow from the current state. The rule is deterministic: for a given time interval only one future state follows from the current state.
Overview
The concept of a dynamical system has its origins in Newtonian mechanics. There, as in other natural sciences and engineering disciplines, the evolution rule of dynamical systems is given implicitly by a relation that gives the state of the system only a short time into the future. (The relation is either a differential equation, difference equation or other time scale.) To determine the state for all future times requires iterating the relation many times—each advancing time a small step. The iteration procedure is referred to as solving the system or integrating the system. Once the system can be solved, given an initial point it is possible to determine all its future points, a collection known as a trajectory or orbit.
Before the advent of fast computing machines, solving a dynamical system required sophisticated mathematical techniques and could be accomplished only for a small class of dynamical systems. Numerical methods implemented on electronic computing machines have simplified the task of determining the orbits of a dynamical system.
For simple dynamical systems, knowing the trajectory is often sufficient, but most dynamical systems are too complicated to be understood in terms of individual trajectories. The difficulties arise because:
* The systems studied may only be known approximately—the parameters of the system may not be known precisely or terms may be missing from the equations. The approximations used bring into question the validity or relevance of numerical solutions. To address these questions several notions of stability have been introduced in the study of dynamical systems, such as Lyapunov stability or structural stability. The stability of the dynamical system implies that there is a class of models or initial conditions for which the trajectories would be equivalent. The operation for comparing orbits to establish their equivalence changes with the different notions of stability.
* The type of trajectory may be more important than one particular trajectory. Some trajectories may be periodic, whereas others may wander through many different states of the system. Applications often require enumerating these classes or maintaining the system within one class. Classifying all possible trajectories has led to the qualitative study of dynamical systems, that is, properties that do not change under coordinate changes. Linear dynamical systems and systems that have two numbers describing a state are examples of dynamical systems where the possible classes of orbits are understood.
* The behavior of trajectories as a function of a parameter may be what is needed for an application. As a parameter is varied, the dynamical systems may have bifurcation points where the qualitative behavior of the dynamical system changes. For example, it may go from having only periodic motions to apparently erratic behavior, as in the transition to turbulence of a fluid.
* The trajectories of the system may appear erratic, as if random. In these cases it may be necessary to compute averages using one very long trajectory or many different trajectories. The averages are well defined for ergodic systems and a more detailed understanding has been worked out for hyperbolic systems. Understanding the probabilistic aspects of dynamical systems has helped establish the foundations of statistical mechanics and of chaos.
It was in the work of Poincaré that these dynamical systems themes developed.
Basic definitions
Main article: Dynamical system (definition)
A dynamical system is a manifold M called the phase (or state) space endowed with a family of smooth evolution functions Φ t that for any element of the time, map a point of the phase space back into the phase space. The notion of smoothness changes with applications and the type of manifold. There are several choices for the set T. When T is taken to be the reals, the dynamical system is called a flow; and if T is restricted to the non-negative reals, then the dynamical system is a semi-flow. When T is taken to be the integers, it is a cascade or a map; and the restriction to the non-negative integers is a semi-cascade.
Examples
The evolution function Φ t is often the solution of a differential equation of motion
The equation gives the time derivative, represented by the dot, of a trajectory x(t) on the phase space starting at some point x0. The vector field v(x) is a smooth function that at every point of the phase space M provides the velocity vector of the dynamical system at that point. (These vectors are not vectors in the phase space M, but in the tangent space TMx of the point x.) Given a smooth Φ t, an autonomous vector field can be derived from it.
There is no need for higher order derivatives in the equation, nor for time dependence in v(x) because these can be eliminated by considering systems of higher dimensions. Other types of differential equations can be used to define the evolution rule:
is an example of an equation that arises from the modeling of mechanical systems with complicated constraints.
The differential equations determining the evolution function Φ t are often ordinary differential equations: in this case the phase space M is a finite dimensional manifold. Many of the concepts in dynamical systems can be extended to infinite-dimensional manifolds—those that are locally Banach spaces—in which case the differential equations are partial differential equations. In the late 20th century the dynamical system perspective to partial differential equations started gaining popularity.
Further examples
* Logistic map
* Tent map
* Double pendulum
* Arnold's cat map
* Horseshoe map
* Baker's map is an example of a chaotic piecewise linear map
* Billiards and outer billiards
* Hénon map
* Lorenz system
* Circle map
* Rössler map
* List of chaotic maps
* Swinging Atwood's machine
* Bouncing ball simulation system
Linear dynamical systems
Main article: Linear dynamical system
Linear dynamical systems can be solved in terms of simple functions and the behavior of all orbits classified. In a linear system the phase space is the N-dimensional Euclidean space, so any point in phase space can be represented by a vector with N numbers. The analysis of linear systems is possible because they satisfy a superposition principle: if u(t) and w(t) satisfy the differential equation for the vector field (but not necessarily the initial condition), then so will u(t) + w(t).
Flows
For a flow, the vector field Φ(x) is a linear function of the position in the phase space, that is,
with A a matrix, b a vector of numbers and x the position vector. The solution to this system can be found by using the superposition principle (linearity). The case b ≠ 0 with A = 0 is just a straight line in the direction of b:
When b is zero and A ≠ 0 the origin is an equilibrium (or singular) point of the flow, that is, if x0 = 0, then the orbit remains there. For other initial conditions, the equation of motion is given by the exponential of a matrix: for an initial point x0,
When b = 0, the eigenvalues of A determine the structure of the phase space. From the eigenvalues and the eigenvectors of A it is possible to determine if an initial point will converge or diverge to the equilibrium point at the origin.
The distance between two different initial conditions in the case A ≠ 0 will change exponentially in most cases, either converging exponentially fast towards a point, or diverging exponentially fast. Linear systems display sensitive dependence on initial conditions in the case of divergence. For nonlinear systems this is one of the (necessary but not sufficient) conditions for chaotic behavior.
Linear vector fields and a few trajectories. (*)
Maps
A discrete-time, affine dynamical system has the form
with A a matrix and b a vector. As in the continuous case, the change of coordinates x → x + (1 - A) –1b removes the term b from the equation. In the new coordinate system, the origin is a fixed point of the map and the solutions are of the linear system A nx0. The solutions for the map are no longer curves, but points that hop in the phase space. The orbits are organized in curves, or fibers, which are collections of points that map into themselves under the action of the map.
As in the continuous case, the eigenvalues and eigenvectors of A determine the structure of phase space. For example, if u1 is an eigenvector of A, with a real eigenvalue smaller than one, then the straight lines given by the points along α u1, with α ∈ R, is an invariant curve of the map. Points in this straight line run into the fixed point.
There are also many other discrete dynamical systems.
Local dynamics
The qualitative properties of dynamical systems do not change under a smooth change of coordinates (this is sometimes taken as a definition of qualitative): a singular point of the vector field (a point where v(x) = 0) will remain a singular point under smooth transformations; a periodic orbit is a loop in phase space and smooth deformations of the phase space cannot alter it being a loop. It is in the neighborhood of singular points and periodic orbits that the structure of a phase space of a dynamical system can be well understood. In the qualitative study of dynamical systems, the approach is to show that there is a change of coordinates (usually unspecified, but computable) that makes the dynamical system as simple as possible.
Rectification
A flow in most small patches of the phase space can be made very simple. If y is a point where the vector field v(y) ≠ 0, then there is a change of coordinates for a region around y where the vector field becomes a series of parallel vectors of the same magnitude. This is known as the rectification theorem.
The rectification theorem says that away from singular points the dynamics of a point in a small patch is a straight line. The patch can sometimes be enlarged by stitching several patches together, and when this works out in the whole phase space M the dynamical system is integrable. In most cases the patch cannot be extended to the entire phase space. There may be singular points in the vector field (where v(x) = 0); or the patches may become smaller and smaller as some point is approached. The more subtle reason is a global constraint, where the trajectory starts out in a patch, and after visiting a series of other patches comes back to the original one. If the next time the orbit loops around phase space in a different way, then it is impossible to rectify the vector field in the whole series of patches.
Near periodic orbits
In general, in the neighborhood of a periodic orbit the rectification theorem cannot be used. Poincaré developed an approach that transforms the analysis near a periodic orbit to the analysis of a map. Pick a point x0 in the orbit γ and consider the points in phase space in that neighborhood that are perpendicular to v(x0). These points are a Poincaré section S(γ, x0), of the orbit. The flow now defines a map, the Poincaré map F : S → S, for points starting in S and returning to S. Not all these points will take the same amount of time to come back, but the times will be close to the time it takes x0.
The intersection of the periodic orbit with the Poincaré section is a fixed point of the Poincaré map F. By a translation, the point can be assumed to be at x = 0. The Taylor series of the map is F(x) = J · x + O(x²), so a change of coordinates h can only be expected to simplify F to its linear part
This is known as the conjugation equation. Finding conditions for this equation to hold has been one of the major tasks of research in dynamical systems. Poincaré first approached it assuming all functions to be analytic and in the process discovered the non-resonant condition. If λ1,…,λν are the eigenvalues of J they will be resonant if one eigenvalue is an integer linear combination of two or more of the others. As terms of the form λi – ∑ (multiples of other eigenvalues) occurs in the denominator of the terms for the function h, the non-resonant condition is also known as the small divisor problem.
Conjugation results
The results on the existence of a solution to the conjugation equation depend on the eigenvalues of J and the degree of smoothness required from h. As J does not need to have any special symmetries, its eigenvalues will typically be complex numbers. When the eigenvalues of J are not in the unit circle, the dynamics near the fixed point x0 of F is called hyperbolic and when the eigenvalues are on the unit circle and complex, the dynamics is called elliptic.
In the hyperbolic case the Hartman-Grobman theorem gives the conditions for the existence of a continuous function that maps the neighborhood of the fixed point of the map to the linear map J · x. The hyperbolic case is also structurally stable. Small changes in the vector field will only produce small changes in the Poincaré map and these small changes will reflect in small changes in the position of the eigenvalues of J in the complex plane, implying that the map is still hyperbolic.
The Kolmogorov-Arnold-Moser (KAM) theorem gives the behavior near an elliptic point.
Bifurcation theory
Main article: Bifurcation theory
When the evolution map Φt (or the vector field it is derived from) depends on a parameter μ, the structure of the phase space will also depend on this parameter. Small changes may produce no qualitative changes in the phase space until a special value μ0 is reached. At this point the phase space changes qualitatively and the dynamical system is said to have gone through a bifurcation.
Bifurcation theory considers a structure in phase space (typically a fixed point, a periodic orbit, or an invariant torus) and studies its behavior as a function of the parameter μ. At the bifurcation point the structure may change its stability, split into new structures, or merge with other structures. By using Taylor series approximations of the maps and an understanding of the differences that may be eliminated by a change of coordinates, it is possible to catalog the bifurcations of dynamical systems.
The bifurcations of a hyperbolic fixed point x0 of a system family Fμ can be characterized by the eigenvalues of the first derivative of the system DFμ(x0) computed at the bifurcation point. For a map, the bifurcation will occur when there are eigenvalues of DFμ on the unit circle. For a flow, it will occur when there are eigenvalues on the imaginary axis. For more information, see the main article on Bifurcation theory.
Some bifurcations can lead to very complicated structures in phase space. For example, the Ruelle-Takens scenario describes how a periodic orbit bifurcates into a torus and the torus into a strange attractor. In another example, Feigenbaum period-doubling describes how a stable periodic orbit goes through a series of period-doubling bifurcations.
Ergodic systems
Main article: Ergodic theory
In many dynamical systems it is possible to choose the coordinates of the system so that the volume (really a ν-dimensional volume) in phase space is invariant. This happens for mechanical systems derived from Newton's laws as long as the coordinates are the position and the momentum and the volume is measured in units of (position) × (momentum). The flow takes points of a subset A into the points Φ t(A) and invariance of the phase space means that
.
In the Hamiltonian formalism, given a coordinate it is possible to derive the appropriate (generalized) momentum such that the associated volume is preserved by the flow. The volume is said to be computed by the Liouville measure.
In a Hamiltonian system not all possible configurations of position and momentum can be reached from an initial condition. Because of energy conservation, only the states with the same energy as the initial condition are accessible. The states with the same energy form an energy shell Ω, a sub-manifold of the phase space. The volume of the energy shell, computed using the Liouville measure, is preserved under evolution.
For systems where the volume is preserved by the flow, Poincaré discovered the recurrence theorem: Assume the phase space has a finite Liouville volume and let F be a phase space volume-preserving map and A a subset of the phase space. Then almost every point of A returns to A infinitely often. The Poincaré recurrence theorem was used by Zermelo to object to Boltzmann's derivation of the increase in entropy in a dynamical system of colliding atoms.
One of the questions raised by Boltzmann's work was the possible equality between time averages and space averages, what he called the ergodic hypothesis. The hypothesis states that the length of time a typical trajectory spends in a region A is vol(A)/vol(Ω).
The ergodic hypothesis turned out not to be the essential property needed for the development of statistical mechanics and a series of other ergodic-like properties were introduced to capture the relevant aspects of physical systems. Koopman approached the study of ergodic systems by the use of functional analysis. An observable a is a function that to each point of the phase space associates a number (say instantaneous pressure, or average height). The value of an observable can be computed at another time by using the evolution function φ t. This introduces an operator U t, the transfer operator,
By studying the spectral properties of the linear operator U it becomes possible to classify the ergodic properties of Φ t. In using the Koopman approach of considering the action of the flow on an observable function, the finite-dimensional nonlinear problem involving Φ t gets mapped into an infinite-dimensional linear problem involving U.
The Liouville measure restricted to the energy surface Ω is the basis for the averages computed in equilibrium statistical mechanics. An average in time along a trajectory is equivalent to an average in space computed with the Boltzmann factor exp(−βH). This idea has been generalized by Sinai, Bowen, and Ruelle (SRB) to a larger class of dynamical systems that includes dissipative systems. SRB measures replace the Boltzmann factor and they are defined on attractors of chaotic systems.
