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http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-doi-10_4064-ba57-3-2
|
PL EN
Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
Czasopismo
## Bulletin of the Polish Academy of Sciences. Mathematics
2009 | 57 | 3 | 199-207
Tytuł artykułu
### L-like Combinatorial Principles and Level by Level Equivalence
Autorzy
Treść / Zawartość
Warianty tytułu
Języki publikacji
EN
Abstrakty
EN
We force and construct a model in which GCH and level by level equivalence between strong compactness and supercompactness hold, along with certain additional "L-like" combinatorial principles. In particular, this model satisfies the following properties:
(1) $♢_δ$ holds for every successor and Mahlo cardinal δ.
(2) There is a stationary subset S of the least supercompact cardinal κ₀ such that for every δ ∈ S, $◻_δ$ holds and δ carries a gap 1 morass.
(3) A weak version of $◻_δ$ holds for every infinite cardinal δ.
(4) There is a locally defined well-ordering of the universe 𝓦, i.e., for all κ ≥ ℵ₂ a regular cardinal, 𝓦 ↾ H(κ⁺) is definable over the structure ⟨H(κ⁺),∈ ⟩ by a parameter free formula.
The model constructed amalgamates and synthesizes results due to the author, the author and Cummings, and Asperó and Sy Friedman. It has no restrictions on the structure of its class of supercompact cardinals and may be considered as part of Friedman's "outer model programme".
Słowa kluczowe
Kategorie tematyczne
Rocznik
Tom
Numer
Strony
199-207
Opis fizyczny
Daty
wydano
2009
Twórcy
autor
• Department of Mathematics, Baruch College of CUNY, New York, NY 10010, U.S.A.
• The CUNY Graduate Center, Mathematics, 365 Fifth Avenue, New York, NY 10016, U.S.A.
Bibliografia
Typ dokumentu
Bibliografia
Identyfikatory
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2021-03-03 00:23:16
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http://mathhelpforum.com/algebra/23218-when-solving-2x3-equation-applying-matrix-system.html
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# Math Help - when solving a 2x3 equation applying the matrix system
1. ## when solving a 2x3 equation applying the matrix system
Hello.
When solving a 2x3 equation with the matrix system what is the goal? is it to end up with a zero like this?
a b c
0 d e
is that right? or should I also end up with a 1?
thank you.
2. Originally Posted by jhonwashington
Hello.
When solving a 2x3 equation with the matrix system what is the goal? is it to end up with a zero like this?
a b c
0 d e
is that right? or should I also end up with a 1?
thank you.
You need to be clearer about the exact nature of your question.
If your mean you have a set of equations:
$
a x + b y = c
d x + e y = d
$
which you are going to solve by reducing the augmented matrix:
$
\left[
\begin{array}
{cc|c} a&b&c\\d&e&f
\end{array}
\right]
$
by using row opperations.
How you do this depends on what you have been taught, but if you reduce
this to the form:
$
\left[
\begin{array}
{cc|c} g&0&h\\0&i&j
\end{array}
\right]
$
then $x=h,\ y=j$ is the solution to the original set of equations.
RonL
|
2016-05-27 05:28:00
|
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http://kea-monad.blogspot.com/2008/07/m-theory-lesson-206.html
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occasional meanderings in physics' brave new world
Name:
Location: New Zealand
Marni D. Sheppeard
## Monday, July 14, 2008
### M Theory Lesson 206
Carl Brannen's new post on 1-circulant and 2-circulant operators extends his previous analysis to the remainder of the fundamental fermions and their quantum numbers. He works with $6 \times 6$ circulants of the form for $(1)$ a 1-circulant and $(2)$ a 2-circulant. Just as for the $2 \times 2$ case with numerical matrix entries, we can think of $(1) \pm (2)$ as the eigenvalues of the $6 \times 6$ operator. Notice that the idempotents obtained have simple 2-circulants $(2)$ of democratic form, which means that adding or subtracting them from $(1)$ results in another 1-circulant. For example, for the $e_{R}^{+}$ quantum numbers one finds that which is a unitary 1-circulant since all entries have norm $\frac{1}{3}$. The same matrix results from $(1) + (2)$ for $\overline{\nu}_{R}$. The democratic matrix with all values equal to $\frac{1}{3}$ comes from, for instance, the $\overline{d}_{L}$ quark idempotent. Tony Smith, who likes to think of the Higgs as a top quark condensate, might like this correspondence between Higgs numbers and quark operators.
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2017-07-26 10:44:05
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|
https://curriculum.illustrativemathematics.org/HS/teachers/2/7/14/index.html
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# Lesson 14
Putting It All Together
## 14.1: Equal Slices (5 minutes)
### Warm-up
This warm-up invites students to reason about areas of circles and their sectors in the context of cost-efficiency. This context will be explored further in the next activity.
### Student Facing
At a pizza restaurant, a personal pizza has a radius of 10 centimeters and costs $5. Another restaurant takes a pizza with radius 30 centimeters, cuts it into 8 slices of equal area, and charges$5 per slice. Which is a better deal? Explain your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
Ask for 1 or 2 brief responses from students before moving on to the next activity. The important point is that for the same price, the large slice gives more pizza, making it a better deal.
## 14.2: Pizza Palooza (25 minutes)
### Activity
In this activity, students are building skills that will help them in mathematical modeling (MP4). They formulate a model of a pizza slice as a sector of a circle. They compute unit costs per square inch of pizza to compare the value of several different vendors’ pizza deals. During the activity synthesis, students report their conclusions and the reasoning behind them, and they have an opportunity to consider how to quantify variables beyond price and area.
Making spreadsheet technology available gives students an opportunity to choose appropriate tools strategically (MP5). Monitor for students who use protractors, and for those who use the string to measure arc length and radius. Identify groups who choose their pizza based strictly on cost, and those who also consider other variables such as number of toppings and crust thickness.
### Launch
Arrange students in groups of 4. Distribute the following to each group: one set of cards from the blackline master, protractors, string, rulers, and, if possible, digital tools that can run spreadsheet technology. Tell students that they can choose from any of these tools as they work.
Ask students, “When a restaurant advertises a 12-inch pizza, what does that mean?” (The diameter of the pizza measures 12 inches.)
Action and Expression: Develop Expression and Communication. Invite students to talk about their ideas with a partner before writing them down. Display sentence frames to support students when they explain their ideas. For example, “We need to know. . . ”, “We can use [this tool] for . . .”, “Can we measure. . . ?”, “Is there a formula for . . . ?” Encourage students to design a table to organize their ideas.
Supports accessibility for: Language; Organization
### Student Facing
Elena was researching offers for the upcoming Pizza Palooza festival. She wants to get a good deal on a single slice of pizza.
Your teacher will give you cards that show the deals offered by 4 vendors. Which vendor should Elena choose? Explain or show your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Students may accidentally use the diameter instead of the radius when calculating circle areas. Ask them what expression they are using to calculate circle areas, and to check if they’ve used correct measurements.
### Activity Synthesis
Select previously identified groups to share their choices in this order: First, choose a group who used string to estimate the angle measures in radians. Next, choose a group who used a protractor to estimate the angle measures in degrees. Ask students to compare and contrast the protractor and string strategies. (String is a more common object and so you’re more likely to have it with you. A protractor gives you the angle measure in one step without needing to calculate a ratio. Both give the same answer, just in different units.)
Then, choose a group who chose vendor A because it represents the least expensive unit cost. Then, choose a group who chose a different vendor based on other variables. Ask students how they could quantify some of these other variables. (Students could calculate the volume of the pizza if crust thickness is important to them. They could try to figure out a calorie count if they want the most energy for their money. They could create an expression that assigns relative importance to toppings, size, and crust type.)
Consider asking students, “The photos didn’t show the actual sizes of the pizza slices—the images were scaled down versions of the real thing. Why could we use them to gather information about the areas of the slices?” (Scaling preserves angle measures, so we could use the photos to find the measures of the sectors’ central angles.)
Representing, Conversing: MLR7 Compare and Connect. Use this routine to help students prepare for the whole-class discussion. After groups decide which vendor has the best deal for a single slice of pizza, invite them to create a visual display of their work. Students should consider how to display their calculations so that another student can interpret them. Students may wish to add notes or details to their drawings to help communicate their thinking. Select and arrange 2–4 displays for all to see, based on the suggestions in the Activity Synthesis. To begin the whole-class discussion, give students 1–2 minutes of quiet think time to interpret the displays. Listen for and amplify the language students use to compare and contrast the use of degrees or radians as the unit of measurement.
Design Principle(s): Cultivate conversation
## 14.3: A Fair Split (15 minutes)
### Optional activity
In this activity, students have the opportunity to use what they know about trigonometry and circles to decide how to divide a pizza slice between two people equally so that one of them doesn’t have to eat the crust. Students build skills that will help them in mathematical modeling (MP4) as they decide how to represent the situation. Students may define variables, draw diagrams, and use trial and error.
### Launch
Arrange students in groups of 2. After quiet work time, ask students to compare their responses to their partner’s and decide if they are both correct, even if they are different. Follow with a whole-class discussion.
Representation: Develop Language and Symbols. Use diagrams to connect symbols to concrete objects or values. If students don’t know how to start, show an image of a pizza slice cut into two regions so that one has no crust. Students can then explore how to place the cut so that those regions have the same area.
Supports accessibility for: Visual-spatial processing; Conceptual processing
### Student Facing
Jada and Andre want to share a big slice of pizza so that each of them gets the same amount, but Andre doesn’t like the crust. The pizza slice is a sector of a circle with a radius of 20 cm and a central angle that measures $$\frac{\pi}{3}$$ radians.
How can Andre and Jada divide the slice of pizza into 2 equal pieces so that Andre doesn’t have to eat any crust?
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
Ask students to share their strategies for deciding where to divide the pizza slice. Ask students, “What would change if the radius and central angle were different? What would stay the same?” (The process would be the same. But when the angle increases past a certain point, the isosceles triangle, whose congruent sides are radii, starts to have a smaller area than the rest of the sector, making the problem impossible.)
Speaking: MLR8 Discussion Supports. As students share their reasoning for how to divide the pizza into two equal pieces, press for details by asking how they know that the cut divides the slice into an isosceles triangle and the rest of the sector. Also, ask how they know that they can use trigonometry to find the base of the triangle in terms of the height. Show concepts multi-modally by drawing the sector with a line that cuts the sector into an isosceles triangle and the rest of the sector. This will help students justify their reasoning for how to divide the pizza into two equal pieces so that Andre doesn’t have to eat any crust.