Nonlinear dynamical systems and chaos
Main article: Chaos theory
Simple nonlinear dynamical systems and even piecewise linear systems can exhibit a completely unpredictable behavior, which might seem to be random. (Remember that we are speaking of completely deterministic systems!). This seemingly unpredictable behavior has been called chaos. Hyperbolic systems are precisely defined dynamical systems that exhibit the properties ascribed to chaotic systems. In hyperbolic systems the tangent space perpendicular to a trajectory can be well separated into two parts: one with the points that converge towards the orbit (the stable manifold) and another of the points that diverge from the orbit (the unstable manifold).
This branch of mathematics deals with the long-term qualitative behavior of dynamical systems. Here, the focus is not on finding precise solutions to the equations defining the dynamical system (which is often hopeless), but rather to answer questions like "Will the system settle down to a steady state in the long term, and if so, what are the possible attractors?" or "Does the long-term behavior of the system depend on its initial condition?"
Note that the chaotic behavior of complicated systems is not the issue. Meteorology has been known for years to involve complicated—even chaotic—behavior. Chaos theory has been so surprising because chaos can be found within almost trivial systems. The logistic map is only a second-degree polynomial; the horseshoe map is piecewise linear.
Geometrical definition
A dynamical system is the tuple \langle , with a manifold (locally a Banach space or Euclidean space), the domain for time (non-negative reals, the integers, ...) and f an evolution rule t→f t (with ) such that f t is a diffeomorphism of the manifold to itself. So, f is a mapping of the time-domain into the space of diffeomorphisms of the manifold to itself. In other terms, f(t) is a diffeomorphism, for every time t in the domain .
Measure theoretical definition
See main article Measure-preserving dynamical system.
A dynamical system may be defined formally, as a measure-preserving transformation of a sigma-algebra, the quadruplet (X,Σ,μ,τ). Here, X is a set, and Σ is a sigma-algebra on X, so that the pair (X,Σ) is a measurable space. μ is a finite measure on the sigma-algebra, so that the triplet (X,Σ,μ) is a probability space. A map is said to be Σ-measurable if and only if, for every a, one has . A map τ is said to preserve the measure if and only if, for every ., one has μ(τ − 1σ) = μ(σ). Combining the above, a map τ is said to be a measure-preserving transformation of X , if it is a map from X to itself, it is Σ-measurable, and is measure-preserving. The quadruple (X,Σ,μ,τ), for such a τ, is then defined to be a dynamical system.
The map τ embodies the time evolution of the dynamical system. Thus, for discrete dynamical systems the iterates \tau^n=\tau for integer n are studied. For continuous dynamical systems, the map τ is understood to be finite time evolution map and the construction is more complicated.
Examples of dynamical systems
* Arnold's cat map
* Baker's map is an example of a chaotic piecewise linear map
* Circle map
* Double pendulum
* Billiards and Outer Billiards
* Henon map
* Horseshoe map
* Irrational rotation
* List of chaotic maps
* Logistic map
* Lorenz system
* Rossler map
* Bouncing Ball
* Mechanical Strings
* Journal of Advanced Research in Dynamical and Control Systems
* Swinging Atwood's Machine (SAM)
* Interactive applet for the Standard and Henon Maps by A. Luhn
Multidimensional generalization
Dynamical systems are defined over a single independent variable, usually thought of as time. A more general class of systems are defined over multiple independent variables and are therefore called Multidimensional systems. Such systems are useful for modeling, for example, Image processing.
* Behavioral modeling
* Dynamical systems theory
* List of dynamical system topics
* Oscillation
* People in systems and control
* Sarkovskii's theorem
* System dynamics
* Systems theory
References
* Ralph Abraham and Jerrold E. Marsden (1978). Foundations of mechanics. Benjamin-Cummings. ISBN 0-8053-0102-X. (available as a reprint: ISBN 0-201-40840-6)
* Encyclopaedia of Mathematical Sciences (ISSN 0938-0396) has a sub-series on dynamical systems with reviews of current research.
* Anatole Katok and Boris Hasselblatt (1996). Introduction to the modern theory of dynamical systems. Cambridge. ISBN 0-521-57557-5.
* Christian Bonatti, Lorenzo J. Díaz, Marcelo Viana (2005). Dynamics Beyond Uniform Hyperbolicity: A Global Geometric and Probabilistic Perspective. Springer. ISBN 3-540-22066-6.
* Diederich Hinrichsen and Anthony J. Pritchard (2005). Mathematical Systems Theory I - Modelling, State Space Analysis, Stability and Robustness. Springer-Verlag. ISBN 978-3-540-44125-0.
* Stephen Smale (1967). "Differential dynamical systems". 73. Bulletin of the American Mathematical Society. pp. 747–817.
Introductory texts with a unique perspective:
* V. I. Arnold (1982). Mathematical methods of classical mechanics. Springer-Verlag. ISBN 0-387-96890-3.
* Jacob Palis and Wellington de Melo (1982). Geometric theory of dynamical systems: an introduction. Springer-Verlag. ISBN 0-387-90668-1.
* David Ruelle (1989). Elements of Differentiable Dynamics and Bifurcation Theory. Academic Press. ISBN 0-12-601710-7.
* Tim Bedford, Michael Keane and Caroline Series, eds. (1991). Ergodic theory, symbolic dynamics and hyperbolic spaces. Oxford University Press. ISBN 0-19-853390-X.
* Ralph H. Abraham and Christopher D. Shaw (1992). Dynamics—the geometry of behavior, 2nd edition. Addison-Wesley. ISBN 0-201-56716-4.
Textbooks
* Steven H. Strogatz (1994). Nonlinear dynamics and chaos: with applications to physics, biology chemistry and engineering. Addison Wesley. ISBN 0-201-54344-3.
* Kathleen T. Alligood, Tim D. Sauer and James A. Yorke (2000). Chaos. An introduction to dynamical systems. Springer Verlag. ISBN 0-387-94677-2.
* Morris W. Hirsch, Stephen Smale and Robert Devaney (2003). Differential Equations, dynamical systems, and an introduction to chaos. Academic Press. ISBN 0-12-349703-5.
* Julien Clinton Sprott (2003). Chaos and time-series analysis. Oxford University Press. ISBN 0-19-850839-5.
Popularizations:
* Florin Diacu and Philip Holmes (1996). Celestial Encounters. Princeton. ISBN 0-691-02743-9.
* James Gleick (1988). Chaos: Making a New Science. Penguin. ISBN 0-14-009250-1.
* Ivar Ekeland (1990). Mathematics and the Unexpected (Paperback). University Of Chicago Press. ISBN 0-226-19990-8.
* Ian Stewart (1997). Does God Play Dice? The New Mathematics of Chaos. Penguin. ISBN 0140256024.
* A collection of dynamic and non-linear system models and demo applets (in Monash University's Virtual Lab)
* Arxiv preprint server has daily submissions of (non-refereed) manuscripts in dynamical systems.
* DSWeb provides up-to-date information on dynamical systems and its applications.
* Encyclopedia of dynamical systems A part of Scholarpedia — peer reviewed and written by invited experts.
* Nonlinear Dynamics. Models of bifurcation and chaos by Elmer G. Wiens
* Oliver Knill has a series of examples of dynamical systems with explanations and interactive controls.
* Sci.Nonlinear FAQ 2.0 (Sept 2003) provides definitions, explanations and resources related to nonlinear science
Online books or lecture notes:
* Geometrical theory of dynamical systems. Nils Berglund's lecture notes for a course at ETH at the advanced undergraduate level.
* Dynamical systems. George D. Birkhoff's 1927 book already takes a modern approach to dynamical systems.
* Chaos: classical and quantum. An introduction to dynamical systems from the periodic orbit point of view.
* Modeling Dynamic Systems. An introduction to the development of mathematical models of dynamic systems.
* Learning Dynamical Systems. Tutorial on learning dynamical systems.
* Ordinary Differential Equations and Dynamical Systems. Lecture notes by Gerald Teschl
Research groups:
* Dynamical Systems Group Groningen, IWI, University of Groningen.
* Chaos @ UMD. Concentrates on the applications of dynamical systems.
* Dynamical Systems, SUNY Stony Brook. Lists of conferences, researchers, and some open problems.
* Center for Dynamics and Geometry, Penn State.
* Control and Dynamical Systems, Caltech.
* Laboratory of Nonlinear Systems, Ecole Polytechnique Fédérale de Lausanne (EPFL).
* Center for Dynamical Systems, University of Bremen
* Systems Analysis, Modelling and Prediction Group, University of Oxford
* Non-Linear Dynamics Group, Instituto Superior Técnico, Technical University of Lisbon
* Dynamical Systems, IMPA, Instituto Nacional de Matemática Pura e Applicada.
* Nonlinear Dynamics Workgroup, Institute of Computer Science, Czech Academy of Sciences.
Simulation software based on Dynamical Systems approach:
* FyDiK
* iDMC, simulation and dynamical analysis of nonlinear models | 6,089 | 28,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-23 | longest | en | 0.919941 |
https://isabelle.in.tum.de/repos/isabelle/rev/00c06f1315d0 | 1,670,503,861,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00587.warc.gz | 342,139,999 | 23,439 | author paulson Tue, 26 May 2015 21:58:04 +0100 changeset 60303 00c06f1315d0 parent 60302 6dcb8aa0966a child 60304 3f429b7d8eb5
New material about paths, and some lemmas
src/HOL/Finite_Set.thy file | annotate | diff | comparison | revisions src/HOL/Fun.thy file | annotate | diff | comparison | revisions src/HOL/Library/Convex.thy file | annotate | diff | comparison | revisions src/HOL/Library/Countable_Set.thy file | annotate | diff | comparison | revisions src/HOL/Multivariate_Analysis/Brouwer_Fixpoint.thy file | annotate | diff | comparison | revisions src/HOL/Multivariate_Analysis/Convex_Euclidean_Space.thy file | annotate | diff | comparison | revisions src/HOL/Multivariate_Analysis/Linear_Algebra.thy file | annotate | diff | comparison | revisions src/HOL/Multivariate_Analysis/Path_Connected.thy file | annotate | diff | comparison | revisions src/HOL/Real_Vector_Spaces.thy file | annotate | diff | comparison | revisions
```--- a/src/HOL/Finite_Set.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Finite_Set.thy Tue May 26 21:58:04 2015 +0100
@@ -277,7 +277,8 @@
then have B_A: "insert x B = f ` A" by simp
then obtain y where "x = f y" and "y \<in> A" by blast
from B_A `x \<notin> B` have "B = f ` A - {x}" by blast
- with B_A `x \<notin> B` `x = f y` `inj_on f A` `y \<in> A` have "B = f ` (A - {y})" by (simp add: inj_on_image_set_diff)
+ with B_A `x \<notin> B` `x = f y` `inj_on f A` `y \<in> A` have "B = f ` (A - {y})"
+ by (simp add: inj_on_image_set_diff Set.Diff_subset)
moreover from `inj_on f A` have "inj_on f (A - {y})" by (rule inj_on_diff)
ultimately have "finite (A - {y})" by (rule insert.hyps)
then show "finite A" by simp```
```--- a/src/HOL/Fun.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Fun.thy Tue May 26 21:58:04 2015 +0100
@@ -478,16 +478,14 @@
lemma inj_on_image_Int:
"[| inj_on f C; A<=C; B<=C |] ==> f`(A Int B) = f`A Int f`B"
-done
+ by (simp add: inj_on_def, blast)
lemma inj_on_image_set_diff:
- "[| inj_on f C; A<=C; B<=C |] ==> f`(A-B) = f`A - f`B"
-done
+ "[| inj_on f C; A-B \<subseteq> C; B \<subseteq> C |] ==> f`(A-B) = f`A - f`B"
+ by (simp add: inj_on_def, blast)
lemma image_Int: "inj f ==> f`(A Int B) = f`A Int f`B"
+ by (simp add: inj_on_def, blast)
lemma image_set_diff: "inj f ==> f`(A-B) = f`A - f`B"
```--- a/src/HOL/Library/Convex.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Library/Convex.thy Tue May 26 21:58:04 2015 +0100
@@ -48,13 +48,13 @@