Design Principle(s): Support sense-making; Optimize output (for justification)
## 14.4: Let Your Light Shine (15 minutes)
### Optional activity
In this activity, students use properties of circumscribed circles and inscribed angles to locate possible positions for a light in a photography session.
### Launch
Reading, Listening, Conversing: MLR6 Three Reads. Use a modified version of this routine to support reading comprehension of this problem. Use the first read to orient students to the situation. Ask students to describe what the situation is about. (Noah is taking photos of a sculpture and wants to try different positions for the light.) Use the second read to identify important features of the diagram. (Points A, B, and C form a triangle. The edges of the light beam must meet the endpoints of the backdrop.) After the third read, ask students to brainstorm strategies to solve the problem. This will help students connect the language in the word problem and the reasoning needed to solve the problem while keeping the cognitive demand of the task.
Design Principle(s): Support sense-making
### Student Facing
Noah is taking photos of a sculpture he made in art class. He will submit the photos to a contest. The sculpture is in front of a backdrop, which is represented from an overhead view in the image by segment $$AB$$. Noah positioned a light at point $$C$$ so that the edges of the light beam meet up exactly with the backdrop at segment $$AB$$.
Noah wants to try different positions for the light to highlight different aspects of the sculpture, but he still wants the edges of the beam to exactly meet the endpoints of the backdrop. Find at least 3 other places Noah can place the light. Explain or show your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
If students aren’t sure how to begin, remind them of the name of the unit: Circles. Are there any circles that relate to triangles? Are there any other theorems about circles that might help?
### Activity Synthesis
Here are some questions for discussion:
• “What are some vocabulary terms from this unit that are relevant to this problem?” (The sides of the beam of light are represented by chords. We needed to find the circumcenter of the triangle to construct the circumscribed circle. The angle made by the sides of the light beam is an inscribed angle.)
• “How many possible positions are there for the light?” (Technically, there are an infinite number of positions. Any point on the circumscribed circle that is in front of the backdrop will work.)
• “What are some other situations where inscribed angles might be relevant?” (A spotlight shining on a stage is similar. Sightlines and viewing angles in auditoriums might involve inscribed angles. There could be applications in design, when creating the spoke pattern of a car wheel or designing a logo for a company.)
## Lesson Synthesis
### Lesson Synthesis
Ask students to think of other situations where each of these concepts might appear in real life: tangent lines, angles circumscribed about circles, circumcenters, incenters, inscribed circles, arc length, or any other topic from this unit. Consider assigning each group of students one particular topic to think about. Give students 2–3 minutes of time to discuss and brainstorm, then follow with a whole-class discussion.
Sample responses:
• A car wheel touches the ground in 1 point, so the ground is tangent to the car. (The wheel actually flattens and touches the ground in a region, not just 1 point, but we may be able to model this situation with a tangent line.)
• The sightlines of a satellite monitoring the earth can be modeled by a circumscribed angle.
• A circumcenter is a point that is the same distance from 3 vertices, so it can be used when finding fair locations for placements of things like a hospital or a playground.
• An incenter is a point that is the same distance from 3 sides of a triangle, so could be used if we needed to locate, say, a fire station an equal distance from 3 straight roads connecting towns.
• Inscribed circles are the largest circles that can be cut from a triangle, so if we need to cut a circle out of a piece of wood or stone, we might use the inscribed circle.
• Arc length could be used to measure the distance a point on a gear travels during rotation.
## Student Lesson Summary
### Student Facing
We can use sector areas to compare the value of product offers. Suppose the manager of a store wants to buy several dozen mirrors shaped like sectors of a circle to decorate the store. The manager is choosing between these 2 brands of mirrors at a tradeshow. Brand A’s mirror has radius 14 inches and costs \$35 for each sector. Brand B’s mirror has radius 15 inches and costs \$42. Which is the better deal in terms of cost per square inch of mirror?
Brand A’s mirror is made from a circle cut into 6 congruent slices. Using the expression $$\pi r^2$$ with radius 14 inches, we find that the area of the full circle is $$196\pi$$ or about 616 square inches. Divide that by 6 to find that each sector-shaped mirror has an area of about 103 square inches. At \$35 per sector, this mirror costs about \$0.34 per square inch.
For brand B, we don’t know how many slices the mirror was cut into. However, we can estimate the measure of the central angle using arc length and radius. Suppose the manager uses a flexible measuring tape to find that the length of the arc around the outside of the sector is 19 inches. The ratio of the arc length to the radius gives the measure of the central angle in radians, $$\frac{19}{15}$$. This is about 1.27 radians, but to avoid rounding errors, let’s use the exact value, $$\frac{19}{15}$$, in our calculations.
Next, we can find the area of the sector with the formula $$\frac12 r^2 \theta$$ where $$r$$ is the radius and $$\theta$$ is the radian measure of the central angle. Substitute in our values to get $$\frac12 (15)^2 \boldcdot \frac{19}{15}$$, or 142.5 square centimeters. At \$42 per sector, this mirror costs about \$0.29 per square inch. Brand B’s mirror is a better deal.
|
2022-05-27 12:21:25
|
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|
https://math.stackexchange.com/questions/927656/integral-int-01-frac-ln-frac3x3-x-sqrtx1-xdx
|
# Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$
I have a problem with the following integral:
$$\int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\, {{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,}$$
The first idea was to use the integration by parts because
$$\int{{\rm d}x \over \,\sqrt{x\left(\,1 - x\,\right)\,}\,} =\arcsin\left(\,2x - 1\,\right) + C$$
but what would be the next step is not clear. Another idea would be expand $\ln\left(\,\cdot\right)$ into Taylor series but it seems to be even worse option.
So, what are the other options?
Let $x = \sin(t)^2$ and $s = 2t$, we have
$$\int_0^1 \log\left(\frac{3+x}{3-x}\right)\frac{dx}{\sqrt{x(1-x)}} = \int_0^{\pi/2} \log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right)\frac{2\sin t\cos tdt}{ \sqrt{\sin(t)^2(1-\sin(t)^2)}}\\ = 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right) dt = \int_0^{\pi}\log\left(\frac{3 + \frac{1-\cos s}{2}}{3-\frac{1-\cos s}{2}}\right) ds\\ = \int_0^{\pi}\left(\log(7-\cos s) - \log(5+\cos s)\right) ds$$
Notice for any $a > 1$, we have
$$\frac{1}{\pi}\int_0^\pi \log(a \pm \cos s)ds = \cosh^{-1}(a) = \log\left(\frac{a + \sqrt{a^2-1}}{2}\right)\tag{*1}$$ The integral we desired is simply $$\pi\left( \cosh^{-1}(7) - \cosh^{-1}(5)\right) = \pi \log\left(\frac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\approx 1.072804016182156$$
I'm sure the identity in $(*1)$ has a name but I can't remember what it is. Let us prove it!
Notice for any $b > 1$, the function $\log(b+z)$ is analytic over and inside the unit circle $S^1$ in $\mathbb{C}$.
By Residue theorem, we have
$$\frac{1}{2\pi i}\int_{S^1} \log(b + z) \frac{dz}{z} = \log(b)$$
If one parametrize the unit circle by $z = e^{i\theta}$, we get
$$\frac{1}{2\pi}\int_0^{2\pi} \log(b + e^{i\theta}) d\theta = \log(b)$$
Take the real part on both sides, this leads to
\begin{align} &\frac{1}{2\pi}\int_0^{2\pi} \log(b^2 + 1 + 2b\cos\theta) d\theta = \log(b^2)\\ \iff & \frac{1}{2\pi}\int_0^{2\pi} \log\left(\frac{b+b^{-1}}{2} + \cos\theta\right)d\theta = \log\left(\frac{b}{2}\right)\end{align} Substitute $\displaystyle\;\frac{b+b^{-1}}{2}\;$ by $a$, we have $\displaystyle\;\frac{b-b^{-1}}{2} = \sqrt{a^2-1}$ and it is clear $(*1)$ follows.