shows "((1 - u) *\<^sub>R a + u *\<^sub>R b) \<in> s"
using assms unfolding convex_alt by auto
-lemma convex_empty[intro]: "convex {}"
+lemma convex_empty[intro,simp]: "convex {}"
unfolding convex_def by simp
-lemma convex_singleton[intro]: "convex {a}"
+lemma convex_singleton[intro,simp]: "convex {a}"
unfolding convex_def by (auto simp: scaleR_left_distrib[symmetric])
-lemma convex_UNIV[intro]: "convex UNIV"
+lemma convex_UNIV[intro,simp]: "convex UNIV"
unfolding convex_def by auto
lemma convex_Inter: "(\<forall>s\<in>f. convex s) \<Longrightarrow> convex(\<Inter> f)"```
```--- a/src/HOL/Library/Countable_Set.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Library/Countable_Set.thy Tue May 26 21:58:04 2015 +0100
@@ -182,6 +182,9 @@
by (blast dest: comp_inj_on subset_inj_on intro: countableI)
qed
+lemma countable_image_inj_on: "countable (f ` A) \<Longrightarrow> inj_on f A \<Longrightarrow> countable A"
+ by (metis countable_image the_inv_into_onto)
+
lemma countable_UN[intro, simp]:
fixes I :: "'i set" and A :: "'i => 'a set"
assumes I: "countable I"
@@ -221,6 +224,9 @@
lemma countable_Diff[intro, simp]: "countable A \<Longrightarrow> countable (A - B)"
by (blast intro: countable_subset)
+lemma countable_insert_eq [simp]: "countable (insert x A) = countable A"
+ by auto (metis Diff_insert_absorb countable_Diff insert_absorb)
+
lemma countable_vimage: "B \<subseteq> range f \<Longrightarrow> countable (f -` B) \<Longrightarrow> countable B"
by (metis Int_absorb2 assms countable_image image_vimage_eq)
@@ -325,4 +331,7 @@
"uncountable A \<Longrightarrow> countable B \<Longrightarrow> uncountable (A - B)"
using countable_Un[of B "A - B"] assms by auto
+lemma countable_Diff_eq [simp]: "countable (A - {x}) = countable A"
+ by (meson countable_Diff countable_empty countable_insert uncountable_minus_countable)
+
end```
```--- a/src/HOL/Multivariate_Analysis/Brouwer_Fixpoint.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Multivariate_Analysis/Brouwer_Fixpoint.thy Tue May 26 21:58:04 2015 +0100
@@ -30,7 +30,7 @@
have "f ` (A - {a}) = g ` (A - {a})"
by (intro image_cong) (simp_all add: eq)
then have "B - {f a} = B - {g a}"
- using f g a by (auto simp: bij_betw_def inj_on_image_set_diff set_eq_iff)
+ using f g a by (auto simp: bij_betw_def inj_on_image_set_diff set_eq_iff Diff_subset)
moreover have "f a \<in> B" "g a \<in> B"
using f g a by (auto simp: bij_betw_def)
ultimately show ?thesis
@@ -211,7 +211,7 @@
moreover obtain a where "rl a = Suc n" "a \<in> s"
by (metis atMost_iff image_iff le_Suc_eq rl)
ultimately have n: "{..n} = rl ` (s - {a})"
- by (auto simp add: inj_on_image_set_diff rl)
+ by (auto simp add: inj_on_image_set_diff Diff_subset rl)
have "{a\<in>s. rl ` (s - {a}) = {..n}} = {a}"
using inj_rl `a \<in> s` by (auto simp add: n inj_on_image_eq_iff[OF inj_rl] Diff_subset)
then show "card ?S = 1"
@@ -234,7 +234,7 @@
{ fix x assume "x \<in> s" "x \<notin> {a, b}"
then have "rl ` s - {rl x} = rl ` ((s - {a}) - {x})"
- by (auto simp: eq inj_on_image_set_diff[OF inj])
+ by (auto simp: eq Diff_subset inj_on_image_set_diff[OF inj])
also have "\<dots> = rl ` (s - {x})"
using ab `x \<notin> {a, b}` by auto
also assume "\<dots> = rl ` s"
@@ -597,8 +597,8 @@
have "\<And>i. Suc ` {..< i} = {..< Suc i} - {0}"
by (auto simp: image_iff Ball_def) arith
then have upd_Suc: "\<And>i. i \<le> n \<Longrightarrow> (upd\<circ>Suc) ` {..< i} = upd ` {..< Suc i} - {n}"
- using `upd 0 = n`
- by (simp add: image_comp[symmetric] inj_on_image_set_diff[OF inj_upd])
+ using `upd 0 = n` upd_inj
+ by (auto simp add: image_comp[symmetric] inj_on_image_set_diff[OF inj_upd])
have n_in_upd: "\<And>i. n \<in> upd ` {..< Suc i}"
using `upd 0 = n` by auto
@@ -773,7 +773,7 @@
by auto
finally have eq: "s - {a} = f' ` {.. n} - {f' n}"
unfolding s_eq `a = enum i` `i = 0`
- by (simp add: inj_on_image_set_diff[OF inj_enum] inj_on_image_set_diff[OF inj_f'])
+ by (simp add: Diff_subset inj_on_image_set_diff[OF inj_enum] inj_on_image_set_diff[OF inj_f'])
have "enum 0 < f' 0"
using `n \<noteq> 0` by (simp add: enum_strict_mono f'_eq_enum)
@@ -887,9 +887,9 @@
by auto
finally have eq: "s - {a} = b.enum ` {.. n} - {b.enum 0}"
unfolding s_eq `a = enum i` `i = n`
- using inj_on_image_set_diff[OF inj_enum order_refl, of "{n}"]
- inj_on_image_set_diff[OF b.inj_enum order_refl, of "{0}"]
+ using inj_on_image_set_diff[OF inj_enum Diff_subset, of "{n}"]
+ inj_on_image_set_diff[OF b.inj_enum Diff_subset, of "{0}"]
+ by (simp add: comp_def )
have "b.enum 0 \<le> b.enum n"
@@ -980,8 +980,8 @@
by (intro image_cong) auto
then have eq: "s - {a} = b.enum ` {.. n} - {b.enum i}"
unfolding s_eq `a = enum i`
- using inj_on_image_set_diff[OF inj_enum order_refl `{i} \<subseteq> {..n}`]
- inj_on_image_set_diff[OF b.inj_enum order_refl `{i} \<subseteq> {..n}`]
+ using inj_on_image_set_diff[OF inj_enum Diff_subset `{i} \<subseteq> {..n}`]
+ inj_on_image_set_diff[OF b.inj_enum Diff_subset `{i} \<subseteq> {..n}`]
have "a \<noteq> b.enum i"```
```--- a/src/HOL/Multivariate_Analysis/Convex_Euclidean_Space.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Multivariate_Analysis/Convex_Euclidean_Space.thy Tue May 26 21:58:04 2015 +0100
@@ -384,13 +384,13 @@
lemma affine_UNIV[intro]: "affine UNIV"
unfolding affine_def by auto
-lemma affine_Inter: "(\<forall>s\<in>f. affine s) \<Longrightarrow> affine (\<Inter> f)"
+lemma affine_Inter[intro]: "(\<forall>s\<in>f. affine s) \<Longrightarrow> affine (\<Inter> f)"
unfolding affine_def by auto
-lemma affine_Int: "affine s \<Longrightarrow> affine t \<Longrightarrow> affine (s \<inter> t)"
+lemma affine_Int[intro]: "affine s \<Longrightarrow> affine t \<Longrightarrow> affine (s \<inter> t)"
unfolding affine_def by auto
-lemma affine_affine_hull: "affine(affine hull s)"
+lemma affine_affine_hull [simp]: "affine(affine hull s)"
unfolding hull_def
using affine_Inter[of "{t. affine t \<and> s \<subseteq> t}"] by auto
@@ -2355,13 +2355,13 @@
lemma affine_hull_translation: "affine hull ((\<lambda>x. a + x) ` S) = (\<lambda>x. a + x) ` (affine hull S)"
proof -
have "affine ((\<lambda>x. a + x) ` (affine hull S))"
- using affine_translation affine_affine_hull by auto
+ using affine_translation affine_affine_hull by blast
moreover have "(\<lambda>x. a + x) ` S \<subseteq> (\<lambda>x. a + x) ` (affine hull S)"
using hull_subset[of S] by auto
ultimately have h1: "affine hull ((\<lambda>x. a + x) ` S) \<subseteq> (\<lambda>x. a + x) ` (affine hull S)"
by (metis hull_minimal)
have "affine((\<lambda>x. -a + x) ` (affine hull ((\<lambda>x. a + x) ` S)))"
- using affine_translation affine_affine_hull by (auto simp del: uminus_add_conv_diff)
+ using affine_translation affine_affine_hull by blast
moreover have "(\<lambda>x. -a + x) ` (\<lambda>x. a + x) ` S \<subseteq> (\<lambda>x. -a + x) ` (affine hull ((\<lambda>x. a + x) ` S))"
using hull_subset[of "(\<lambda>x. a + x) ` S"] by auto
moreover have "S = (\<lambda>x. -a + x) ` (\<lambda>x. a + x) ` S"
@@ -2933,7 +2933,7 @@
have "dim (affine hull S) \<ge> dim S"
using dim_subset by auto
moreover have "dim (span S) \<ge> dim (affine hull S)"
- using dim_subset affine_hull_subset_span by auto
+ using dim_subset affine_hull_subset_span by blast
moreover have "dim (span S) = dim S"
using dim_span by auto
ultimately show ?thesis by auto
@@ -3178,13 +3178,13 @@
"rel_interior S = {x \<in> S. \<exists>e. e > 0 \<and> cball x e \<inter> affine hull S \<subseteq> S}"
using mem_rel_interior_cball [of _ S] by auto
-lemma rel_interior_empty: "rel_interior {} = {}"
+lemma rel_interior_empty [simp]: "rel_interior {} = {}"
-lemma affine_hull_sing: "affine hull {a :: 'n::euclidean_space} = {a}"
+lemma affine_hull_sing [simp]: "affine hull {a :: 'n::euclidean_space} = {a}"
by (metis affine_hull_eq affine_sing)
-lemma rel_interior_sing: "rel_interior {a :: 'n::euclidean_space} = {a}"
+lemma rel_interior_sing [simp]: "rel_interior {a :: 'n::euclidean_space} = {a}"
unfolding rel_interior_ball affine_hull_sing
apply auto
apply (rule_tac x = "1 :: real" in exI)
@@ -3218,6 +3218,12 @@
unfolding rel_interior interior_def by auto
qed
+lemma rel_interior_interior:
+ fixes S :: "'n::euclidean_space set"
+ assumes "affine hull S = UNIV"
+ shows "rel_interior S = interior S"
+ using assms unfolding rel_interior interior_def by auto
+
lemma rel_interior_open:
fixes S :: "'n::euclidean_space set"
assumes "open S"
@@ -3694,7 +3700,7 @@
unfolding affine_hull_insert_span_gen span_substd_basis[OF assms,symmetric] * ..
qed
-lemma affine_hull_convex_hull: "affine hull (convex hull S) = affine hull S"
+lemma affine_hull_convex_hull [simp]: "affine hull (convex hull S) = affine hull S"
by (metis Int_absorb1 Int_absorb2 convex_hull_subset_affine_hull hull_hull hull_mono hull_subset)
@@ -8040,22 +8046,14 @@
and "convex T"
shows "rel_interior (S \<times> T) = rel_interior S \<times> rel_interior T"
proof -
- {
- assume "S = {}"
+ { assume "S = {}"
then have ?thesis
- apply auto
- using rel_interior_empty
- apply auto
- done
+ by auto
}
moreover
- {
- assume "T = {}"
+ { assume "T = {}"
then have ?thesis
- apply auto
- using rel_interior_empty
- apply auto
- done
+ by auto
}
moreover
{```
```--- a/src/HOL/Multivariate_Analysis/Linear_Algebra.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Multivariate_Analysis/Linear_Algebra.thy Tue May 26 21:58:04 2015 +0100
@@ -29,12 +29,7 @@
fixes e :: real
shows "e > 0 \<Longrightarrow> \<exists>d. 0 < d \<and> (\<forall>y. \<bar>y - x\<bar> < d \<longrightarrow> \<bar>y * y - x * x\<bar> < e)"
using isCont_power[OF continuous_ident, of x, unfolded isCont_def LIM_eq, rule_format, of e 2]
- apply (auto simp add: power2_eq_square)
- apply (rule_tac x="s" in exI)
- apply auto
- apply (erule_tac x=y in allE)
- apply auto
- done
+ by (force simp add: power2_eq_square)
text{* Hence derive more interesting properties of the norm. *}
@@ -1594,6 +1589,12 @@
shows "independent S \<Longrightarrow> finite S \<and> card S \<le> DIM('a)"
using independent_span_bound[OF finite_Basis, of S] by auto
+corollary
+ fixes S :: "'a::euclidean_space set"
+ assumes "independent S"
+ shows independent_imp_finite: "finite S" and independent_card_le:"card S \<le> DIM('a)"
+using assms independent_bound by auto
+
lemma dependent_biggerset:
fixes S :: "'a::euclidean_space set"
shows "(finite S \<Longrightarrow> card S > DIM('a)) \<Longrightarrow> dependent S"
@@ -1785,7 +1786,7 @@
shows "(finite S \<Longrightarrow> card S > dim S) \<Longrightarrow> dependent S"
using independent_bound_general[of S] by (metis linorder_not_le)
-lemma dim_span:
+lemma dim_span [simp]:
fixes S :: "'a::euclidean_space set"
shows "dim (span S) = dim S"
proof -```
```--- a/src/HOL/Multivariate_Analysis/Path_Connected.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Multivariate_Analysis/Path_Connected.thy Tue May 26 21:58:04 2015 +0100
@@ -1,5 +1,5 @@
(* Title: HOL/Multivariate_Analysis/Path_Connected.thy
- Author: Robert Himmelmann, TU Muenchen
+ Author: Robert Himmelmann, TU Muenchen, and LCP with material from HOL Light
*)
section {* Continuous paths and path-connected sets *}
@@ -8,7 +8,73 @@
imports Convex_Euclidean_Space
begin
-subsection {* Paths. *}
+(*FIXME move up?*)
+ fixes d::"'a::linordered_idom" shows "(op + d ` {a..b}) = {a+d..b+d}"
+ apply auto
+ apply (rule_tac x="x-d" in rev_image_eqI, auto)
+ done
+
+lemma image_diff_atLeastAtMost [simp]:
+ fixes d::"'a::linordered_idom" shows "(op - d ` {a..b}) = {d-b..d-a}"
+ apply auto
+ apply (rule_tac x="d-x" in rev_image_eqI, auto)
+ done
+
+lemma image_mult_atLeastAtMost [simp]:
+ fixes d::"'a::linordered_field"
+ assumes "d>0" shows "(op * d ` {a..b}) = {d*a..d*b}"
+ using assms
+ apply auto
+ apply (rule_tac x="x/d" in rev_image_eqI)
+ apply (auto simp: field_simps)
+ done
+
+lemma image_affinity_interval:
+ fixes c :: "'a::ordered_real_vector"
+ shows "((\<lambda>x. m *\<^sub>R x + c) ` {a..b}) = (if {a..b}={} then {}
+ else if 0 <= m then {m *\<^sub>R a + c .. m *\<^sub>R b + c}
+ else {m *\<^sub>R b + c .. m *\<^sub>R a + c})"
+ apply (case_tac "m=0", force)
+ apply (auto simp: scaleR_left_mono)
+ apply (rule_tac x="inverse m *\<^sub>R (x-c)" in rev_image_eqI, auto simp: pos_le_divideR_eq le_diff_eq scaleR_left_mono_neg)
+ apply (metis diff_le_eq inverse_inverse_eq order.not_eq_order_implies_strict pos_le_divideR_eq positive_imp_inverse_positive)
+ apply (rule_tac x="inverse m *\<^sub>R (x-c)" in rev_image_eqI, auto simp: not_le neg_le_divideR_eq diff_le_eq)
+ using le_diff_eq scaleR_le_cancel_left_neg
+ apply fastforce
+ done
+
+lemma image_affinity_atLeastAtMost:
+ fixes c :: "'a::linordered_field"
+ shows "((\<lambda>x. m*x + c) ` {a..b}) = (if {a..b}={} then {}
+ else if 0 \<le> m then {m*a + c .. m *b + c}
+ else {m*b + c .. m*a + c})"
+ apply (case_tac "m=0", auto)
+ apply (rule_tac x="inverse m*(x-c)" in rev_image_eqI, auto simp: field_simps)
+ apply (rule_tac x="inverse m*(x-c)" in rev_image_eqI, auto simp: field_simps)
+ done
+
+lemma image_affinity_atLeastAtMost_diff:
+ fixes c :: "'a::linordered_field"
+ shows "((\<lambda>x. m*x - c) ` {a..b}) = (if {a..b}={} then {}
+ else if 0 \<le> m then {m*a - c .. m*b - c}
+ else {m*b - c .. m*a - c})"
+ using image_affinity_atLeastAtMost [of m "-c" a b]
+ by simp
+
+lemma image_affinity_atLeastAtMost_div_diff:
+ fixes c :: "'a::linordered_field"
+ shows "((\<lambda>x. x/m - c) ` {a..b}) = (if {a..b}={} then {}
+ else if 0 \<le> m then {a/m - c .. b/m - c}
+ else {b/m - c .. a/m - c})"
+ using image_affinity_atLeastAtMost_diff [of "inverse m" c a b]
+ by (simp add: field_class.field_divide_inverse algebra_simps)
+
+lemma closed_segment_real_eq:
+ fixes u::real shows "closed_segment u v = (\<lambda>x. (v - u) * x + u) ` {0..1}"
+
+subsection {* Paths and Arcs *}
definition path :: "(real \<Rightarrow> 'a::topological_space) \<Rightarrow> bool"
where "path g \<longleftrightarrow> continuous_on {0..1} g"
@@ -31,17 +97,134 @@
definition simple_path :: "(real \<Rightarrow> 'a::topological_space) \<Rightarrow> bool"
where "simple_path g \<longleftrightarrow>
- (\<forall>x\<in>{0..1}. \<forall>y\<in>{0..1}. g x = g y \<longrightarrow> x = y \<or> x = 0 \<and> y = 1 \<or> x = 1 \<and> y = 0)"
+ path g \<and> (\<forall>x\<in>{0..1}. \<forall>y\<in>{0..1}. g x = g y \<longrightarrow> x = y \<or> x = 0 \<and> y = 1 \<or> x = 1 \<and> y = 0)"
-definition injective_path :: "(real \<Rightarrow> 'a::topological_space) \<Rightarrow> bool"
- where "injective_path g \<longleftrightarrow> (\<forall>x\<in>{0..1}. \<forall>y\<in>{0..1}. g x = g y \<longrightarrow> x = y)"
+definition arc :: "(real \<Rightarrow> 'a :: topological_space) \<Rightarrow> bool"
+ where "arc g \<longleftrightarrow> path g \<and> inj_on g {0..1}"
-subsection {* Some lemmas about these concepts. *}
+subsection{*Invariance theorems*}
+
+lemma path_eq: "path p \<Longrightarrow> (\<And>t. t \<in> {0..1} \<Longrightarrow> p t = q t) \<Longrightarrow> path q"
+ using continuous_on_eq path_def by blast
+
+lemma path_continuous_image: "path g \<Longrightarrow> continuous_on (path_image g) f \<Longrightarrow> path(f o g)"
+ unfolding path_def path_image_def
+ using continuous_on_compose by blast
+
+lemma path_translation_eq:
+ fixes g :: "real \<Rightarrow> 'a :: real_normed_vector"
+ shows "path((\<lambda>x. a + x) o g) = path g"
+proof -
+ have g: "g = (\<lambda>x. -a + x) o ((\<lambda>x. a + x) o g)"
+ by (rule ext) simp
+ show ?thesis
+ unfolding path_def
+ apply safe
+ apply (subst g)
+ apply (rule continuous_on_compose)
+ apply (auto intro: continuous_intros)
+ done
+qed
+
+lemma path_linear_image_eq:
+ fixes f :: "'a::euclidean_space \<Rightarrow> 'b::euclidean_space"
+ assumes "linear f" "inj f"
+ shows "path(f o g) = path g"
+proof -
+ from linear_injective_left_inverse [OF assms]
+ obtain h where h: "linear h" "h \<circ> f = id"
+ by blast
+ then have g: "g = h o (f o g)"
+ by (metis comp_assoc id_comp)
+ show ?thesis
+ unfolding path_def
+ using h assms
+ by (metis g continuous_on_compose linear_continuous_on linear_conv_bounded_linear)
+qed
+
+lemma pathstart_translation: "pathstart((\<lambda>x. a + x) o g) = a + pathstart g"
+
+lemma pathstart_linear_image_eq: "linear f \<Longrightarrow> pathstart(f o g) = f(pathstart g)"
+
+lemma pathfinish_translation: "pathfinish((\<lambda>x. a + x) o g) = a + pathfinish g"
+
+lemma pathfinish_linear_image: "linear f \<Longrightarrow> pathfinish(f o g) = f(pathfinish g)"
+
+lemma path_image_translation: "path_image((\<lambda>x. a + x) o g) = (\<lambda>x. a + x) ` (path_image g)"
+ by (simp add: image_comp path_image_def)
+
+lemma path_image_linear_image: "linear f \<Longrightarrow> path_image(f o g) = f ` (path_image g)"
+ by (simp add: image_comp path_image_def)
+
+lemma reversepath_translation: "reversepath((\<lambda>x. a + x) o g) = (\<lambda>x. a + x) o reversepath g"
+ by (rule ext) (simp add: reversepath_def)
-lemma injective_imp_simple_path: "injective_path g \<Longrightarrow> simple_path g"
- unfolding injective_path_def simple_path_def
- by auto
+lemma reversepath_linear_image: "linear f \<Longrightarrow> reversepath(f o g) = f o reversepath g"
+ by (rule ext) (simp add: reversepath_def)
+
+lemma joinpaths_translation:
+ "((\<lambda>x. a + x) o g1) +++ ((\<lambda>x. a + x) o g2) = (\<lambda>x. a + x) o (g1 +++ g2)"
+ by (rule ext) (simp add: joinpaths_def)
+
+lemma joinpaths_linear_image: "linear f \<Longrightarrow> (f o g1) +++ (f o g2) = f o (g1 +++ g2)"
+ by (rule ext) (simp add: joinpaths_def)
+
+lemma simple_path_translation_eq:
+ fixes g :: "real \<Rightarrow> 'a::euclidean_space"
+ shows "simple_path((\<lambda>x. a + x) o g) = simple_path g"
+ by (simp add: simple_path_def path_translation_eq)
+
+lemma simple_path_linear_image_eq:
+ fixes f :: "'a::euclidean_space \<Rightarrow> 'b::euclidean_space"
+ assumes "linear f" "inj f"
+ shows "simple_path(f o g) = simple_path g"
+ using assms inj_on_eq_iff [of f]
+ by (auto simp: path_linear_image_eq simple_path_def path_translation_eq)
+
+lemma arc_translation_eq:
+ fixes g :: "real \<Rightarrow> 'a::euclidean_space"
+ shows "arc((\<lambda>x. a + x) o g) = arc g"
+ by (auto simp: arc_def inj_on_def path_translation_eq)
+
+lemma arc_linear_image_eq:
+ fixes f :: "'a::euclidean_space \<Rightarrow> 'b::euclidean_space"
+ assumes "linear f" "inj f"
+ shows "arc(f o g) = arc g"
+ using assms inj_on_eq_iff [of f]
+ by (auto simp: arc_def inj_on_def path_linear_image_eq)
+
+
+lemma arc_imp_simple_path: "arc g \<Longrightarrow> simple_path g"
+ by (simp add: arc_def inj_on_def simple_path_def)
+
+lemma arc_imp_path: "arc g \<Longrightarrow> path g"
+ using arc_def by blast
+
+lemma simple_path_imp_path: "simple_path g \<Longrightarrow> path g"
+ using simple_path_def by blast
+
+lemma simple_path_cases: "simple_path g \<Longrightarrow> arc g \<or> pathfinish g = pathstart g"
+ unfolding simple_path_def arc_def inj_on_def pathfinish_def pathstart_def
+ by (force)
+
+lemma simple_path_imp_arc: "simple_path g \<Longrightarrow> pathfinish g \<noteq> pathstart g \<Longrightarrow> arc g"
+ using simple_path_cases by auto
+
+lemma arc_distinct_ends: "arc g \<Longrightarrow> pathfinish g \<noteq> pathstart g"
+ unfolding arc_def inj_on_def pathfinish_def pathstart_def
+ by fastforce
+
+lemma arc_simple_path: "arc g \<longleftrightarrow> simple_path g \<and> pathfinish g \<noteq> pathstart g"
+ using arc_distinct_ends arc_imp_simple_path simple_path_cases by blast
+
+lemma simple_path_eq_arc: "pathfinish g \<noteq> pathstart g \<Longrightarrow> (simple_path g = arc g)"
lemma path_image_nonempty: "path_image g \<noteq> {}"
unfolding path_image_def image_is_empty box_eq_empty
@@ -57,15 +240,11 @@
lemma connected_path_image[intro]: "path g \<Longrightarrow> connected (path_image g)"
unfolding path_def path_image_def
- apply (erule connected_continuous_image)
- apply (rule convex_connected, rule convex_real_interval)
- done
+ using connected_continuous_image connected_Icc by blast
lemma compact_path_image[intro]: "path g \<Longrightarrow> compact (path_image g)"
unfolding path_def path_image_def
- apply (erule compact_continuous_image)
- apply (rule compact_Icc)
- done
+ using compact_continuous_image connected_Icc by blast
lemma reversepath_reversepath[simp]: "reversepath (reversepath g) = g"
unfolding reversepath_def
@@ -91,12 +270,7 @@
proof -
have *: "\<And>g. path_image (reversepath g) \<subseteq> path_image g"
unfolding path_image_def subset_eq reversepath_def Ball_def image_iff
- apply rule
- apply rule
- apply (erule bexE)
- apply (rule_tac x="1 - xa" in bexI)
- apply auto
- done
+ by force
show ?thesis
using *[of g] *[of "reversepath g"]
unfolding reversepath_reversepath
@@ -119,6 +293,28 @@
by (rule iffI)
qed
+lemma arc_reversepath:
+ assumes "arc g" shows "arc(reversepath g)"
+proof -
+ have injg: "inj_on g {0..1}"
+ using assms
+ have **: "\<And>x y::real. 1-x = 1-y \<Longrightarrow> x = y"
+ by simp
+ show ?thesis
+ apply (auto simp: arc_def inj_on_def path_reversepath)
+ apply (simp add: arc_imp_path assms)
+ apply (rule **)
+ apply (rule inj_onD [OF injg])
+ apply (auto simp: reversepath_def)
+ done
+qed
+
+lemma simple_path_reversepath: "simple_path g \<Longrightarrow> simple_path (reversepath g)"
+ apply (force simp: reversepath_def)
+ done
+
lemmas reversepath_simps =
path_reversepath path_image_reversepath pathstart_reversepath pathfinish_reversepath
@@ -162,6 +358,44 @@
done
qed
+section {*Path Images*}
+
+lemma bounded_path_image: "path g \<Longrightarrow> bounded(path_image g)"
+ by (simp add: compact_imp_bounded compact_path_image)
+
+lemma closed_path_image:
+ fixes g :: "real \<Rightarrow> 'a::t2_space"
+ shows "path g \<Longrightarrow> closed(path_image g)"
+ by (metis compact_path_image compact_imp_closed)
+
+lemma connected_simple_path_image: "simple_path g \<Longrightarrow> connected(path_image g)"
+ by (metis connected_path_image simple_path_imp_path)
+
+lemma compact_simple_path_image: "simple_path g \<Longrightarrow> compact(path_image g)"
+ by (metis compact_path_image simple_path_imp_path)
+
+lemma bounded_simple_path_image: "simple_path g \<Longrightarrow> bounded(path_image g)"
+ by (metis bounded_path_image simple_path_imp_path)
+
+lemma closed_simple_path_image:
+ fixes g :: "real \<Rightarrow> 'a::t2_space"
+ shows "simple_path g \<Longrightarrow> closed(path_image g)"
+ by (metis closed_path_image simple_path_imp_path)
+
+lemma connected_arc_image: "arc g \<Longrightarrow> connected(path_image g)"
+ by (metis connected_path_image arc_imp_path)
+
+lemma compact_arc_image: "arc g \<Longrightarrow> compact(path_image g)"
+ by (metis compact_path_image arc_imp_path)
+
+lemma bounded_arc_image: "arc g \<Longrightarrow> bounded(path_image g)"
+ by (metis bounded_path_image arc_imp_path)
+
+lemma closed_arc_image:
+ fixes g :: "real \<Rightarrow> 'a::t2_space"
+ shows "arc g \<Longrightarrow> closed(path_image g)"
+ by (metis closed_path_image arc_imp_path)
+
lemma path_image_join_subset: "path_image (g1 +++ g2) \<subseteq> path_image g1 \<union> path_image g2"
unfolding path_image_def joinpaths_def
by auto
@@ -174,34 +408,16 @@
by auto
lemma path_image_join:
- assumes "pathfinish g1 = pathstart g2"
- shows "path_image (g1 +++ g2) = path_image g1 \<union> path_image g2"
- apply rule
- apply (rule path_image_join_subset)
- apply rule
- unfolding Un_iff
-proof (erule disjE)
- fix x
- assume "x \<in> path_image g1"
- then obtain y where y: "y \<in> {0..1}" "x = g1 y"
- unfolding path_image_def image_iff by auto