• (+1). Nice. Actually we performed the same calculations by two different approaches: the Residue theorem and the Riemann sums. – Jack D'Aurizio Sep 11 '14 at 17:45
• Thanks for the answer! Clear! – Martin Gales Sep 15 '14 at 14:06
We have:
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx = 2\pi\log\frac{2+\sqrt{3}}{2}.\tag{1}$$
This happens because: $$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=2\int_{0}^{1}\frac{\log(3+x^2)}{\sqrt{1-x^2}}\,dx =\int_{-\pi/2}^{\pi/2}\log(3+\cos^2\theta)\,d\theta,$$ $$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=\frac{1}{2}\int_{-\pi}^{\pi}\log\left(\frac{7+\cos\theta}{2}\right)d\theta=-\pi\log 2+\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta.$$ Now comes an interesting technique - we have: $$\begin{eqnarray*}\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta &=& \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n}\log\left(7+\cos\frac{k\pi}{n}\right)\\&=&\lim_{n\to +\infty}\frac{\pi}{n}\log\prod_{k=1}^{n}\left(7+\cos\frac{k\pi}{n}\right)\end{eqnarray*}$$ but since: $$z^{2n}-1 = \prod_{k=1}^{2n}\left(z-e^{\frac{\pi i k }{n}}\right)=(z^2-1)\prod_{k=1}^{n-1}\left(z^2+1-2z\cos\frac{k\pi}{n}\right)$$ and the solutions of $$\frac{z^2+1}{2z}=-7$$ are $z=\pm 4\sqrt{3}-7$, it follows that: $$\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta=\pi\log\left(\frac{7}{2}+2\sqrt{3}\right).$$ With the same technique we can prove:
$$\int_{0}^{1}\frac{\log(3-x)}{\sqrt{x(1-x)}}\,dx = \pi\log\left(\frac{5}{4}+\sqrt{\frac{3}{2}}\right),\tag{2}$$
hence we have:
$$\int_{0}^{1}\frac{\log\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\,dx = \pi\log\left((7+4\sqrt{3})(5-2\sqrt{6})\right).\tag{3}$$
• This just proves that, sometimes, the Residue theorem can be replaced by a Riemann sums argument. – Jack D'Aurizio Sep 11 '14 at 17:42
• (+1) Actually, I first derive the result by a third method (differentiate under integral sign) but the derivation along that direction is pretty clumsy.... – achille hui Sep 11 '14 at 18:16
• @achillehui I've just wanted to use that and I'm typing the answer. – Tunk-Fey Sep 11 '14 at 18:19
• Mr. @achillehui, Tunk-Fey & Jack D'Aurizio: Could you please help me? I'm stuck （ つ﹏╰） See my answer below – Anastasiya-Romanova 秀 Sep 12 '14 at 10:13
• Thank you! The product formula is especially nice! – Martin Gales Sep 15 '14 at 14:09
Split the integral into two forms by expanding the logarithm function $$\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^1\frac{\ln(3+x)}{\sqrt{x(1-x)}}\ dx-\int_{0}^1\frac{\ln(3-x)}{\sqrt{x(1-x)}}\ dx$$ Let $t=\sqrt{x}\ \rightarrow\ dt=\dfrac{dx}{2\sqrt{x}}$, we have $$2\int_{0}^1\frac{\ln(3+t^2)}{\sqrt{1-t^2}}\ dt-2\int_{0}^1\frac{\ln(3-t^2)}{\sqrt{1-t^2}}\ dt$$ Let $t=\sin\theta\ \rightarrow\ dt=\cos\theta\ d\theta$, we have $$2\int_{0}^{\pi/2}\ln(3+\sin^2\theta)\ d\theta-2\int_{0}^{\pi/2}\ln(3-\sin^2\theta)\ d\theta$$ Using identity $\sin^2\theta=\dfrac12(1-\cos2\theta)$ and setting $y=2\theta$, we have $$\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^{\pi}\ln\left(7-\cos y\right)\ dy-\int_{0}^{\pi}\ln\left(5+\cos y\right)\ dy$$ We will use Feynman's way to evaluate integral above. Consider $$I(k)=\int_{0}^{\large\pi}\ln\left(k\pm\cos y\right)\ dy$$ then \begin{align} I'(k)&=\int_{0}^{\large\pi}\frac{1}{k\pm\cos y}\ dy \end{align} Using formula $$\int_0^\pi\frac{1}{a^2+b^2-2ab\cos x}dx=\frac{\pi}{a^2-b^2}$$ Setting $b=\pm\dfrac{1}{2a}$, we have $$\int_0^\pi\frac{1}{a^2+\frac{1}{4a^2}\pm\cos x}dx=\frac{4\pi a^2}{4a^4-1}$$ Clearly $k=a^2+\dfrac{1}{4a^2}\ \rightarrow\ a^2=\dfrac{k+\sqrt{k^2-1}}{2}$ and $dk=\dfrac{4a^4-1}{2a^3}da$, then \begin{align} I(k)&=\int\int_{0}^{\large\pi}\frac{1}{k\pm\cos y}\,dy\,dk\\ &=\int\frac{4\pi a^2}{4a^4-1}\cdot\dfrac{4a^4-1}{2a^3}da\\ &=2\pi\int\frac{1}{a}\,da\\ &=2\pi\ln(a)+C\\ &=\pi\ln(a^2)+C\\ &=\pi\ln\left(\dfrac{k+\sqrt{k^2-1}}{2}\right)+C\\ \end{align} Finally \begin{align} \int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx&=I(7)-I(5)\\ &=\pi\ln\left(\dfrac{7+\sqrt{7^2-1}}{5+\sqrt{5^2-1}}\right)\\ &=\pi\ln\left(\dfrac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\\ \end{align} Yeayyy, I'm done! (>‿◠)✌
• Stuck? $\displaystyle\;\frac{4\pi a^2}{4a^4-1} = 2\pi \frac{k+\sqrt{k^2-1}}{2(k^2-1)+2k\sqrt{k^2-1}} = \frac{\pi}{\sqrt{k^2-1}}$ – achille hui Sep 12 '14 at 10:35
• Mr. @achillehui Sorry for bothering you, I've just gotten another idea. Thanks anyway for your help (ô‿ô) – Anastasiya-Romanova 秀 Sep 12 '14 at 10:36
• Many thanks! Differentiation of the integrand is a cool thing! – Martin Gales Sep 15 '14 at 14:13
• @MartinGales You're welcome (ô‿ô) – Anastasiya-Romanova 秀 Sep 15 '14 at 14:32
|
2019-06-25 18:20:43
|
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|
https://plotly.com/matlab/3d-isosurface-plots/
|
# 3D Isosurface Plots in MATLAB®
How to make 3D Isosurface Plots in MATLAB® with Plotly.
## Basic Isosurface plot
[x,y,z] = meshgrid([-3:0.25:3]);
V = x.*exp(-x.^2 -y.^2 -z.^2);
isosurface(x,y,z,V,1e-4);
fig2plotly(gcf, 'TreatAs', 'isosurface');
[x,y,z] = meshgrid([-3:0.25:3]);
V = x.*exp(-x.^2 -y.^2 -z.^2);
[faces,verts,colors] = isosurface(x,y,z,V,1e-4,x);
patch('Vertices',verts,'Faces',faces,'FaceVertexCData',colors,...
'FaceColor','interp','EdgeColor','interp')
view(3)
colormap copper
fig2plotly(gcf, 'TreatAs', 'isosurface');
## Draw Isosurface with Lighting
Load the flow data set, which represents the speed profile of a submerged jet within an infinite tank. Draw the isosurface at the data value of -3 and prepare the isosurface for lighting by:
• Recalculating the isosurface normals based on the volume data.
• Setting the face and edge color.
• Specifying the view.
[x,y,z,v] = flow;
p = patch(isosurface(x,y,z,v,-3));
isonormals(x,y,z,v,p)
p.FaceColor = 'red';
p.EdgeColor = 'none';
daspect([1 1 1])
view(3);
axis tight
camlight
lighting gouraud
fig2plotly(gcf, 'TreatAs', 'isosurface');
[x,y,z] = meshgrid([-3:0.25:3]);
V = x.*exp(-x.^2 -y.^2 -z.^2);
s = isosurface(x,y,z,V,1e-4);
p = patch(s);
isonormals(x,y,z,V,p)
view(3);
set(p,'FaceColor',[0.5 1 0.5]);
set(p,'EdgeColor','none');
camlight;
lighting gouraud;
fig2plotly(gcf, 'TreatAs', 'isosurface');
## Set Isosurface Colors
Visualize the flow data, but color-code the surface to indicate magnitude along the x-axis. Use a sixth argument to isosurface, which provides a means to overlay another data set by coloring the resulting isosurface. The colors variable is a vector containing a scalar value for each vertex in the isosurface, to be portrayed with the current color map. In this case, it is one of the variables that define the surface, but it could be entirely independent. You can apply a different color scheme by changing the current figure color map.
[x,y,z,v] = flow;
[faces,verts,colors] = isosurface(x,y,z,v,-3,x);
patch('Vertices',verts,'Faces',faces,'FaceVertexCData',colors,...
'FaceColor','interp','EdgeColor','interp')
view(30,-15)
axis vis3d
colormap copper
fig2plotly(gcf, 'TreatAs', 'isosurface');
|
2022-05-23 22:52:33
|
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|
https://calculatores.com/wheel-torque-calculator
|
# Wheel Torque Calculator
## Introduction to Wheel Torque Calculator
A torque calculator is an online tool that can calculate the torque of any automobile wheels to adjust them according to the wheel torque. It calculates the torque of the front, rear, and all wheels using the engine wheel of the automobile.
As you already learned, torque in physics has many applications in our daily life as it is a necessary term in automobile-like care. The wheel torque calculations are essential to adjust the wheel lug nuts using a special socket. Therefore, we introduce you to a tool that helps you to calculate the torque of any wheel easily.
## The Formula used by Wheel Torque Formula Calculator
Since the wheels of an automobile produce torque, which is the turning effect of force, it plays an important role in the movement of an automobile. A proper adjustment of torque wheels can save you from injury or breakage of wheels. Calculate wheel torque according to the torque of the engine by using this wheel torque calculator. The formula to calculate the torque of a wheel is:
$$WT \;=\; \frac{ET}{DL}$$
Where,
WT = is the Wheel Torque.
ET = is the Engine Torque.
DL = is the Drive Train Loss.
The wheel calculator uses the above formula to calculate the torque of a wheel. So you can easily estimate how much torque is required to adjust wheel lug nuts properly.
## How to use Front Wheel Calculator?
When calculating a wheel's torque, you want to find any easy and quick way to calculate it. Here the wheel torque tool can be your smart way to get wheel torque. You need to follow the given steps:
• Enter the torque of the automobile engine.
• Select the wheel from the given drop-down list.
• Click on the calculate button.
The wheel torque will be calculated within a few seconds as you click the calculate button.
## Why use a Rear Wheel Calculator?
When you are calculating wheel torque for your car, you need a tool that can quickly tell you how much wheel torque will be required according to your automobile engine. So the main reason to use this tool is that you can find the exact torque value.
While calculating the torque of a wheel manually, you may forget to use an exact formula, which may result in wrong calculations. The wrong calculation may lead to an unfortunate incident for you. That's why you need to use the wheel torque calculator.
## Benefits of using Torque Calculator
It is easy to calculate the wheel torque with this tool because it has many benefits that you can get. Some of those benefits are:
1. It is easy to use because you only need to write input values and then hit the calculate button.
2. It can tell you the exact wheel torque with no time waste.
3. It is free to use because it does not demand you to pay a registration fee.
4. You can use it anytime, anywhere, without restrictions because it allows you to use it multiple times to calculate wheel torque.