- then show "x \<in> path_image (g1 +++ g2)"
- unfolding joinpaths_def path_image_def image_iff
- apply (rule_tac x="(1/2) *\<^sub>R y" in bexI)
- apply auto
- done
-next
- fix x
- assume "x \<in> path_image g2"
- then obtain y where y: "y \<in> {0..1}" "x = g2 y"
- unfolding path_image_def image_iff by auto
- then show "x \<in> path_image (g1 +++ g2)"
- unfolding joinpaths_def path_image_def image_iff
- apply (rule_tac x="(1/2) *\<^sub>R (y + 1)" in bexI)
- using assms(1)[unfolded pathfinish_def pathstart_def]
- done
-qed
+ "pathfinish g1 = pathstart g2 \<Longrightarrow> path_image(g1 +++ g2) = path_image g1 \<union> path_image g2"
+ apply (rule subset_antisym [OF path_image_join_subset])
+ apply (auto simp: pathfinish_def pathstart_def path_image_def joinpaths_def image_def)
+ apply (drule sym)
+ apply (rule_tac x="xa/2" in bexI, auto)
+ apply (rule ccontr)
+ apply (drule_tac x="(xa+1)/2" in bspec)
+ apply (auto simp: field_simps)
+ apply (drule_tac x="1/2" in bspec, auto)
+ done
lemma not_in_path_image_join:
assumes "x \<notin> path_image g1"
@@ -210,166 +426,380 @@
using assms and path_image_join_subset[of g1 g2]
by auto
-lemma simple_path_reversepath:
- assumes "simple_path g"
- shows "simple_path (reversepath g)"
- using assms
- unfolding simple_path_def reversepath_def
- apply -
- apply (rule ballI)+
- apply (erule_tac x="1-x" in ballE)
- apply (erule_tac x="1-y" in ballE)
- apply auto
+lemma pathstart_compose: "pathstart(f o p) = f(pathstart p)"
+
+lemma pathfinish_compose: "pathfinish(f o p) = f(pathfinish p)"
+
+lemma path_image_compose: "path_image (f o p) = f ` (path_image p)"
+ by (simp add: image_comp path_image_def)
+
+lemma path_compose_join: "f o (p +++ q) = (f o p) +++ (f o q)"
+ by (rule ext) (simp add: joinpaths_def)
+
+lemma path_compose_reversepath: "f o reversepath p = reversepath(f o p)"
+ by (rule ext) (simp add: reversepath_def)
+
+lemma join_paths_eq:
+ "(\<And>t. t \<in> {0..1} \<Longrightarrow> p t = p' t) \<Longrightarrow>
+ (\<And>t. t \<in> {0..1} \<Longrightarrow> q t = q' t)
+ \<Longrightarrow> t \<in> {0..1} \<Longrightarrow> (p +++ q) t = (p' +++ q') t"
+ by (auto simp: joinpaths_def)
+
+lemma simple_path_inj_on: "simple_path g \<Longrightarrow> inj_on g {0<..<1}"
+ by (auto simp: simple_path_def path_image_def inj_on_def less_eq_real_def Ball_def)
+
+
+subsection{*Simple paths with the endpoints removed*}
+
+lemma simple_path_endless:
+ "simple_path c \<Longrightarrow> path_image c - {pathstart c,pathfinish c} = c ` {0<..<1}"
+ apply (auto simp: simple_path_def path_image_def pathstart_def pathfinish_def Ball_def Bex_def image_def)
+ apply (metis eq_iff le_less_linear)
+ apply (metis leD linear)
+ using less_eq_real_def zero_le_one apply blast
+ using less_eq_real_def zero_le_one apply blast
done
+lemma connected_simple_path_endless:
+ "simple_path c \<Longrightarrow> connected(path_image c - {pathstart c,pathfinish c})"
+apply (rule connected_continuous_image)
+apply (meson continuous_on_subset greaterThanLessThan_subseteq_atLeastAtMost_iff le_numeral_extra(3) le_numeral_extra(4) path_def simple_path_imp_path)
+by auto
+
+lemma nonempty_simple_path_endless:
+ "simple_path c \<Longrightarrow> path_image c - {pathstart c,pathfinish c} \<noteq> {}"
+
+
+subsection{* The operations on paths*}
+
+lemma path_image_subset_reversepath: "path_image(reversepath g) \<le> path_image g"
+ by (auto simp: path_image_def reversepath_def)
+
+lemma continuous_on_op_minus: "continuous_on (s::real set) (op - x)"
+ by (rule continuous_intros | simp)+
+
+lemma path_imp_reversepath: "path g \<Longrightarrow> path(reversepath g)"
+ apply (auto simp: path_def reversepath_def)
+ using continuous_on_compose [of "{0..1}" "\<lambda>x. 1 - x" g]
+ apply (auto simp: continuous_on_op_minus)
+ done
+
+lemma forall_01_trivial: "(\<forall>x\<in>{0..1}. x \<le> 0 \<longrightarrow> P x) \<longleftrightarrow> P (0::real)"
+ by auto
+
+lemma forall_half1_trivial: "(\<forall>x\<in>{1/2..1}. x * 2 \<le> 1 \<longrightarrow> P x) \<longleftrightarrow> P (1/2::real)"
+ by auto (metis add_divide_distrib mult_2_right real_sum_of_halves)
+
+lemma continuous_on_joinpaths:
+ assumes "continuous_on {0..1} g1" "continuous_on {0..1} g2" "pathfinish g1 = pathstart g2"
+ shows "continuous_on {0..1} (g1 +++ g2)"
+proof -
+ have *: "{0..1::real} = {0..1/2} \<union> {1/2..1}"
+ by auto
+ have gg: "g2 0 = g1 1"
+ by (metis assms(3) pathfinish_def pathstart_def)
+ have 1: "continuous_on {0..1 / 2} (g1 +++ g2)"
+ apply (rule continuous_on_eq [of _ "g1 o (\<lambda>x. 2*x)"])
+ apply (rule continuous_intros | simp add: assms)+
+ done
+ have 2: "continuous_on {1 / 2..1} (g1 +++ g2)"
+ apply (rule continuous_on_eq [of _ "g2 o (\<lambda>x. 2*x-1)"])
+ apply (rule continuous_intros | simp add: forall_half1_trivial gg)+
+ apply (rule continuous_on_subset)
+ apply (rule assms, auto)
+ done
+ show ?thesis
+ apply (subst *)
+ apply (rule continuous_on_union)
+ using 1 2
+ apply auto
+ done
+qed
+
+lemma path_join_imp: "\<lbrakk>path g1; path g2; pathfinish g1 = pathstart g2\<rbrakk> \<Longrightarrow> path(g1 +++ g2)"
+
+lemmas join_paths_simps = path_join path_image_join pathstart_join pathfinish_join
+
lemma simple_path_join_loop:
- assumes "injective_path g1"
- and "injective_path g2"
- and "pathfinish g2 = pathstart g1"
- and "path_image g1 \<inter> path_image g2 \<subseteq> {pathstart g1, pathstart g2}"
- shows "simple_path (g1 +++ g2)"
- unfolding simple_path_def
-proof (intro ballI impI)
- let ?g = "g1 +++ g2"
- note inj = assms(1,2)[unfolded injective_path_def, rule_format]
- fix x y :: real
- assume xy: "x \<in> {0..1}" "y \<in> {0..1}" "?g x = ?g y"
- show "x = y \<or> x = 0 \<and> y = 1 \<or> x = 1 \<and> y = 0"
- proof (cases "x \<le> 1/2", case_tac[!] "y \<le> 1/2", unfold not_le)
- assume as: "x \<le> 1 / 2" "y \<le> 1 / 2"
- then have "g1 (2 *\<^sub>R x) = g1 (2 *\<^sub>R y)"
- using xy(3)
- unfolding joinpaths_def
- by auto
- moreover have "2 *\<^sub>R x \<in> {0..1}" "2 *\<^sub>R y \<in> {0..1}"
- using xy(1,2) as
- by auto
- ultimately show ?thesis
- using inj(1)[of "2*\<^sub>R x" "2*\<^sub>R y"]
- by auto
- next
- assume as: "x > 1 / 2" "y > 1 / 2"
- then have "g2 (2 *\<^sub>R x - 1) = g2 (2 *\<^sub>R y - 1)"
- using xy(3)
- unfolding joinpaths_def
- by auto
- moreover have "2 *\<^sub>R x - 1 \<in> {0..1}" "2 *\<^sub>R y - 1 \<in> {0..1}"
- using xy(1,2) as
- by auto
- ultimately show ?thesis
- using inj(2)[of "2*\<^sub>R x - 1" "2*\<^sub>R y - 1"] by auto
- next
- assume as: "x \<le> 1 / 2" "y > 1 / 2"
- then have "?g x \<in> path_image g1" "?g y \<in> path_image g2"
- unfolding path_image_def joinpaths_def
- using xy(1,2) by auto
- moreover have "?g y \<noteq> pathstart g2"
- using as(2)
- unfolding pathstart_def joinpaths_def
- using inj(2)[of "2 *\<^sub>R y - 1" 0] and xy(2)
- by (auto simp add: field_simps)
- ultimately have *: "?g x = pathstart g1"
- using assms(4)
- unfolding xy(3)
- by auto
- then have "x = 0"
- unfolding pathstart_def joinpaths_def
- using as(1) and xy(1)
- using inj(1)[of "2 *\<^sub>R x" 0]
- by auto
- moreover have "y = 1"
- using *
- unfolding xy(3) assms(3)[symmetric]
- unfolding joinpaths_def pathfinish_def
- using as(2) and xy(2)
- using inj(2)[of "2 *\<^sub>R y - 1" 1]
- by auto
- ultimately show ?thesis
- by auto
- next
- assume as: "x > 1 / 2" "y \<le> 1 / 2"
- then have "?g x \<in> path_image g2" and "?g y \<in> path_image g1"
- unfolding path_image_def joinpaths_def
- using xy(1,2) by auto
- moreover have "?g x \<noteq> pathstart g2"
- using as(1)
- unfolding pathstart_def joinpaths_def
- using inj(2)[of "2 *\<^sub>R x - 1" 0] and xy(1)
- by (auto simp add: field_simps)
- ultimately have *: "?g y = pathstart g1"
- using assms(4)
- unfolding xy(3)
- by auto
- then have "y = 0"
- unfolding pathstart_def joinpaths_def
- using as(2) and xy(2)
- using inj(1)[of "2 *\<^sub>R y" 0]
- by auto
- moreover have "x = 1"
- using *
- unfolding xy(3)[symmetric] assms(3)[symmetric]
- unfolding joinpaths_def pathfinish_def using as(1) and xy(1)
- using inj(2)[of "2 *\<^sub>R x - 1" 1]
- by auto
- ultimately show ?thesis
- by auto
- qed
+ assumes "arc g1" "arc g2"
+ "pathfinish g1 = pathstart g2" "pathfinish g2 = pathstart g1"
+ "path_image g1 \<inter> path_image g2 \<subseteq> {pathstart g1, pathstart g2}"
+ shows "simple_path(g1 +++ g2)"
+proof -
+ have injg1: "inj_on g1 {0..1}"
+ using assms
+ have injg2: "inj_on g2 {0..1}"
+ using assms
+ have g12: "g1 1 = g2 0"
+ and g21: "g2 1 = g1 0"
+ and sb: "g1 ` {0..1} \<inter> g2 ` {0..1} \<subseteq> {g1 0, g2 0}"
+ using assms
+ by (simp_all add: arc_def pathfinish_def pathstart_def path_image_def)
+ { fix x and y::real
+ assume xyI: "x = 1 \<longrightarrow> y \<noteq> 0"
+ and xy: "x \<le> 1" "0 \<le> y" " y * 2 \<le> 1" "\<not> x * 2 \<le> 1" "g2 (2 * x - 1) = g1 (2 * y)"
+ have g1im: "g1 (2 * y) \<in> g1 ` {0..1} \<inter> g2 ` {0..1}"
+ using xy
+ apply simp
+ apply (rule_tac x="2 * x - 1" in image_eqI, auto)
+ done
+ have False
+ using subsetD [OF sb g1im] xy
+ apply auto
+ apply (drule inj_onD [OF injg1])
+ using g21 [symmetric] xyI
+ apply (auto dest: inj_onD [OF injg2])
+ done
+ } note * = this
+ { fix x and y::real
+ assume xy: "y \<le> 1" "0 \<le> x" "\<not> y * 2 \<le> 1" "x * 2 \<le> 1" "g1 (2 * x) = g2 (2 * y - 1)"
+ have g1im: "g1 (2 * x) \<in> g1 ` {0..1} \<inter> g2 ` {0..1}"
+ using xy
+ apply simp
+ apply (rule_tac x="2 * x" in image_eqI, auto)
+ done
+ have "x = 0 \<and> y = 1"
+ using subsetD [OF sb g1im] xy
+ apply auto
+ apply (force dest: inj_onD [OF injg1])
+ using g21 [symmetric]
+ apply (auto dest: inj_onD [OF injg2])
+ done
+ } note ** = this
+ show ?thesis
+ using assms
+ apply (simp add: arc_def simple_path_def path_join, clarify)
+ apply (simp add: joinpaths_def split: split_if_asm)
+ apply (force dest: inj_onD [OF injg1])
+ apply (metis *)
+ apply (metis **)
+ apply (force dest: inj_onD [OF injg2])
+ done
+qed
+
+lemma arc_join:
+ assumes "arc g1" "arc g2"
+ "pathfinish g1 = pathstart g2"
+ "path_image g1 \<inter> path_image g2 \<subseteq> {pathstart g2}"
+ shows "arc(g1 +++ g2)"
+proof -
+ have injg1: "inj_on g1 {0..1}"
+ using assms
+ have injg2: "inj_on g2 {0..1}"
+ using assms
+ have g11: "g1 1 = g2 0"
+ and sb: "g1 ` {0..1} \<inter> g2 ` {0..1} \<subseteq> {g2 0}"
+ using assms
+ by (simp_all add: arc_def pathfinish_def pathstart_def path_image_def)
+ { fix x and y::real
+ assume xy: "x \<le> 1" "0 \<le> y" " y * 2 \<le> 1" "\<not> x * 2 \<le> 1" "g2 (2 * x - 1) = g1 (2 * y)"
+ have g1im: "g1 (2 * y) \<in> g1 ` {0..1} \<inter> g2 ` {0..1}"
+ using xy
+ apply simp
+ apply (rule_tac x="2 * x - 1" in image_eqI, auto)
+ done
+ have False
+ using subsetD [OF sb g1im] xy
+ by (auto dest: inj_onD [OF injg2])
+ } note * = this
+ show ?thesis
+ apply (simp add: arc_def inj_on_def)
+ apply (clarsimp simp add: arc_imp_path assms path_join)
+ apply (simp add: joinpaths_def split: split_if_asm)
+ apply (force dest: inj_onD [OF injg1])
+ apply (metis *)
+ apply (metis *)
+ apply (force dest: inj_onD [OF injg2])
+ done
+qed
+
+lemma reversepath_joinpaths:
+ "pathfinish g1 = pathstart g2 \<Longrightarrow> reversepath(g1 +++ g2) = reversepath g2 +++ reversepath g1"
+ unfolding reversepath_def pathfinish_def pathstart_def joinpaths_def
+ by (rule ext) (auto simp: mult.commute)
+
+
+subsection{* Choosing a subpath of an existing path*}
+
+definition subpath :: "real \<Rightarrow> real \<Rightarrow> (real \<Rightarrow> 'a) \<Rightarrow> real \<Rightarrow> 'a::real_normed_vector"
+ where "subpath a b g \<equiv> \<lambda>x. g((b - a) * x + a)"
+
+lemma path_image_subpath_gen [simp]:
+ fixes g :: "real \<Rightarrow> 'a::real_normed_vector"
+ shows "path_image(subpath u v g) = g ` (closed_segment u v)"
+ apply (simp add: closed_segment_real_eq path_image_def subpath_def)
+ apply (subst o_def [of g, symmetric])
+ apply (simp add: image_comp [symmetric])
+ done
+
+lemma path_image_subpath [simp]:
+ fixes g :: "real \<Rightarrow> 'a::real_normed_vector"
+ shows "path_image(subpath u v g) = (if u \<le> v then g ` {u..v} else g ` {v..u})"
+
+lemma path_subpath [simp]:
+ fixes g :: "real \<Rightarrow> 'a::real_normed_vector"
+ assumes "path g" "u \<in> {0..1}" "v \<in> {0..1}"
+ shows "path(subpath u v g)"
+proof -
+ have "continuous_on {0..1} (g o (\<lambda>x. ((v-u) * x+ u)))"
+ apply (rule continuous_intros | simp)+
+ apply (simp add: image_affinity_atLeastAtMost [where c=u])
+ using assms
+ apply (auto simp: path_def continuous_on_subset)
+ done
+ then show ?thesis
+ by (simp add: path_def subpath_def)
qed
-lemma injective_path_join:
- assumes "injective_path g1"
- and "injective_path g2"
- and "pathfinish g1 = pathstart g2"
- and "path_image g1 \<inter> path_image g2 \<subseteq> {pathstart g2}"
- shows "injective_path (g1 +++ g2)"
- unfolding injective_path_def
-proof (rule, rule, rule)
- let ?g = "g1 +++ g2"
- note inj = assms(1,2)[unfolded injective_path_def, rule_format]
- fix x y
- assume xy: "x \<in> {0..1}" "y \<in> {0..1}" "(g1 +++ g2) x = (g1 +++ g2) y"
- show "x = y"
- proof (cases "x \<le> 1/2", case_tac[!] "y \<le> 1/2", unfold not_le)
- assume "x \<le> 1 / 2" and "y \<le> 1 / 2"
- then show ?thesis
- using inj(1)[of "2*\<^sub>R x" "2*\<^sub>R y"] and xy
- unfolding joinpaths_def by auto
- next
- assume "x > 1 / 2" and "y > 1 / 2"
- then show ?thesis
- using inj(2)[of "2*\<^sub>R x - 1" "2*\<^sub>R y - 1"] and xy
- unfolding joinpaths_def by auto
- next
- assume as: "x \<le> 1 / 2" "y > 1 / 2"
- then have "?g x \<in> path_image g1" and "?g y \<in> path_image g2"
- unfolding path_image_def joinpaths_def
- using xy(1,2)
- by auto
- then have "?g x = pathfinish g1" and "?g y = pathstart g2"
- using assms(4)
- unfolding assms(3) xy(3)
- by auto
- then show ?thesis
- using as and inj(1)[of "2 *\<^sub>R x" 1] inj(2)[of "2 *\<^sub>R y - 1" 0] and xy(1,2)
- unfolding pathstart_def pathfinish_def joinpaths_def
- by auto
- next
- assume as:"x > 1 / 2" "y \<le> 1 / 2"
- then have "?g x \<in> path_image g2" and "?g y \<in> path_image g1"
- unfolding path_image_def joinpaths_def
- using xy(1,2)
- by auto
- then have "?g x = pathstart g2" and "?g y = pathfinish g1"
- using assms(4)
- unfolding assms(3) xy(3)
- by auto
- then show ?thesis using as and inj(2)[of "2 *\<^sub>R x - 1" 0] inj(1)[of "2 *\<^sub>R y" 1] and xy(1,2)
- unfolding pathstart_def pathfinish_def joinpaths_def
- by auto
- qed
+lemma pathstart_subpath [simp]: "pathstart(subpath u v g) = g(u)"
+ by (simp add: pathstart_def subpath_def)
+
+lemma pathfinish_subpath [simp]: "pathfinish(subpath u v g) = g(v)"
+ by (simp add: pathfinish_def subpath_def)
+
+lemma subpath_trivial [simp]: "subpath 0 1 g = g"
+
+lemma subpath_reversepath: "subpath 1 0 g = reversepath g"
+ by (simp add: reversepath_def subpath_def)
+
+lemma reversepath_subpath: "reversepath(subpath u v g) = subpath v u g"
+ by (simp add: reversepath_def subpath_def algebra_simps)
+
+lemma subpath_translation: "subpath u v ((\<lambda>x. a + x) o g) = (\<lambda>x. a + x) o subpath u v g"
+ by (rule ext) (simp add: subpath_def)
+
+lemma subpath_linear_image: "linear f \<Longrightarrow> subpath u v (f o g) = f o subpath u v g"
+ by (rule ext) (simp add: subpath_def)
+
+lemma affine_ineq:
+ fixes x :: "'a::linordered_idom"
+ assumes "x \<le> 1" "v < u"
+ shows "v + x * u \<le> u + x * v"
+proof -
+ have "(1-x)*(u-v) \<ge> 0"
+ using assms by auto
+ then show ?thesis
qed
-lemmas join_paths_simps = path_join path_image_join pathstart_join pathfinish_join
+lemma simple_path_subpath_eq:
+ "simple_path(subpath u v g) \<longleftrightarrow>
+ path(subpath u v g) \<and> u\<noteq>v \<and>
+ (\<forall>x y. x \<in> closed_segment u v \<and> y \<in> closed_segment u v \<and> g x = g y
+ \<longrightarrow> x = y \<or> x = u \<and> y = v \<or> x = v \<and> y = u)"
+ (is "?lhs = ?rhs")
+proof (rule iffI)
+ assume ?lhs
+ then have p: "path (\<lambda>x. g ((v - u) * x + u))"
+ and sim: "(\<And>x y. \<lbrakk>x\<in>{0..1}; y\<in>{0..1}; g ((v - u) * x + u) = g ((v - u) * y + u)\<rbrakk>
+ \<Longrightarrow> x = y \<or> x = 0 \<and> y = 1 \<or> x = 1 \<and> y = 0)"
+ by (auto simp: simple_path_def subpath_def)
+ { fix x y
+ assume "x \<in> closed_segment u v" "y \<in> closed_segment u v" "g x = g y"
+ then have "x = y \<or> x = u \<and> y = v \<or> x = v \<and> y = u"
+ using sim [of "(x-u)/(v-u)" "(y-u)/(v-u)"] p
+ by (auto simp: closed_segment_real_eq image_affinity_atLeastAtMost divide_simps
+ split: split_if_asm)
+ } moreover
+ have "path(subpath u v g) \<and> u\<noteq>v"
+ using sim [of "1/3" "2/3"] p
+ by (auto simp: subpath_def)
+ ultimately show ?rhs
+ by metis
+next
+ assume ?rhs
+ then
+ have d1: "\<And>x y. \<lbrakk>g x = g y; u \<le> x; x \<le> v; u \<le> y; y \<le> v\<rbrakk> \<Longrightarrow> x = y \<or> x = u \<and> y = v \<or> x = v \<and> y = u"
+ and d2: "\<And>x y. \<lbrakk>g x = g y; v \<le> x; x \<le> u; v \<le> y; y \<le> u\<rbrakk> \<Longrightarrow> x = y \<or> x = u \<and> y = v \<or> x = v \<and> y = u"
+ and ne: "u < v \<or> v < u"
+ and psp: "path (subpath u v g)"
+ by (auto simp: closed_segment_real_eq image_affinity_atLeastAtMost)
+ have [simp]: "\<And>x. u + x * v = v + x * u \<longleftrightarrow> u=v \<or> x=1"
+ by algebra
+ show ?lhs using psp ne
+ unfolding simple_path_def subpath_def
+ by (fastforce simp add: algebra_simps affine_ineq mult_left_mono crossproduct_eq dest: d1 d2)
+qed
+
+lemma arc_subpath_eq:
+ "arc(subpath u v g) \<longleftrightarrow> path(subpath u v g) \<and> u\<noteq>v \<and> inj_on g (closed_segment u v)"
+ (is "?lhs = ?rhs")
+proof (rule iffI)
+ assume ?lhs
+ then have p: "path (\<lambda>x. g ((v - u) * x + u))"
+ and sim: "(\<And>x y. \<lbrakk>x\<in>{0..1}; y\<in>{0..1}; g ((v - u) * x + u) = g ((v - u) * y + u)\<rbrakk>
+ \<Longrightarrow> x = y)"
+ by (auto simp: arc_def inj_on_def subpath_def)
+ { fix x y
+ assume "x \<in> closed_segment u v" "y \<in> closed_segment u v" "g x = g y"
+ then have "x = y"
+ using sim [of "(x-u)/(v-u)" "(y-u)/(v-u)"] p
+ by (force simp add: inj_on_def closed_segment_real_eq image_affinity_atLeastAtMost divide_simps
+ split: split_if_asm)
+ } moreover
+ have "path(subpath u v g) \<and> u\<noteq>v"
+ using sim [of "1/3" "2/3"] p
+ by (auto simp: subpath_def)
+ ultimately show ?rhs
+ unfolding inj_on_def
+ by metis
+next
+ assume ?rhs
+ then
+ have d1: "\<And>x y. \<lbrakk>g x = g y; u \<le> x; x \<le> v; u \<le> y; y \<le> v\<rbrakk> \<Longrightarrow> x = y"
+ and d2: "\<And>x y. \<lbrakk>g x = g y; v \<le> x; x \<le> u; v \<le> y; y \<le> u\<rbrakk> \<Longrightarrow> x = y"
+ and ne: "u < v \<or> v < u"
+ and psp: "path (subpath u v g)"
+ by (auto simp: inj_on_def closed_segment_real_eq image_affinity_atLeastAtMost)
+ show ?lhs using psp ne
+ unfolding arc_def subpath_def inj_on_def
+ by (auto simp: algebra_simps affine_ineq mult_left_mono crossproduct_eq dest: d1 d2)
+qed
+
+
+lemma simple_path_subpath:
+ assumes "simple_path g" "u \<in> {0..1}" "v \<in> {0..1}" "u \<noteq> v"
+ shows "simple_path(subpath u v g)"
+ using assms
+ apply (simp add: simple_path_subpath_eq simple_path_imp_path)
+ apply (simp add: simple_path_def closed_segment_real_eq image_affinity_atLeastAtMost, fastforce)
+ done
+
+lemma arc_simple_path_subpath:
+ "\<lbrakk>simple_path g; u \<in> {0..1}; v \<in> {0..1}; g u \<noteq> g v\<rbrakk> \<Longrightarrow> arc(subpath u v g)"
+ by (force intro: simple_path_subpath simple_path_imp_arc)
+
+lemma arc_subpath_arc:
+ "\<lbrakk>arc g; u \<in> {0..1}; v \<in> {0..1}; u \<noteq> v\<rbrakk> \<Longrightarrow> arc(subpath u v g)"
+ by (meson arc_def arc_imp_simple_path arc_simple_path_subpath inj_onD)
+
+lemma arc_simple_path_subpath_interior:
+ "\<lbrakk>simple_path g; u \<in> {0..1}; v \<in> {0..1}; u \<noteq> v; \<bar>u-v\<bar> < 1\<rbrakk> \<Longrightarrow> arc(subpath u v g)"
+ apply (rule arc_simple_path_subpath)
+ apply (force simp: simple_path_def)+
+ done
+
+lemma path_image_subpath_subset:
+ "\<lbrakk>path g; u \<in> {0..1}; v \<in> {0..1}\<rbrakk> \<Longrightarrow> path_image(subpath u v g) \<subseteq> path_image g"
+ apply (simp add: closed_segment_real_eq image_affinity_atLeastAtMost)
+ apply (auto simp: path_image_def)
+ done
+
+lemma join_subpaths_middle: "subpath (0) ((1 / 2)) p +++ subpath ((1 / 2)) 1 p = p"
+ by (rule ext) (simp add: joinpaths_def subpath_def divide_simps)
subsection {* Reparametrizing a closed curve to start at some chosen point *}
@@ -500,22 +930,15 @@
lemma path_image_linepath[simp]: "path_image (linepath a b) = closed_segment a b"
unfolding path_image_def segment linepath_def
- apply (rule set_eqI)
- apply rule
- defer
- unfolding mem_Collect_eq image_iff
- apply (erule exE)
- apply (rule_tac x="u *\<^sub>R 1" in bexI)
- apply auto
- done
+ by auto
lemma reversepath_linepath[simp]: "reversepath (linepath a b) = linepath b a"
unfolding reversepath_def linepath_def
by auto
-lemma injective_path_linepath:
+lemma arc_linepath:
assumes "a \<noteq> b"
- shows "injective_path (linepath a b)"
+ shows "arc (linepath a b)"
proof -
{
fix x y :: "real"
@@ -526,12 +949,12 @@
by simp
}
then show ?thesis
- unfolding injective_path_def linepath_def
- by (auto simp add: algebra_simps)
+ unfolding arc_def inj_on_def
+ by (simp add: path_linepath) (force simp: algebra_simps linepath_def)
qed
lemma simple_path_linepath[intro]: "a \<noteq> b \<Longrightarrow> simple_path (linepath a b)"
- by (auto intro!: injective_imp_simple_path injective_path_linepath)
+ by (simp add: arc_imp_simple_path arc_linepath)
subsection {* Bounding a point away from a path *}
@@ -623,29 +1046,16 @@
lemma path_component_set:
"{y. path_component s x y} =
{y. (\<exists>g. path g \<and> path_image g \<subseteq> s \<and> pathstart g = x \<and> pathfinish g = y)}"
- apply (rule set_eqI)
- unfolding mem_Collect_eq
- unfolding path_component_def
+ unfolding mem_Collect_eq path_component_def
apply auto
done
lemma path_component_subset: "{y. path_component s x y} \<subseteq> s"
- apply rule
- apply (rule path_component_mem(2))
- apply auto
- done
+ by (auto simp add: path_component_mem(2))
lemma path_component_eq_empty: "{y. path_component s x y} = {} \<longleftrightarrow> x \<notin> s"
- apply rule
- apply (drule equals0D[of _ x])
- defer
- apply (rule equals0I)
- unfolding mem_Collect_eq
- apply (drule path_component_mem(1))
- using path_component_refl
- apply auto
- done
-
+ using path_component_mem path_component_refl_eq
+ by fastforce
subsection {* Path connectedness of a space *}
@@ -656,15 +1066,9 @@
unfolding path_connected_def path_component_def by auto
lemma path_connected_component_set: "path_connected s \<longleftrightarrow> (\<forall>x\<in>s. {y. path_component s x y} = s)"
- unfolding path_connected_component
- apply rule
- apply rule
- apply rule
- apply (rule path_component_subset)
- unfolding subset_eq mem_Collect_eq Ball_def
+ unfolding path_connected_component path_component_subset
apply auto
- done
-
+ using path_component_mem(2) by blast
subsection {* Some useful lemmas about path-connectedness *}
```
```--- a/src/HOL/Real_Vector_Spaces.thy Mon May 25 22:52:17 2015 +0200
+++ b/src/HOL/Real_Vector_Spaces.thy Tue May 26 21:58:04 2015 +0100
@@ -518,6 +518,13 @@
end
+lemma neg_le_divideR_eq:
+ fixes a :: "'a :: ordered_real_vector"
+ assumes "c < 0"
+ shows "a \<le> b /\<^sub>R c \<longleftrightarrow> b \<le> c *\<^sub>R a"
+ using pos_le_divideR_eq [of "-c" a "-b"] assms
+ by simp
+
lemma scaleR_nonneg_nonneg: "0 \<le> a \<Longrightarrow> 0 \<le> (x::'a::ordered_real_vector) \<Longrightarrow> 0 \<le> a *\<^sub>R x"