5. It is reliable because it provides highly accurate results; that is its main ability.
### Shaun Murphy
Last Updated March 28, 2022
A professional content writer who likes to write on science, technology and education.
|
2022-09-26 05:42:35
|
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|
https://labs.tib.eu/arxiv/?author=E.%20H.%20Neilsen
|
• ### The Electromagnetic Counterpart of the Binary Neutron Star Merger LIGO/VIRGO GW170817. II. UV, Optical, and Near-IR Light Curves and Comparison to Kilonova Models(1710.05840)
Oct. 16, 2017 astro-ph.HE
We present UV, optical, and NIR photometry of the first electromagnetic counterpart to a gravitational wave source from Advanced LIGO/Virgo, the binary neutron star merger GW170817. Our data set extends from the discovery of the optical counterpart at $0.47$ days to $18.5$ days post-merger, and includes observations with the Dark Energy Camera (DECam), Gemini-South/FLAMINGOS-2 (GS/F2), and the {\it Hubble Space Telescope} ({\it HST}). The spectral energy distribution (SED) inferred from this photometry at $0.6$ days is well described by a blackbody model with $T\approx 8300$ K, a radius of $R\approx 4.5\times 10^{14}$ cm (corresponding to an expansion velocity of $v\approx 0.3c$), and a bolometric luminosity of $L_{\rm bol}\approx 5\times10^{41}$ erg s$^{-1}$. At $1.5$ days we find a multi-component SED across the optical and NIR, and subsequently we observe rapid fading in the UV and blue optical bands and significant reddening of the optical/NIR colors. Modeling the entire data set we find that models with heating from radioactive decay of $^{56}$Ni, or those with only a single component of opacity from $r$-process elements, fail to capture the rapid optical decline and red optical/NIR colors. Instead, models with two components consistent with lanthanide-poor and lanthanide-rich ejecta provide a good fit to the data, the resulting "blue" component has $M_\mathrm{ej}^\mathrm{blue}\approx 0.01$ M$_\odot$ and $v_\mathrm{ej}^\mathrm{blue}\approx 0.3$c, and the "red" component has $M_\mathrm{ej}^\mathrm{red}\approx 0.04$ M$_\odot$ and $v_\mathrm{ej}^\mathrm{red}\approx 0.1$c. These ejecta masses are broadly consistent with the estimated $r$-process production rate required to explain the Milky Way $r$-process abundances, providing the first evidence that BNS mergers can be a dominant site of $r$-process enrichment.
• ### The Dark Energy Camera(1504.02900)
April 11, 2015 astro-ph.IM
The Dark Energy Camera is a new imager with a 2.2-degree diameter field of view mounted at the prime focus of the Victor M. Blanco 4-meter telescope on Cerro Tololo near La Serena, Chile. The camera was designed and constructed by the Dark Energy Survey Collaboration, and meets or exceeds the stringent requirements designed for the wide-field and supernova surveys for which the collaboration uses it. The camera consists of a five element optical corrector, seven filters, a shutter with a 60 cm aperture, and a CCD focal plane of 250 micron thick fully-depleted CCDs cooled inside a vacuum Dewar. The 570 Mpixel focal plane comprises 62 2kx4k CCDs for imaging and 12 2kx2k CCDs for guiding and focus. The CCDs have 15 microns x15 microns pixels with a plate scale of 0.263 arc sec per pixel. A hexapod system provides state-of-the-art focus and alignment capability. The camera is read out in 20 seconds with 6-9 electrons readout noise. This paper provides a technical description of the camera's engineering, construction, installation, and current status.
• ### The Sloan Digital Sky Survey Monitor Telescope Pipeline(astro-ph/0608575)
Aug. 26, 2006 astro-ph
The photometric calibration of the Sloan Digital Sky Survey (SDSS) is a multi-step process which involves data from three different telescopes: the 1.0-m telescope at the US Naval Observatory (USNO), Flagstaff Station, Arizona (which was used to establish the SDSS standard star network); the SDSS 0.5-m Photometric Telescope (PT) at the Apache Point Observatory (APO), New Mexico (which calculates nightly extinctions and calibrates secondary patch transfer fields); and the SDSS 2.5-m telescope at APO (which obtains the imaging data for the SDSS proper). In this paper, we describe the Monitor Telescope Pipeline, MTPIPE, the software pipeline used in processing the data from the single-CCD telescopes used in the photometric calibration of the SDSS (i.e., the USNO 1.0-m and the PT). We also describe transformation equations that convert photometry on the USNO-1.0m u'g'r'i'z' system to photometry the SDSS 2.5m ugriz system and the results of various validation tests of the MTPIPE software. Further, we discuss the semi-automated PT factory, which runs MTPIPE in the day-to-day standard SDSS operations at Fermilab. Finally, we discuss the use of MTPIPE in current SDSS-related projects, including the Southern u'g'r'i'z' Standard Star project, the u'g'r'i'z' Open Star Clusters project, and the SDSS extension (SDSS-II).
• ### Candidate spectroscopic binaries in the Sloan Digital Sky Survey(astro-ph/0508605)
Aug. 29, 2005 astro-ph
We have examined the radial velocity data for stars spectroscopically observed by the Sloan Digital Sky Survey (SDSS) more than once to investigate the incidence of spectroscopic binaries, and to evaluate the accuracy of the SDSS stellar radial velocities. We find agreement between the fraction of stars with significant velocity variations and the expected fraction of binary stars in the halo and thick disk populations. The observations produce a list of 675 possible new spectroscopic binary stars and orbits for eight of them.
• ### The Sloan Digital Sky Survey Quasar Catalog III. Third Data Release(astro-ph/0503679)
March 30, 2005 astro-ph
We present the third edition of the Sloan Digital Sky Survey (SDSS) Quasar Catalog. The catalog consists of the 46,420 objects in the SDSS Third Data Release that have luminosities larger than M_i = -22 (in a cosmology with H_0 = 70 km/s/Mpc, Omega_M = 0.3, and Omega_Lambda = 0.7), have at least one emission line with FWHM larger than 1000 km/s or are unambiguously broad absorption line quasars, are fainter than i = 15.0, and have highly reliable redshifts. The area covered by the catalog is 4188 sq. deg. The quasar redshifts range from 0.08 to 5.41, with a median value of 1.47; the high-redshift sample includes 520 quasars at redshifts greater than four, of which 17 are at redshifts greater than five. For each object the catalog presents positions accurate to better than 0.2 arcsec. rms per coordinate, five-band (ugriz) CCD-based photometry with typical accuracy of 0.03 mag, and information on the morphology and selection method. The catalog also contains radio, near-infrared, and X-ray emission properties of the quasars, when available, from other large-area surveys. The calibrated digital spectra cover the wavelength region 3800--9200A at a spectral resolution about 2000; the spectra can be retrieved from the public database using the information provided in the catalog. A total of 44,221 objects in the catalog were discovered by the SDSS; 28,400 of the SDSS discoveries are reported here for the first time.
• ### LOTIS, Super-LOTIS, SDSS and Tautenburg Observations of GRB 010921(astro-ph/0112397)
Dec. 17, 2001 astro-ph
We present multi-instrument optical observations of the High Energy Transient Explorer (HETE-2)/Interplanetary Network (IPN) error box of GRB 010921. This event was the first gamma ray burst (GRB) localized by HETE-2 which has resulted in the detection of an optical afterglow. In this paper we report the earliest known observations of the GRB010921 field, taken with the 0.11-m Livermore Optical Transient Imaging System (LOTIS) telescope, and the earliest known detection of the GRB010921 optical afterglow, using the 0.5-m Sloan Digital Sky Survey Photometric Telescope (SDSS PT). Observations with the LOTIS telescope began during a routine sky patrol 52 minutes after the burst. Observations were made with the SDSS PT, the 0.6-m Super-LOTIS telescope, and the 1.34-m Tautenburg Schmidt telescope at 21.3, 21.8, and 37.5 hours after the GRB, respectively. In addition, the host galaxy was observed with the USNOFS 1.0-m telescope 56 days after the burst. We find that at later times (t > 1 day after the burst), the optical afterglow exhibited a power-law decline with a slope of $\alpha = 1.75 \pm 0.28$. However, our earliest observations show that this power-law decline can not have extended to early times (t < 0.035 day).
• ### Cataclysmic Variables from SDSS I. The First Results(astro-ph/0110291)
Oct. 12, 2001 astro-ph
The commissioning year of the Sloan Digital Sky Survey has demonstrated that many cataclysmic variables have been missed in previous surveys with brighter limits. We report the identification of 22 cataclysmic variables, of which 19 are new discoveries and 3 are known systems (SW UMa, BH Lyn and Vir4). A compendium of positions, colors and characteristics of these systems obtained from the SDSS photometry and spectroscopy is presented along with data obtained during follow-up studies with the Apache Point Observatory (APO) and Manastash Ridge Observatory (MRO) telescopes. We have determined orbital periods for 3 of the new systems: two show dwarf nova outbursts, and the third is a likely magnetic system with eclipses of its region of line emission. Based on these results, we expect the completed survey to locate at least 400 new CVs. Most of these will be faint systems with low accretion rates that will provide new constraints on binary evolution models.
|
2020-11-30 17:45:07
|
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|
https://codereview.stackexchange.com/questions/78712/efficiency-of-calling-other-classes/78721
|
# Efficiency of calling other classes
After much trial and error, I finally came up with this code to call on other classes to perform their function. This code works and does what I want it to do, so I am looking for some comments on the code efficiency.
What I want to happen is have the user input which calculator they would like to use and have that calculator come up presenting its own variables and calculations. The code for the calculators works just fine, so I don't need help with that. Just the efficiency of my calling from other classes.
My first iteration of this was to have the other calculators inside methods within the class. When I tried to call them up, it was nothing but problems, but this way seems to work. (I only mention this because my last question got knocked out because the code didn't work, so I found a way that worked, albeit not how I intended.)
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter 2 for Compound Interest Calculator");
int ans = sc.nextInt();
if (ans == 1) {
} else if (ans == 2){
CompInt.CalcInt();
} else {
System.out.println("That was not an option.");
}
sc.close();
}
• It makes little sense to worry about the microseconds it takes to execute these method calls, when the human takes much much longer to make a choice. – 200_success Jan 27 '15 at 7:09
• Not worried so much about how long it takes, but rather how clean/proper my code is to do what I want it to do. Probably my fault in the wording. Apologies. – PALADIN 458S Jan 27 '15 at 8:47
• Not only am I still barely getting into programming...I'm still learning HOW to ask questions about programming the right way. – PALADIN 458S Jan 27 '15 at 8:47
• I see. "Efficiency" probably isn't the right word then — the connotation is that it's about runtime performance. You're probably more concerned with whether your code is compact, idiomatic, or expressive. – 200_success Jan 27 '15 at 8:56
### Simplify multiple conditions using a switch
Whenever multiple conditions depend only on the value of a single variable, consider using a switch instead to simplify:
switch (ans) {
case 1:
break;
case 2:
CompInt.CalcInt();
break;
default:
System.out.println("That was not an option.");
}
### Naming
The common convention is to use camelCase for function names. The method names CalcQuad and CalcInt violate that.
It also seems that the class names could be better. If you spell out their purpose, the names won't be much longer but a lot more intuitive and unambiguous, for example:
• Quadratic -> QuadraticCalculator
• CompInt -> CompoundInterestCalculator
And then the methods that do the calculation can be simply calculate in both classes, without including the type of calculation, which is already included in the class name.
Alternatively, you could use a single Calculator class with calculateQuadratic and calculateCompoundInterest methods.