using scaleR_left_mono [of 0 x a]
by simp``` | 17,195 | 51,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-49 | latest | en | 0.782705 |
https://nl.mathworks.com/matlabcentral/cody/problems/44492-approximate-the-cosine-function/solutions/1559061 | 1,591,267,686,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00064.warc.gz | 453,531,358 | 15,830 | Cody
Problem 44492. Approximate the cosine function
Solution 1559061
Submitted on 13 Jun 2018 by yuval arad
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
filetext = fileread('myCos.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden')
2 Pass
filetext = fileread('myCos.m'); trigUsed = any(strfind(filetext, 'cos')) || any(strfind(filetext, 'sin')) ||... any(strfind(filetext, 'exp')); assert(~trigUsed, 'Cannot use MATLAB trigonometric functions')
3 Pass
x = 0; assert(abs(myCos(x)-cos(x)) < 0.0001)
4 Pass
x = pi; assert(abs(myCos(x)-cos(x)) < 0.0001)
5 Pass
x = pi/2; assert(abs(myCos(x)-cos(x)) < 0.0001)
6 Pass
x = 5*pi/3; assert(abs(myCos(x)-cos(x)) < 0.0001) | 266 | 825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-24 | latest | en | 0.377736 |
https://computergraphics.stackexchange.com/questions/218/overlaying-graphics-on-particle-simulation | 1,701,160,595,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099281.67/warc/CC-MAIN-20231128083443-20231128113443-00704.warc.gz | 211,358,710 | 43,707 | # Overlaying graphics on particle simulation
How can I overlay graphics on particle simulations? For instance if I am representing a liquid or gas with tons of little points how can I make those points look like a liquid or gas? Take Nvidia Flex for instance, they simulate liquids and gases with tons of small points but can render them as realistic liquids and gases instead of small balls.
• Have you tried a closed surface with control points, i.e. nurbs or bspline? or free form deformation stuff? both of these are described by using point, but moving a point you deforms the surface described. (In the mean while i try to gather more info on the problem). I was even thinking to convex hull, but i'm not sure of the result, since a liquid deformation could be not convex at all. Aug 12, 2015 at 10:08
For rendering of gases, I think the usual approach is to simply render each particle as a tiny disc. Gases don't really coalesce into surfaces like liquids do, so this should produce acceptable results. You could perhaps apply a light blur over the gas layer afterwards to soften it and hide the fact that it is made of discrete elements.
Liquids, on the other hand, tend to coalesce together to form droplets and smooth surfaces, so you need to derive a surface from the particles somehow. One way to do this is to use Metaballs, which also display this behavior and can be tweaked to suit different liquids and particle densities. By interpreting each particle as a metaball, you will have an implicit equation representing your liquid surface. To render this implicit surface you can then use an algorithm like Marching Tetrahedra to convert it to triangles, or utilize Ray Marching to directly render it. (Ray marching can be easily done in realtime in a fragment shader these days.) You can also use this approach for gases if you want a somewhat softer look.
A good reference for game physics is this chapter 4 describes the basic free deformable surfaces (nurbs and bspline are of course cited and treated enough well) fluid and gases are instead treated in chapter five (basically the author derivate simplified model of navier stokes equations, suitable for real time applications).
So actually i guess what i've written commenting your post was correct, i.e. combine the physical deformation of the point that controls the shape of the surface with a ffd computations.
The book i cited should provide somewhere/somehow on the web source code, and itself it cites some example involving the technique i cited. If that is what you need let me know for a more specific explanation. | 540 | 2,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-50 | longest | en | 0.938108 |
https://www.javaguides.net/2024/06/java-math-floordiv-method.html | 1,721,927,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00012.warc.gz | 702,789,944 | 43,148 | Java Math floorDiv() Method
The `Math.floorDiv()` method in Java is used to perform integer division and return the largest (closest to positive infinity) value that is less than or equal to the algebraic quotient.
1. Introduction
2. `floorDiv()` Method Syntax
3. Overloaded `floorDiv()` Methods
4. Examples
• `floorDiv(int x, int y)`
• `floorDiv(long x, int y)`
• `floorDiv(long x, long y)`
5. Real-World Use Case
6. Conclusion
Introduction
The `Math.floorDiv()` method provides a way to perform integer division and ensure that the result is the largest integer that is less than or equal to the algebraic quotient. This method is useful in scenarios where you need to perform division and ensure that the result is rounded towards negative infinity.
floorDiv() Method Syntax
The syntax for the `floorDiv()` method varies depending on the types of the arguments:
`floorDiv(int x, int y)`
``````public static int floorDiv(int x, int y)
``````
`floorDiv(long x, int y)`
``````public static long floorDiv(long x, int y)
``````
`floorDiv(long x, long y)`
``````public static long floorDiv(long x, long y)
``````
Parameters:
• `x`: The dividend.
• `y`: The divisor.
Returns:
• The largest integer value that is less than or equal to the algebraic quotient.
Throws:
• `ArithmeticException` if the divisor `y` is zero.
The `Math.floorDiv()` method is overloaded to handle different combinations of primitive data types: `int` and `long`. Each version returns the floor value of the division.
Examples
`floorDiv(int x, int y)`
The `floorDiv(int x, int y)` method returns the largest integer value that is less than or equal to the algebraic quotient of two `int` values.
Example
``````public class FloorDivIntExample {
public static void main(String[] args) {
int x1 = 7, y1 = 3;
int x2 = -7, y2 = 3;
int result1 = Math.floorDiv(x1, y1);
int result2 = Math.floorDiv(x2, y2);
System.out.println("Floor division of " + x1 + " / " + y1 + " is " + result1);
System.out.println("Floor division of " + x2 + " / " + y2 + " is " + result2);
}
}
``````
Output:
``````Floor division of 7 / 3 is 2
Floor division of -7 / 3 is -3
``````
`floorDiv(long x, int y)`
The `floorDiv(long x, int y)` method returns the largest integer value that is less than or equal to the algebraic quotient of a `long` and an `int` value.
Example
``````public class FloorDivLongIntExample {
public static void main(String[] args) {
long x1 = 10L, y1 = 3;
long x2 = -10L, y2 = 3;
long result1 = Math.floorDiv(x1, y1);
long result2 = Math.floorDiv(x2, y2);
System.out.println("Floor division of " + x1 + " / " + y1 + " is " + result1);
System.out.println("Floor division of " + x2 + " / " + y2 + " is " + result2);
}
}
``````
Output:
``````Floor division of 10 / 3 is 3
Floor division of -10 / 3 is -4
``````
`floorDiv(long x, long y)`
The `floorDiv(long x, long y)` method returns the largest integer value that is less than or equal to the algebraic quotient of two `long` values.
Example
``````public class FloorDivLongLongExample {
public static void main(String[] args) {
long x1 = 20L, y1 = 4L;
long x2 = -20L, y2 = 4L;
long result1 = Math.floorDiv(x1, y1);
long result2 = Math.floorDiv(x2, y2);
System.out.println("Floor division of " + x1 + " / " + y1 + " is " + result1);
System.out.println("Floor division of " + x2 + " / " + y2 + " is " + result2);
}
}
``````
Output:
``````Floor division of 20 / 4 is 5
Floor division of -20 / 4 is -5
``````
Real-World Use Case
Calculating Page Numbers in Pagination
In real-world scenarios, the `Math.floorDiv()` method can be used to calculate page numbers in pagination, ensuring that the result is always rounded towards negative infinity.
Example
``````public class PaginationExample {
public static void main(String[] args) {
int totalItems = 23;
int itemsPerPage = 5;
int totalPages = Math.floorDiv(totalItems, itemsPerPage) + (totalItems % itemsPerPage == 0 ? 0 : 1);
System.out.println("Total pages needed: " + totalPages);
}
}
``````
Output:
``````Total pages needed: 5
``````
Conclusion
The `Math.floorDiv()` method in Java provides a way to perform integer division with rounding towards negative infinity, ensuring that the result is the largest integer less than or equal to the algebraic quotient.
By understanding how to use this method and its overloaded versions, you can handle various division operations and solve problems that require floor division.
Whether you are working with integers or long integers, the `floorDiv()` method offers a reliable tool for ensuring correct division results. | 1,245 | 4,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-30 | latest | en | 0.684305 |
http://slidegur.com/doc/1779/systems--3-var- | 1,503,517,066,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123359.11/warc/CC-MAIN-20170823190745-20170823210745-00236.warc.gz | 396,187,421 | 7,805 | ### systems (3 var)
```3.6 Solving Systems of Linear
Equations in 3 Variables
p. 177
A system of lin. eqns. in 3 variables
Looks something like:
x+3y-z=-11
2x+y+z=1
5x-2y+3z=21
A solution is an ordered triple (x,y,z) that
makes all 3 equations true.
Steps for solving in 3 variables
1. Using the 1st 2 equations, cancel one of
the variables.
2. Using the last 2 equations, cancel the
same variable from step 1.
3. Use the results of steps 1 & 2 to solve for
the 2 remaining variables.
4. Plug the results from step 3 into one of
the original 3 equations and solve for the
3rd remaining variable.
5. Write the solution as an ordered triple
(x,y,z).
Solve the system.
1. x+3y-z=-11
2x+y+z=1
z’s are easy to cancel!
3x+4y=-10
2. 2x+y+z=1
5x-2y+3z=21
Must cancel z’s again!
-6x-3y-3z=-3
5x-2y+3z=21
-x-5y=18
2(2)+(-4)+z=1
4-4+z=1
z=1
x+3y-z=-11
2x+y+z=1
5x-2y+3z=21
3. 3x+4y=-10
-x-5y=18
Solve for x & y.
3x+4y=-10
-3x-15y+54
-11y=44
y=- 4
3x+4(-4)=-10
x=2
(2, - 4, 1)
Solve the system.
-x+2y+z=3
2x+2y+z=5
4x+4y+2z=6
2. 2x+2y+z=5
1. -x+2y+z=3
4x+4y+2z=6
2x+2y+z=5
z’s are easy to cancel! Cancel z’s again.
-4x-4y-2z=-10
-x+2y+z=3
4x+4y+2z=6
-2x-2y-z=-5
0=- 4
-3x=-2
Doesn’t make sense!
x=2/3
No solution
Solve the system.
1. -2x+4y+z=1
3x-3y-z=2
z’s are easy to cancel!
x+y=3
2. 3x-3y-z=2
5x-y-z=8
Cancel z’s again.
-3x+3y+z=-2
5x-y-z=8
2x+2y=6
-2x+4y+z=1
3x-3y-z=2
5x-y-z=8
3. x+y=3
2x+2y=6
Cancel the x’s.
-2x-2y=-6
2x+2y=6
0=0
This is true.
¸ many solutions
Assignment
``` | 740 | 1,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2017-34 | longest | en | 0.638848 |
https://puzzling.stackexchange.com/questions/78866/discrete-peaceful-encampments-player-4-has-entered-the-game | 1,726,624,840,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00386.warc.gz | 430,164,707 | 43,178 | # Discrete Peaceful Encampments: Player 4 has entered the game!
Here's a variation of Discrete Peaceful Encampments: Player 3 has entered the game! (which itself is a variation of Peaceful Encampments).
You have 3 white queens, 3 black queens, 3 red queens, and 3 green queens. Place all these pieces onto a normal 8x8 chessboard in such a way that no queen threatens a queen of a different color.
Okay, that was easier than the previous variations, right? You can probably use the pattern you found to solve these problems as well:
Place 5 queens of each of four different colors onto a 10x10 checkerboard so that no queen threatens a queen of a different color.
Place 7 queens of each of four different colors onto a 12x12 checkerboard so that no queen threatens a queen of a different color.