Thought you meant efficiency as in runtime performance. Woops.
Disclaimer: I am no expert by any stretch. My answer is based on a very basic understanding of Java and the JVM. Use my (possibly incorrect) answer at your own peril.
Since you're simply calling a function, the efficiency of that call (I would think) would rely on the JVM.
I'm willing to bet (but I'm no expert, so I may very well be wrong) that the execution time of, say, Quadratic.CalcQuad(); would be very similar, if not identical to the time it would take if you computed whatever is in that function, in your main class.
You can check the execution time by getting the current time right before the function call, as well as immediately after, and then subtract the two. For example:
/*
The beginning of your program here
*/
if (ans == 1) {
int startTime = System.currentTimeMillis();
int endTime = System.currentTimeMillis();
System.out.print("It took ");
System.out.print(endTime - startTime);
System.out.print(" milliseconds to calculate a quadratic using a function call.");
//This will output how long it takes to run the function (In milliseconds).
startTime = System.currentTimeMillis();
calculation without calling the function.**/
endTime = System.currentTimeMillis();
System.out.print("It took ");
System.out.println(endTime - startTime);
System.out.print(" milliseconds to calculate the same quadratic WITHOUT the function call.");
//This will output how long it takes to run the same function,
//only without the call (also in milliseconds).
} else if (ans == 2){
/*
The rest of your program here
*/
|
2019-12-06 04:14:56
|
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|
https://exam.mangrovebd.org/jkg0b/cross-product-example-dfe539
|
# cross product example
Sciences, Culinary Arts and Personal Find \vec{T}, \vec{N} and \vec{B} for the curve \vec{r}(t) = \langle 3 cos(3t), 3sin(3t), 5t \rangle at the point t = 0. courses that prepare you to earn Find the length and direction? first two years of college and save thousands off your degree. Example #2. In the above example, Z is the cross product of the two input arrays x and y. p=1, which means that the output is perpendicular to the inputs x and y. lessons in math, English, science, history, and more. Enrolling in a course lets you earn progress by passing quizzes and exams. (when defined) of u v. u = 4i + 9j v = i - j. You'll see that the product 6 is larger than the product of 5, so 3/5 is larger than 1/2. Get access risk-free for 30 days, a) Find the curl of the vector field. Is There Too Much Technology in the Classroom? And it all happens in 3 dimensions! Also find the volume of the parallelepiped with PQ, PR, and PQ, 1.a)Find a non-zero vector orthogonal to the plane through the points P\left ( 1,0,0 \right ),Q\left ( 4,-2,3 \right ),R\left ( -1,2,1 \right ) b)Find the volume of a parallelpiped through the positio. The cross product or vector product is a binary operation on two vectors in three-dimensional space (R3) and is denoted by the symbol x.Two linearly independent vectors a and b, the cross product, a x b, is a vector that is perpendicular to both a and b and therefore normal to the plane containing them. Find the magnitudes of these vectors and the angles between them. Examples of calculating the cross product. As a member, you'll also get unlimited access to over 83,000 Anyone can earn study If the denominators are the same, the fraction with the largest numerator is larger. Working Scholars® Bringing Tuition-Free College to the Community. Already registered? To unlock this lesson you must be a Study.com Member. Get the unbiased info you need to find the right school. Being a vector operation, the cross product is extremely important in all sorts of sciences (particularly physics), engineering, and mathematics. There are two vector A and B and we have to find the dot product and cross product of two vector array. We can only compare two fractions at a time with this method, so if you're trying to compare more than two fractions, you'll need to repeat these steps using two fractions at a time. Here's an example: If the numerators are the same, the fraction with the smallest denominator is larger. Be sure to write your products down in such a way that you can identify which fraction they each represent. Dot product is also known as scalar product and cross product also known as vector product. Services. She is currently working on a Doctoral Degree in Education. An error occurred trying to load this video. | {{course.flashcardSetCount}} {{courseNav.course.mDynamicIntFields.lessonCount}} lessons All rights reserved. Calculate the cross product between $\vc{a} = (3, -3, 1)$ and $\vc{b} = (4,9,2)$. 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Let a, b, c be vectors such that a \times b = c, b \times c = a and c \times a = b. © copyright 2003-2020 Study.com. succeed. Log in here for access. It's critical that you multiply upward diagonally to get accurate products for the corresponding fractions. You can test out of the Dot Product – Let we have given two vector A = a1 * i + a2 * j + a3 * k and B = b1 * i + b2 * j + b3 * k. Where i, j and k are the unit vector along the x, y and z directions. The cross product method is used to compare two fractions. Write the fractions vertically, next to each other. If it is not perpendicular, then p will be 0. Plus, get practice tests, quizzes, and personalized coaching to help you b) Find two unit vectors perpendicular to both \vec{v} = \langle 3, 1, -1 \rangle and \vec{w} = \langle 0, 1, 2 \rangle, For P(1,0,2), Q(2,1,-1), and R(-1,1,0) Find PQ \times PR, PQ \cdot (PQ \times PR), and the area of the triangle using P,Q,R as vertices. 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2021-05-08 12:25:19
|
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https://andrew-johnson-4.github.io/lsts-tutorial/examples/batteries_not_included.html
|
## Batteries Not Included
The title of this example is a pun on the Python motto "batteries included". LSTS is a type-checker. LSTS does nothing else.
LSTS may connect with a bunch of other software products that include batteries. However, LSTS by itself is very minimal. LSTS does not even define "if", although it is given special syntax.
In this example we define "if". An if statement has three arguments. One argument for the Boolean branching condition. Two arguments for each conditional branch. The if statement returns the value of one of the branches. All branches and the return type are parameterized. The shared parameter will become the greatest-common-denominator of both branches.
let $"if"(condition: Boolean, branch1: A, branch2: A): A; The same lack of features applies to the type system as well. The Boolean type referenced above must be defined by the user. Boolean types are somewhat special in that they also have dependent types carried along with them. To denote this special relation, we mark the type as Constant by adding the constant keyword. type constant Boolean = True | False; Numbers are user defined as well. type Number; type constant Integer: Number = /^[0-9][_0-9]*([eE][_0-9]+)?$/;
type Real : Number = /^[0-9][_.0-9]*([eE][-]?[_0-9]+)?\$/;
|
2022-12-09 09:50:29
|
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|
https://archive.lib.msu.edu/crcmath/math/math/t/t435.htm
|
Twin Peaks
For an Integer , let denote the Least Prime Factor of . A Pair of Integers is called a twin peak if
1. ,
2. ,
3. For all , Implies .
A broken-line graph of the least prime factor function resembles a jagged terrain of mountains. In terms of this terrain, a twin peak consists of two mountains of equal height with no mountain of equal or greater height between them. Denote the height of twin peak by . By definition of the Least Prime Factor function, must be Prime.
Call the distance between two twin peaks
Then must be an Even multiple of ; that is, where is Even. A twin peak with is called a -twin peak. Thus we can speak of -twin peaks, -twin peaks, etc. A -twin peak is fully specified by , , and , from which we can easily compute .
The set of -twin peaks is periodic with period , where is the Primorial of . That is, if is a -twin peak, then so is . A fundamental -twin peak is a twin peak having in the fundamental period . The set of fundamental -twin peaks is symmetric with respect to the fundamental period; that is, if is a twin peak on , then so is .
The question of the Existence of twin peaks was first raised by David Wilson in the math-fun mailing list on Feb. 10, 1997. Wilson already had privately showed the Existence of twin peaks of height to be unlikely, but was unable to rule them out altogether. Later that same day, John H. Conway, Johan de Jong, Derek Smith, and Manjul Bhargava collaborated to discover the first twin peak. Two hours at the blackboard revealed that admits the -twin peak
which settled the Existence question. Immediately thereafter, Fred Helenius found the smaller -twin peak with and
The effort now shifted to finding the least Prime admitting a -twin peak. On Feb. 12, 1997, Fred Helenius found , which admits 240 fundamental -twin peaks, the least being
Helenius's results were confirmed by Dan Hoey, who also computed the least -twin peak and number of fundamental -twin peaks for , 79, and 83. His results are summarized in the following table.
71 7310131732015251470110369 240 73 2061519317176132799110061 40296 79 3756800873017263196139951 164440 83 6316254452384500173544921 6625240
The -twin peak of height is the smallest known twin peak. Wilson found the smallest known -twin peak with , as well as another very large -twin peak with . Richard Schroeppel noted that the latter twin peak is at the high end of its fundamental period and that its reflection within the fundamental period is smaller.
Many open questions remain concerning twin peaks, e.g.,
1. What is the smallest twin peak (smallest )?
2. What is the least Prime admitting a -twin peak?
3. Do -twin peaks exist?
4. Is there, as Conway has argued, an upper bound on the span of twin peaks?
5. Let be Prime. If and each admit -twin peaks, does then necessarily admit a -twin peak?
See also Andrica's Conjecture, Divisor Function, Least Common Multiple, Least Prime Factor
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2021-11-27 08:06:45
|
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|
https://eprints.soton.ac.uk/261596/
|
The University of Southampton
University of Southampton Institutional Repository
Coded Modulation Assisted Radial Basis Function Aided Turbo Equalisation for Dispersive Rayleigh Fading Channels
Ng, S. X., Yee, M. S. and Hanzo, L. (2004) Coded Modulation Assisted Radial Basis Function Aided Turbo Equalisation for Dispersive Rayleigh Fading Channels IEEE Transactions on Wireless Communications, 3, (6), pp. 2198-2206.
Record type: Article
Abstract
In this contribution a range of Coded Modulation (CM) assisted Radial Basis Function (RBF) based Turbo Equalisation (TEQ) schemes are investigated when communicating over dispersive Rayleigh fading channels. Specifically, 16QAM based Trellis Coded Modulation (TCM), Turbo TCM (TTCM), Bit-Interleaved Coded Modulation (BICM) and iteratively decoded BICM (BICM-ID) are evaluated in the context of an RBF based TEQ scheme and a reduced-complexity RBF based In-phase/Quadrature-phase (I/Q) TEQ scheme. The Least Mean Square (LMS) algorithm was employed for channel estimation, where the initial estimation step-size used was 0.05, which was reduced to 0.01 for the second and the subsequent TEQ iterations. The achievable coding gain of the various CM schemes was significantly increased, when employing the proposed RBF-TEQ or RBF-I/Q-TEQ rather than the conventional non-iterative Decision Feedback Equaliser~(DFE). Explicitly, the reduced-complexity RBF-I/Q-TEQ-CM achieved a similar performance to the full-complexity RBF-TEQ-CM, while attaining a significant complexity reduction. The best overall performer was the RBF-I/Q-TEQ-TTCM scheme, requiring only 1.88~dB higher SNR at BER=10$^{-5}$, than the identical throughput 3~BPS uncoded 8PSK scheme communicating over an AWGN channel. The coding gain of the scheme was 16.78~dB.