Which leads to the real puzzle:
At what point does it become possible to place more than $$N-5$$ queens of each of four different colors peacefully onto an $$N\times N$$ checkerboard?
• I thought I had it :P lichess.org/editor/2QQ4/2Q5/6qq/6q1/QQ6/Q7/4qq2/5q2_w_-_- Commented Jan 26, 2019 at 1:16
• @Yout Ried, D8-G5? Commented Jan 26, 2019 at 3:16
• If I pruned the output right, there are $13$ distinct solutions to four armies of $3$ on the $8\times8$ board, including a couple where one army gets a fourth queen. Some are deceptively easy! Commented Jan 26, 2019 at 6:23
• @athin I know... Commented Jan 26, 2019 at 16:07
• I solved for 8x8 lichess.org/editor/2Q5/2QQ4/6qq/6q1/1Q6/QQ6/4qq2/5q2_w_-_- Commented Jan 26, 2019 at 16:15
While I feel it could be, this may not be optimal. However it is, at least, an upper bound...
$$N=14$$.
I first note that it must be possible with:
$$N=25$$.
Since $$N-4=21=(6+5+4+3+2+1)$$
and $$\frac{N}{4}\ge 6$$
it follows that we can build a symmetric solution where each army occupies a right-isosceles-triangle of side $$6$$.
Like so (the green shading shows the locations under attack by army A):
...and the layout has $$4$$-fold rotational symmetry,
so rotating a quarter or half-turn, other armies are just like A.
...and then note that:
we can squeeze the same armies of $$21$$ onto a $$24 \times 24$$ board, by removing the central column & middle row:
and that we can
...remove the bottom-right soldier of army A (and equivalents), squeeze that into a $$23 \times 23$$ as above, then remove the outermost three rings of locations
...for four armies of $$14$$ on a $$17 \times 17$$ board:
Similarly
...remove the three bottom-right soldiers of army A (and equivalents) and squeeze to make armies of $$11$$ on a $$15 \times 15$$ board:
...and (thanks to Daniel Mathias!)
...remove the top-right soldier of army A (and equivalents) and squeeze to make armies of $$10$$ on a $$14 \times 14$$ board:
There exists no solution for $$4$$ armies of $$9$$ queens each on a $$13\times13$$ board, so Jonathan's $$10$$ on $$14$$ is the first with armies of $$N-4$$ queens.
Here are all of the distinct solutions for armies of $$4$$ queens on a $$9\times9$$ board and armies of $$5$$ queens on a $$10\times10$$ board. (Exception: any queen can be moved to the center square of the fourth 9x9 board.)
These are the only solutions for armies of $$6$$ queens on an $$11\times11$$ board. For the $$12\times12$$ and larger boards, the search was modified to find only rotationally symmetrical solutions.
Optimal solutions on $$16\times16$$ and $$18\times18$$ boards:
These solutions can be easily expanded to ever larger boards, with armies of $$18$$ queens on a $$19\times19$$ board and armies of $$20$$ queens on a $$20\times20$$ board. The claim that these are optimal is based on the fact that there were no centrally located queens in any solution on $$10\times10$$ or $$11\times11$$ boards. This suggests that your general solution for $$4$$ armies is itself optimal.
I haven’t completed my proof yet, but I believe from preliminary results:
12 is the lower bound. Whether 12 or 13 works is yet to be seen. | 1,099 | 3,964 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 39, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-38 | latest | en | 0.89153 |
https://edurev.in/course/quiz/attempt/15659_Test-Bills-Of-Exchange-And-Promissory-3/506290c7-4d53-46b8-bc5d-673e31966801 | 1,653,076,261,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534669.47/warc/CC-MAIN-20220520191810-20220520221810-00109.warc.gz | 287,265,301 | 47,190 | Test: Bills Of Exchange And Promissory - 3
40 Questions MCQ Test Crash Course of Accountancy - Class 11 | Test: Bills Of Exchange And Promissory - 3
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Attempt Test: Bills Of Exchange And Promissory - 3 | 40 questions in 40 minutes | Mock test for CA Foundation preparation | Free important questions MCQ to study Crash Course of Accountancy - Class 11 for CA Foundation Exam | Download free PDF with solutions
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
On 16.6.05 X draws a bill on Y for Rs 25,000 for 30 days. 19th July is a public holiday, due date of the bill will be:
Solution:
QUESTION: 4
X draws a bill on Y for Rs 30,000 on 1.1.05. X accepts the same on 4.1.05. Period of the bill 3 months after date. What will be the due date of the bill:
Solution:
QUESTION: 5
X draws a bill on Y. X endorsed the bill to Z. The payee of the bill will be
Solution:
X draws a bill on Y. X endorsed the bill to Z. Z will be the payee of the bill.
QUESTION: 6
A bill of 12,000 was discounted by A with the banker for 11,880. At maturity, the bill returned dishonoured, noting charges Rs 20. How much amount will the bank deduct from A’s bank balance at the time of such dishonour?
Solution:
QUESTION: 7
X draws a bill on Y for Rs 20,000 on 1.7.05 for 3 months after sight, date of acceptance is 6.1.05. Due date of the bill will be:
Solution:
QUESTION: 8
X sold goods to Y for Rs 1,00,000. Y paid cash Rs 30,000. X will grant 2% discount on balance, and Y request X to draw a bill for balance, the amount of bill will be:
Solution:
QUESTION: 9
On 1.1.05 X draws a bill on Y for Rs 50,000 for 3 months. X got the bill discounted 4.1.05 at 12% rate. The amount of discount on bill will be:
Solution:
QUESTION: 10
Mr. A draws a bill on Mr. Y for Rs 30,000 on 1.1.06 for 3 months. On 4.2.06. X got the bill discounted at 12% rate. The amount of discount will be:
Solution:
QUESTION: 11
X draws a bill on Y for Rs 20,000 for 3 months on 1.1.05. The bill is discounted with banker at a charge of Rs 100. At maturity the bill return dishonoured. In the books of X, for dishonour, the bank account will be credited by:
Solution:
QUESTION: 12
On 1.1.05 X draws a bill on Y for Rs 10,000. At maturity Y request X to renew the bill for 2 month at 12% p.a. interest. Amount of interest will be:
Solution:
QUESTION: 13
On 1.1.05 X draws a bill on Y for Rs 15000 for 3 months. At maturity Y request X to accept Rs 5000 in cash and for balance to draw a fresh bill for 2 months together with 12% p.a. interest, amount of interest will be:
Solution:
QUESTION: 14
On 1.8.05 X draw a bill on Y “for 30 days after sight”. The date of acceptance is 8.8.05. The due date of the bill will be:
Solution:
QUESTION: 15
On 1.6.05 X draw a bill on Y for Rs. 25,000. At maturity Y request X to accept Rs. 5,000 in cash and noting charges incurred Rs. 100 and for the balance X draw a bill on Y for 2 months at 12% p.a. Interest amount will be:
Solution:
QUESTION: 16
On 1.1.05 X draw a bill on Y for Rs. 50,000. At maturity, the bill returned dishonoured as Y become insolvent and 40 paise per rupee is recovered from his estate. The amount recovered is:
Solution:
QUESTION: 17
X draws a bill on Y for Rs 3000. X endorsed to Z. Y will pay the amount of the bill to:
Solution:
QUESTION: 18
On 1.1.05 X draw a bill on Y for 3 months for Rs 10,000. On 4.3.05 Y pay the bill to X at 12% discount, the amount of discount will be:
Solution:
QUESTION: 19
Ram draws on Aslam a bill for Rs. 60,000 on 1.4.01 for 2 months. Aslam accepts the bill and sends it to Ram who gets it discounted for Rs. 58,800. Ram immediately remits Rs. 19,600 to Aslam. On due date, Ram being unable to remit the amount due accepts a bill for Rs. 84,000 for 2 months which is discounted by Aslam for Rs. 82,200. Aslam sends Rs. 14,800 to Ram out of the same. How much discount will be borne by Ram at the time of 14,800 remittances.
Solution:
QUESTION: 20
Mr Bobby sold goods worth Rs 25,000 to Mr Bonny. Bonny immediately accepted a bill on 1.11.01, payable after 2 months. Bobby discounted this bill @ 18% p.a. on 15.11.01. On the due date Bonny failed to discharge the bill. Later on Bonny became insolvent and 50 paise is recovered from Bonny’s estate. How much amount of bad debt will be recorded in the books of Bobby:
Solution:
QUESTION: 21
The purpose of accommodation bill is:
Solution:
QUESTION: 22
M sold goods worth of Rs 50,000 to N. On 1.10.05, N immediately accepted a three month bill. On due date N requested that the bill be renewed for a fresh period of 3 months. N agrees to pay interest @ 18% p.a. in cash. How much interest to be paid in cash by N?
Solution:
QUESTION: 23
On 1.1.05 X draws a bill on Y for Rs 30,000. At maturity Y request X to draw a fresh bill for 2 months together with 12% pa. interest. Noting charges Rs 100. The amount of interest will be:
Solution:
QUESTION: 24
On 18.2.05 A draw a bill on B for Rs 10,000. B accepted the bill on 21.2.05. The bill is drawn for 30 days after sight. The due date of the bill will be:
Solution:
QUESTION: 25
X sold goods to Y for Rs 3,00,000. ½ of the amount will be received in cash and balance in B/R. For what amount X should draw the bill on Y.
Solution:
QUESTION: 26
A draws a bill on B for Rs 50,000 for 3 months. At maturity, the bill returned dishonoured, noting charges Rs 500. 40 paise in a rupee is recovered from B’s estate. The amount of deficiency to be recorded on insolvency in the books of B will be:
Solution:
QUESTION: 27
A sold goods to B for Rs 20,000. A will grant 5% discount to B. B requested A to draw a bill. The amount of bills will be:
Solution:
QUESTION: 28
Fees paid in cash to Notary Public is charged by:
Solution:
QUESTION: 29
A draws a bill on B for Rs 50,000. A endorsed it to C in full settlement of Rs 50,500. Noting charges of Rs 200 as the bill returned dishonoured. A want to pay the amount to C at 2 % discount. The amount to be paid by A to C will be:
Solution:
QUESTION: 30
A draws a bill on B for Rs 1,00,000. A endorsed the bill to C. The bill return dishonoured.Noting charges Rs 1000. B request A to accept the amount at 2% discount by a single cheque. The cheque amount will be:
Solution:
QUESTION: 31
S draws 2 bills of exchange on 1.1.06 for Rs 3,000 and Rs 5,000 respectively. The bill of exchange for Rs 3,000 is for 2 months, while the bill of exchange for Rs 5,000 is for 3 months. These bills are accepted by K. On 4.3.06 K requests S to renew the first bill with interest at 18% p.a. for a period of 2 months. S agrees to this proposal. On 20.3.06 K retires the acceptance for Rs 5,000 the interest rebate i.e., discount being Rs 50.Before the due date of the renewed bill K becomes insolvent and only 60 paise in a rupee can be recovered from his estate. How much bad debt will be recorded in the books of S:
Solution:
QUESTION: 32
The promissory note should be signed by:
Solution:
QUESTION: 33
Kuntal draws a bill on Shyam for Rs 3000. Kuntal endorsed it to Ram. Ram endorsed it to Rahim. The payee of the bill will be:
Solution:
QUESTION: 34
A bill is drawn on 29th Jan’ 06 for one month after date. The date of acceptance is 2nd Feb’06. The bill is drawn on one month after date basis. The due date of the bill will be:
Solution:
QUESTION: 35
Mr. Rex accepted a bill drawn by Mr. Rabin. Mr. Rabin endorsed the bill to Mr Shekar. On the due date, the bill is dishonored as Mr Rex became insolvent. To record the dishonor of the bill in the books of Mr. Rabin, which of the following accounts should be credited?
Solution:
QUESTION: 36
Which of the following statements is true?
Solution:
QUESTION: 37
For mutual accommodation of A and B, B accepted a bill drawn on him by A for 2 months Rs 6000. The said bill is discounted at 12% pa. and remitted 1/3rd of the proceeds to B. The amount remitted by A to B will be:
Solution:
QUESTION: 38
Lara draws an accommodation bill on Sachin. The proceeds are to be borne between Sachin and Lara in the ratio of 3:1. The amount of bill Rs 6000, discounting charges Rs 120. Discount borne by Sachin will be:
Solution:
QUESTION: 39
A draws a bill on B for Rs 4500 for mutual accommodation in the ratio 2:1. A got it discounted at 4230 and remitted 1/3rd of the proceeds to B. At the time of maturity, how much amount A should remit to B such that B can pay off the bill?
Solution:
QUESTION: 40
Suman drew a bill on Sonu for Rs 4500 for mutual accommodation in the ratio 2:1. Sonu accepted the bill and returned to Suman. Suman discounted the bill for Rs 4230 and remitted 1/3rd proceeds to Sonu. Before the due date, not having funds to meet the bill, Sonu drew a bill on Suman for Rs 6,300 on the same terms as to mutual accommodation. The second bill was discounted for Rs 6120. The first bill was honored on the due date and a net amount of Rs 1080 was remitted to Suman by Sonu. The proportionate discount charge on both the bills is to be borne by Suman is:
Solution:
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code | 2,627 | 9,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-21 | latest | en | 0.88138 |
https://wiki-helper.com/if-the-roots-of-the-equation-6-square-13-m-equal-to-zero-receive-true-of-each-other-find-the-val-37429217-11/ | 1,701,428,606,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00539.warc.gz | 708,079,326 | 12,953 | # if the roots of the equation 6 x square – 13 x + m equal to zero receive true of each other find the value of m
if the roots of the equation 6 x square – 13 x + m equal to zero receive true of each other find the value of m
### 1 thought on “if the roots of the equation 6 x square – 13 x + m equal to zero receive true of each other find the value of m”
Quadratic Equation 6x² – 13x + m= 0
Roots are reciprocal to each other
To Find : Value of m
m ની કિંમત શોધો.
Solution:
6x² – 13x + m= 0
Roots are reciprocal to each other
Let say α , 1/α
=> Product of Roots = α * 1/α = 1
Product of Roots = m/6
m/6 = 1
=> m = 6
Value of m = 6 | 236 | 655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-50 | latest | en | 0.806692 |
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