PDF 35_TOW_04.PDF - Other
Published date: November 2004
Keywords: RBF, I/Q, TEQ, CM, TCM, TTCM, BICM, BICM-ID
Organisations: Southampton Wireless Group
Identifiers
Local EPrints ID: 261596
URI: https://eprints.soton.ac.uk/id/eprint/261596
PURE UUID: d7ec9968-b722-4f49-88d7-08be0c067188
ORCID for S. X. Ng: orcid.org/0000-0002-0930-7194
ORCID for L. Hanzo: orcid.org/0000-0002-2636-5214
Catalogue record
Date deposited: 25 Nov 2005
Contributors
Author: S. X. Ng
Author: M. S. Yee
Author: L. Hanzo
University divisions
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Contact ePrints Soton: [email protected]
ePrints Soton supports OAI 2.0 with a base URL of https://eprints.soton.ac.uk/cgi/oai2
This repository has been built using EPrints software, developed at the University of Southampton, but available to everyone to use.
We use cookies to ensure that we give you the best experience on our website. If you continue without changing your settings, we will assume that you are happy to receive cookies on the University of Southampton website.
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|
2018-02-22 16:51:56
|
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https://www.documentation.jijzept.com/docs/jijmodeling_docs/4jijmodeling_transpiler/
|
# Introduction of JijModeling-Transpiler
JijModeling-Transpiler is sub-package of jijmodeling. jijmodeling.transpiler provides transpiling methods to other modeling tools such as PyQUBO and Python-MIP. We can use JijModeling-Transpiler to run optimization of our model which we construct with JijModeling locally without JijZept. jijmodeling.transpiler can be installed through pip command.
pip install jijmodeling-transpiler -U
In this tutorial, we would like to take the Knapsack problem as an example to illustrate the usage of jijmodeling.transpiler.
Before we explain the usage of jijmodeling.transpiler, we need to construct the mathematical model with JijModeling.
import jijmodeling as jm# define variablesv = jm.Placeholder('v', dim=1)N = v.shape[0]w = jm.Placeholder('w', shape=(N))W = jm.Placeholder('W')x = jm.Binary('x', shape=(N))i = jm.Element('i', (0, N))# set problemproblem = jm.Problem('Knapsack') # set objective functionobj = - jm.Sum(i, v[i]*x[i])problem += obj# set total weight constraintconst = jm.Sum(i, w[i]*x[i])problem += jm.Constraint('weight', const<=W)problem
\begin{alignat*}{4}\text{Problem} & \text{: Knapsack} \\\min & \quad - \sum_{ i = 0 }^{ v_{\mathrm{shape}(0)} - 1 } v_{i} \cdot x_{i} \\\text{s.t.} & \\& \text{weight} :\\ &\quad \quad \sum_{ i = 0 }^{ v_{\mathrm{shape}(0)} - 1 } w_{i} \cdot x_{i} \leq W,\\[8pt]& x_{i_{0}} \in \{0, 1\}\end{alignat*}
We also need to prepare the problem instance data.
# set a list of values & weights inst_v = [5, 7, 2, 1, 4, 3]inst_w = [8, 10, 6, 4, 5, 3]# set maximum weightinst_W = 20instance_data = {'v': inst_v, 'w': inst_w, 'W': inst_W}
## convert to PyQUBO model
We can use to_pyqubo to convert jijmodeling to PyQUBO model.
from jijmodeling.transpiler.pyqubo import to_pyqubo# convert to pyqubopyq_model, pyq_chache = to_pyqubo(problem, instance_data, {})
We can create QUBO from PyQUBO model and run annealing using openjij.
import openjij as oj# set multiplierslam1 = 1.0multipliers = {'weight': lam1} # create quboqubo, bias = pyq_model.compile().to_qubo(feed_dict=multipliers)# set samplersampler = oj.SASampler(num_reads = 10)# solve problemresponse = sampler.sample_qubo(qubo)
.decode can be used to obtain results from openjij's result in a more user-friendly manner.
# decode solutionresult = pyq_chache.decode(response)
Finally, let us check the results we obtained.
import numpy as nplowest_result = result.lowest()[0]indices, _, _ = lowest_result.record.solution['x'][0]sum_w = np.sum([instance_data['w'][i] for i in indices[0]])print('Indices of x = 1: ', indices[0])print('Value of objective function: ', lowest_result.evaluation.objective)print('Value of constraint term: ', lowest_result.evaluation.constraint_violations['weight'])print('Total weight: ', sum_w)
Indices of x = 1: [1, 4, 5]Value of objective function: [-14.0]Value of constraint term: [0.0]Total weight: 18
## Convert to Python-MIP
While constructing a mathematical model, you may want to solve the problem with an exact solver to verify your model. JijModeling-Transpiler provides the conversion function to_mip to Python-MIP model which provides modeling and solving tools for Mixed Interger Linearr Problem. The usage is almost same as to_pyqubo.
from jijmodeling.transpiler.mip import to_mip# convert to mipmip_model, mip_cache = to_mip(problem, instance_data, {})
We can solve the problem with .optimize. If you would like to know more detail usage, so please check the Python-MIP document.
status = mip_model.optimize()
Welcome to the CBC MILP Solver Version: TrunkBuild Date: Oct 24 2021 Starting solution of the Linear programming relaxation problem using Primal SimplexCoin0506I Presolve 1 (0) rows, 6 (0) columns and 6 (0) elementsClp1000I sum of infeasibilities 0 - average 0, 5 fixed columnsCoin0506I Presolve 0 (-1) rows, 0 (-6) columns and 0 (-6) elementsClp0000I Optimal - objective value -15.25Clp0000I Optimal - objective value -15.25Coin0511I After Postsolve, objective -15.25, infeasibilities - dual 0 (0), primal 0 (0)Clp0000I Optimal - objective value -15.25Clp0000I Optimal - objective value -15.25Clp0000I Optimal - objective value -15.25Clp0032I Optimal objective -15.25 - 0 iterations time 0.002, Idiot 0.00Starting MIP optimization
We can also use .decode function to obtain the results in a more user-friendly manner.
result = mip_cache.decode((status,mip_model))result
SampleSet(record=Record(solution={'x': [(([1, 4, 5],), [1.0, 1.0, 1.0], ())]}, num_occurrences=[1]), evaluation=Evaluation(energy=[], objective=[-14.0], constraint_violations={}, penalty=[]), measuring_time=MeasuringTime(solve=SolvingTime(preprocess=None, solve=None, postprocess=None), system=SystemTime(post_problem_and_instance_data=None, request_queue=None, fetch_problem_and_instance_data=None, fetch_result=None, deserialize_solution=None), total=None))
indices, _, _ = result.record.solution['x'][0]sum_w = np.sum([instance_data['w'][i] for i in indices[0]])print('Indices of x = 1: ', indices[0])print('Value of objective function: ', result.evaluation.objective)print('Total weight: ', sum_w)
Indices of x = 1: [1, 4, 5]Value of objective function: [-14.0]Total weight: 18
In this tutorial, we explain the basics how to use JijModeling-Transpiler. You can easily convert your model written in jijmodeling to other modeling tools.
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2023-02-02 11:43:43
|
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|
http://oejournal.org/mv_html/j00001/2019-05/A190514000002_WEB.htm
|
光电工程 2019, Vol. 46 Issue (5): 180306 DOI: 10.12086/oee.2019.180306
Two input one output visible light communication system based on pulse amplitude modulation
Shi Meng, Zhang Mengjie, Chi Nan
Key Laboratory of Electromagnetic Wave Information Science, Ministry of Education, Department of Communication Science and Engineering, Fudan University, Shanghai 200433, China
Abstract: To improve the data transmission rate of the conventional point-to-point single input single output (SISO) visible light communication system, a multiple input multiple output (MIMO) visible light communication system is proposed. Considering the complexity of the receiver system, multiple input single output (MISO) visible light communication systems have attracted attention. This paper studies the MISO visible light communication system based on pulse amplitude modulation (PAM), and experimentally proves the advantages of this system in specific scenes. In addition, there are non-linear effects for key devices such as LED light sources and power amplifiers in visible light communication systems. Based on 2×1 MISO visible light communication system, this paper reports a novel equal probability coding mapping scheme for high-order PAM signals with two low-order PAM signals superposition in the optical domain. The system verification is performed through a net data-rate of 700 Mb/s transmission experiment through a red chip of RGB-LED, which proves the feasibility and superiority of this scheme in practice.
Keywords: optical communications visible light communication multiple input single output (MISO) equal probability coding pulse amplitude modulation (PAM)
1 引言
2 两发一收可见光通信系统中光生PAM7信号的编码映射方案
图 1 两路PAM4信号叠加生成PAM7信号的编码示意图。(a)不等概率PAM7信号；(b)等概率PAM7信号 Fig. 1 The schematic diagram of PAM7 signals generated by two PAM4 signals superposition. (a) Unequal probability PAM7 signals; (b) Equal probability PAM7 signals
Signal Equal probability PAM4 Unequal probability PAM4 Equal probability PAM7 Unequal probability PAM7 PAPR 1.80 1.34 2.25 3.6
3 基于PAM调制的两发一收可见光通信系统结构
图 2 基于PAM调制的两发一收可见光通信系统框图及实验装置图 Fig. 2 The block diagram and experimental setup of two input one output visible light communication system based on PAM modulation
4 单发单收和两发一收PAM7系统实验结果比较
$H(x) = H = - \sum\nolimits_{x \in \chi } {{P_X}(x)\log _2^{}} {P_X}(x),$ (1)
4.1 单发单收发射端生成不等概率PAM7电信号
$s(t) = \operatorname{Re} [{A_m}g(t){{\rm{e}}^{{\rm{j}}2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t}}] = {A_m}g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t,$ (2)
${s_{{\rm{Tx}}1}}(t) + {s_{{\rm{Tx}}2}}(t) = ({A_{m1}} + {A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t,$ (3)
$P({A_{m1}}) = P({A_{m2}}) = 1/4,$ (4)
$r(t) = [{s_{{\rm{Tx}}1}}(t) + {s_{{\rm{Tx2}}}}(t)] + {n_0}(t)\\ \;\;\;\;\;= ({A_{m1}} + {A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t + {n_0}(t)。$ (5)
$I(t) = |r(t){|^2} + {n_{\rm{r}}}(t) \\ \;\;\;\;\;= {[({A_{m1}} + {A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t]^2} + {n_0}^2(t) \\ \;\;\;\;\;+ 2{n_0}(t)({A_{m1}} + {A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t{\rm{ + }}{n_{\rm{r}}}(t) 。$ (6)
图 3 单发单收发射端不等概率PAM7信号情况下。(a) BER随Vled和Vpp变化关系；(b)不同Vpp下不等概率PAM7信号星座点和BER Fig. 3 Unequal probability PAM7 signal at transmitter in single input single output system. (a) The BER performance with Vled and Vpp; (b) The constellation points and BER performance under different Vpp
4.2 单发单收发射端生成等概率PAM7电信号
$P({A_{mi}} + {A_{mj}}) = 1/7。$ (7)
图 4 单发单收发射端等概率PAM7信号情况下。(a) BER随Vled和Vpp变化关系；(b)不同Vpp下等概率PAM7信号星座点和BER Fig. 4 Equal probability PAM7 signal at transmitter in single input single output system. (a) The BER performance with Vled and Vpp; (b) The constellation points and BER performance under different Vpp
4.3 两发一收接收端生成不等概率PAM7光信号
$r(t) = {s_{{\rm{Tx}}1}}(t) + {n_{{0_{{\rm{Tx}}1}}}}(t) + {s_{{\rm{Tx2}}}}(t) + {n_{{0_{{\rm{Tx}}2}}}}(t)\\ \;\;\;\;\;= ({A_{m1}}{\rm{ + }}{A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t + {n_{{0_{{\rm{Tx}}1}}}}(t) + {n_{{0_{{\rm{Tx}}2}}}}(t),$ (8)
$I(t) = |r(t){|^2} + {n_{\rm{r}}}(t) \\ \;\;\;\;\; {\rm{ = }}{[({A_{m1}} + {A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t]^2} \\\;\;\;\;\; + {[{n_{{0_{{\rm{Tx}}1}}}}(t) + {n_{{0_{{\rm{Tx}}2}}}}(t)]^2} \\\;\;\;\;\; + 2[{n_{{0_{{\rm{Tx}}1}}}}(t) + {n_{{0_{{\rm{Tx}}2}}}}(t)]({A_{m1}} + {A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t \\ \;\;\;\;\; {\rm{ = }}{[({A_{m1}} + {A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t]^2} \\ \;\;\;\;\; + n_{{0_{{\rm{Tx}}1}}}^2(t) + n_{{0_{{\rm{Tx}}2}}}^2(t){\rm{ + 2}}{n_{{0_{{\rm{Tx}}1}}}}(t){n_{{0_{{\rm{Tx}}2}}}}(t) \\ \;\;\;\;\; + 2[{n_{{0_{{\rm{Tx}}1}}}}(t) + {n_{{0_{{\rm{Tx}}2}}}}(t)] \\\;\;\;\;\; \cdot ({A_{m1}} + {A_{m2}})g(t)\cos 2{\rm{ \mathsf{ π} }}{f_{\rm{c}}}t + {n_{\rm{r}}}(t)。$ (9)
图 5 两发一收接收端不等概率PAM7信号情况下。(a) BER随Vled和Vpp变化关系；(b)不同Vpp下不等概率PAM7信号星座点和BER Fig. 5 Unequal probability PAM7 signal at receiver in two input one output system. (a) The BER performance with Vled and Vpp; (b) The constellation points and BER performance under different Vpp
4.4 两发一收接收端生成等概率PAM7光信号
图 6 两发一收接收端等概率PAM7信号情况下。(a) BER随Vled和Vpp变化关系；(b)不同Vpp下等概率PAM7信号星座点和BER Fig. 6 Equal probability PAM7 signal at receiver in two input one output system. (a) The BER performance with Vled and Vpp; (b) The constellation points and BER performance under different Vpp
图 7 系统动态工作范围比较。(a)单发单收发射端生成不等概率PAM7；(b)单发单收发射端生成等概率PAM7；(c)两发一收接收端生成接收不等概率PAM7；(d)两发一收接收端接收生成等概率PAM7 Fig. 7 The comparison of system's dynamic working area. (a) Unequal probability PAM7 signal at transmitter in SISO system; (b) Equal probability PAM7 signal at transmitter in SISO system; (c) Unequal probability PAM7 signal at receiver in two input one output system; (d) Equal probability PAM7 signal at receiver in two input one output system
5 结论
[1] Zeng L B, O'Brien D C, Le Minh H, et al. High data rate multiple input multiple output (MIMO) optical wireless communications using white LED lighting[J]. IEEE Journal on Selected Areas in Communications, 2009, 27(9): 1654-1662. [Crossref] [2] Mesleh R, Mehmood R, Elgala H, et al. Indoor MIMO optical wireless communication using spatial modulation[C]//Proceedings of 2010 IEEE International Conference on Communications, 2010: 1–5. [3] Yu Z H, Baxley R J, Zhou G T. Multi-user MISO broadcasting for indoor visible light communication[C]//Proceedings of 2013 IEEE International Conference on Acoustics, Speech and Signal Processing, 2013: 4849–4853. [4] Qian H, Yao S J, Cai S Z, et al. Adaptive postdistortion for nonlinear LEDs in visible light communications[J]. IEEE Photonics Journal, 2014. [Crossref] [5] Wang C, Zhou Y J, Chi N. Research of LED's nonlinear distortion compensation algorithm in visible light communications[J]. China Light & Lighting, 2017(7): 9-15, 26. 王灿, 周盈君, 迟楠. 可见光通信中抗非线性方法的比较研究[J]. 中国照明电器, 2017(7): 9-15, 26 [Crossref] [6] Zhang M J, Shi M, Wang F M, et al. 4.05-Gb/s RGB LED-based VLC system utilizing PS-Manchester coded Nyquist PAM-8 modulation and hybrid time-frequency domain equalization[C]//Proceedings of the Optical Fiber Communication Conference, 2017: 1–3.
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2019-06-25 22:35:54
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http://universeinproblems.com/index.php/Characteristic_Parameters_and_Scales
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# Characteristic Parameters and Scales
### Problem 1
Calculate the dark energy density and the cosmological constant value.
### Problem 2
Estimate total number of baryons in the Universe.
### Problem 3
Find dependence of relative density of dark energy $\Omega_\Lambda$ on the redshift. Plot $\Omega_\Lambda(z)$.
### Problem 4
Estimate total number of stars in the Universe.
### Problem 5
Find the ratio of dark energy density to the energy density of electric field of intensity $1\,V/m$. Compare the dark energy density with gravitational field energy density on the Earth surface.
### Problem 6
Estimate the distance between two neutral hydrogen atoms at which the gravitational force of their attraction is balanced by the repulsion force generated by dark energy in the form of cosmological constant. Make the same estimates for the Sun-Earth system.
### Problem 7
Calculate magnitude of physical acceleration.
### Problem 8
How far can one see in the Universe?
### Problem 9
Find age of the Universe.
### Problem 10
Give a qualitative explanation why the age of the Universe in SCM is considerably larger than the age of matter dominated Universe (the Einstein-de Sitter model).
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2017-03-28 23:33:23
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http://math.stackexchange.com/questions/334341/prod-k-0-infty-p-k-0-rightarrow-sum-k-0-infty-1-p-k-infty
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# $\prod_{k=0}^\infty p_k > 0 \Rightarrow \sum_{k=0}^\infty (1-p_k) < \infty$
Let $\{p_k\}$ be a probability mass sequence. Is it true that if $\prod_{k=0}^\infty p_k > 0$ then $\sum_{k=0}^\infty (1-p_k) < \infty$?
-
Actually, I just noticed that my question does not make sense, because if $\{p_k\}$ is a probability mass sequence, then $\sum_{k=0}^\infty p_k = 1$, but then $\prod_{k=0}^\infty p_k$ cannot be greater than $0$. I'm sorry. – user67398 Mar 20 '13 at 1:59
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2016-05-01 23:54:12
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https://www.studentsroom.in/introduction-of-common-emitter-amplifier/
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# introduction of common emitter amplifier
## Aim:
Design and set up the BJT common emitter amplifier using voltage divider bias with and without feedback and determine the gain- bandwidth product from its frequency response.
## Components and equipments required:
Transistor – SL100, Resistors – 470 Ω, 1KΩ, 10KΩ – 2nos, and 33KΩ, Capacitors 100µf, 0.22µf and 0.47µf, Power Supply, 10Hz –3MHz Signal generator, CRO, Connecting wires and Bread board/Spring board with spring terminals.
## Design:
Transistor: SL100
Let VCC = 12V; IC = 4.5 mA; VE = 1.2V; VCE = 6V; hFE = 100.
Given VE = 1.2V. Therefore RE = VE / IE » VE / IC =266.67W; RE =270Ω
Writing KVL for the Collector loop we get, VCC = ICRC + VCE + VE
RC = (VCC – VCE – VE) / IC = (12-6-1.2)V/4mA=1.06KW; RC = 1 KW
hFE RE = 10R2
Assume R2=2.7KΩ,
VB = (Vcc x R2 ) / (R1 + R2)
Hence R1 = 14.14 KW ; R1 = 15 KW
Use CC1 = 0.47mF
Use CC2 = 0.47mF
Use CE =47mF
## Procedure:
Follow the same procedure for both circuits
• After making the connections, switch on the D.C. power supply and check the D.C. conditions without any input signal and record in table below:
Parameter VRC VCE VE ICQ VBE Assumed 4.8V 6V 1.2V 4.5mA 0.6V Practical
• Select sine wave input and set the input signal frequency ≥10f1 (Say = 10 KHz. This will be a convenient ‘Mid – frequency’).
• Observe the input wave form and output wave form on a dual channel CRO.
• Adjust the input amplitude such that the output waveform is just undistorted (or in the verge of becoming distorted). Measure the amplitude of the Input Signal now. This amplitude is the Maximum Signal Handling Capacity of your amplifier.
• Decrease the input voltage to a convenient value such that the output is undistorted. Say 20mV. Measure the corresponding o/p voltage. Calculate mid-band gain, AM = Vo (p-p) / Vin (p-p).
• Keeping the input voltage constant, go on reducing the frequency until the output voltage reduces to 0.707 times its value at 10 KHz. The frequency at which this happens gives you the Lower Cut-off frequency (f1).
• Keeping the input voltage constant, go on increasing the frequency until the output voltage decreases to 0.707 times its value at 10 KHz. The frequency at which this happens gives you the Upper Cut-off frequency (f2).
• Thus you have pre-determined f1 and f2. Find the amplifier band width, BW = f2 f1
• Determine Gain Bandwidth product (GBW product) which is a Figure of Merit of your amplifier as GBW = AM x BW.
• Now repeat the experiment by recording values of output voltage versus frequency keeping the input voltage at a constant value convenient to you. You should take at least 5 readings below f1 and 5 readings above f1, at least 5 readings in the mid band, at least 5 readings below f2 and 5 readings above f2.
• Plot graphs of AV versus Frequency, f and /or M, dB versus Frequency, f on a semi log graph paper. From the graph determine: Mid –band – gain, Lower and Upper Cut-off frequencies and Band width. Compute the GBW product and verify with answer obtained earlier.
## Observation:
Use the tabular column separately for each circuit
Vin (P-P) = …….. V
AV= VO(P-P)/Vin(P-P)
M = 20log (AV) dB
Frequenc y in Hz 100 200 300 350 400 500 600 700 800 VO(P-P) in volts Av M dB(Av in dB)
Frequenc y in Hz 1k 2k 3k 5k 8k 10k 20k 30k 50k VO(P-P) in volts Av M dB(Av in dB)
Frequenc y in Hz 100k 200k 300k 400k 500k 600k 700k 800k 900k VO(P-P) in volts Av M dB(Av in dB)
## Result:
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Insert math as
$${}$$
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2018-12-19 14:42:20
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http://fivethirtyeight.com/features/kentuckyoregon-primary-thread/
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# FiveThirtyEight
## Politics
9:28 PM. It’s an excellent speech on its own merits, but emotionally, I just think you get a much higher high if you wait the hour until Oregon comes in.
9:00 PM. The thing that I find intriguing about Sirota’s Race Chasm theory is that it implies a certain amount of non-linearity in the way that people vote. It may not be simply a matter of adding and subtracting constituencies but rather there are various sorts of tipping points and network effects.
It’s hard to believe that only 5 percent of the Democrats in Floyd County, Kentucky came out of the womb ready to vote for Barack Obama. But maybe the number is 20 percent, and the 20 percent talk to their neighbors in the 80 percent, and before long, they become part of that majority as well.
Toward that end, one thing that may be particuarly relevant in Appalachia is that it tends to have very low rates of mobility. People from outside the region don’t tend to move there, and people from inside the region don’t tend to leave. As such, any sort of network effects might tend to be magnified there.
8:38 PM. The operative question tonight is not so much whether Clinton can get close enough to have a path to the nomination, but whether she can get close enough that she thinks she has a path to the nomination.
Her tone in her victory speech tonight was different than in West Virginia — and seemed to suggest that she might think she does have such a path. Howard Fineman seems to think she might think so.
She certainly came pretty close to her best-case scenario tonight in terms of her popular vote gain in Kentucky — not so much because of the margin (which everyone except my model seemed to get about right) but because of the relatively high turnout. Conversely, the magnitude of Obama’s margin in Oregon might be fairly important in terms of disabusing her of that notion.
8:19 PM. Floyd County, Kentucky: Clinton 11,215 votes, Obama 653.
8:08 PM. When they’re finished counting Kentucky’s votes, Obama will hold a lead in the + Florida popular vote count of about 180,000 votes. Obama is liable to get anywhere from 50,000 to 150,000 of those votes back in Oregon. Erring slightly to the lower side of that estimate, he’ll probably exit the evening about 250,000 votes ahead in the +Florida count.
I have absolutely no idea how to project the results in Puerto Rico. A 25-point win for Clinton on turnout of 1 million would get her those 250,000 votes. A 13-point win (the margin in the only public poll of Puerto Rico) for Clinton on turnout of 600,000 would net her 78,000 votes. Either of these results are plausible. An Obama victory or a very close result also is plausible.
The narrative issue that Clinton faces, I think, is that her argument loses moral force if it all boils down to what turnout in Puerto Rico is liable to be.
7:45 PM. To follow up on my previous thought. If you Google the phrase: “as _____ goes, so goes the nation”, here are the states that come up most often.
1. Maine 2,8702. Utah 1,1003. Ohio 9654. California 6925. Iowa 3526. Florida 2217. Missouri 2168. Texas 1549. West Virginia 11010. Michigan 103
Maine was the subject of the original phrase, whereas Utah was the subject of a Mitt Romney joke. Nobody in the history of modern society has apparently used the phrases “As Oklahoma goes, so goes the nation”, or, “As Alabama goes, so goes the nation”. Until now.
7:28 PM. Clinton: “It has often been said: As Kentucky Goes, So Goes the Nation”.
Actually, if you Google that phrase, it comes up just 16 times.
7:15 PM. To answer SPorcupine’s question: you can infer those numbers from the exit polls, and 15 percent of Clinton supporters who stated a preference said that they’d vote for McCain in November.
6:42 PM. Chuck Todd clearly reads David Sirota. For what it’s worth, I’m not completely convinced by the Race Chasm theory, but I think it makes for an interesting discussion.
FWIW, Obama has opened up a lead of a couple of points in Jefferson County (Louisville) and turnout there is enormous. Although, there’s something funny going on with the reporting right now and I’m taking everything with a slight grain of salt at this point.
6:32 PM. It sure looks to me like Obama is not going to achieve viability in KY-5. He’s losing some counties like 91-7.
6:16 PM. It’s one thing for a campaign surrogate to spin information, but Terry McAuliffe just flat out made something up on Hardball, claiming that there was a general election poll in Kentucky that showed Hillary Clinton ahead of John McCain. If such a poll exists, there is no evidence of it anywhere on the Internet. I also heard him make the same claim about a week ago, so it wasn’t any kind of misspeak.
6:02 PM. Kentucky called by all networks for Clinton. Exit polls suggest a big win, about 64-29. The exit poll also suggests that only about 9 percent of the electorate was black. Based on typical patterns in other states, you’d expect black turnout to equal about 150 percent of the state’s African-American population, or 12 percent. So it could be that Obama’s lackluster campaign in Kentucky meant that he didn’t turn out his base. Or it could mean that the exit poll was a little off and Obama might slightly outperform those numbers.
5:53 PM. I don’t want to call this a “liveblog” because it will be updated fairly sporadically, but if I have any thoughts worth sharing, they will go here.
Don’t be fooled by the early results in Kentucky: almost everything is from Louisville. We had projected Louisville at 42/54/4 (Clinton/Obama/Other) and it’s actually coming in at 48/49/3. So, Clinton is overperforming our estimates in this district by 11 points, which would imply about a 30-point win overall.
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All Nate Silver
All Politics
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2014-08-30 22:31:52
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https://andthentheresphysics.wordpress.com/
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## The social construction of science
Richard Dawkins posted a tweet that cause a bit of a furore in some sectors of Twitter. He did try to clarify, but it still didn’t go down well.
The problem with his tweet is that science clearly is socially constructed. It’s done by people who make decisions that are strongly influenced by our social norms. In some cases, as discussed in Angela Saini’s book Superior, this not only influences how we do science, but also influences the results of some research activities, or how we interpret research results.
If we want to deal with the issues highlighted in Angela Saini’s book, and also improve diversity and inclusion in science, then we need to recognise that science is socially constructed.
However, I also understand why some scientist push back against this framing. It’s either because they think it’s implying that scientific results are social constructs, or that they will be interpreted as being social constructs. The concern being that this can imply that scientific results are constructed (made up) by people, rather than them tending towards properly representing whatever system is being studied.
I realise that the latter is not what those who highlight the social construction of science are actually suggesting, but I do get why it might sometimes seem that way. However, I do think it’s worth scientists trying to understand why it is important to recognise that societal factors play a big in role in determining how we do science, and can – in some circumstances – influence how we interpret scientific results.
However, I also think it’s worth Humanities scholars understanding why there can be push back from scientists. It’s not that they think societal factors play no role in science, it’s more that they think that this doesn’t necessarily imply that societal factors will have a big influence on scientific results, or how we interpret these results. There’s a concern that this can lead to people undermining scientific results when these results seem inconvenient. I think this is a valid concern, even if this isn’t what the social construction of science actually implies.
## ‘Net zero’
There’s been some recent debate about the term ‘net-zero’. Just to give some basic background, given that the zero emission commitment is close to zero (i.e., when we get anthropogenic emissions to zero, global surface temperatures should soon stabilise) means we can define a carbon budget. This tells us how much more we can emit if we want some chance of staying below some temperature target. It also tells us that our emissions must go to zero. The complication is that this could occur through emissions actually going to zero, or through some kind of negative emission technology offseting some continued human-caused emissions (this could include some active land management).
What some are concerned about is the possibility that some future negative emission technology could allow some to make ‘net zero’ promises that they may not be able to keep, or never actually intended to keep. In other words, they will claim that they’re aiming to get to a stage where they are offsetting all of their emissions despite it not yet being known if such technologies can actually operate at a suitable scale. Essentially, this becomes a form of greenwashing.
Another problem, though, is that some are interpreting ‘net zero’ in ways that aren’t consistent with what is intended. Mark Carney, for example, claimed that a company for which he is vice-Chair was ‘net zero’ because their enormous renewables business had avoided emissions that were comparable to their actual emissions. However, this isn’t ‘net zero’, it just means that they’ve ended up emitting about half of what they might have emitted.
‘Net zero’ requires actively sequestering an amount comparable to the amount actually emitted, not avoiding emitting an amount comparable to how much was actually emitted. I managed to come up with a reasonably popular tweet that illustrated the problem with Mark Carney’s suggestion.
I also came across another complication today, where someone interpreted ‘net zero’ as being the point when actual emissions are offset by negative emission technologies and by natural sinks. If this means that there is no net emission of any kind (anthropogenic, or natural), then it would imply constant concentrations. However, the reason that the zero emission commitment is probably small is because the natural sinks continue to take up some of our emissions after our emissions have gone to zero so that atmospheric CO2 concentrations actually decrease.
If we get to a stage where atmospheric concentrations stabilise, then we would actually continue to warm (this is the constant concentration commitment, which is different to the zero emission commitment). Hence, ‘net zero’ in this context means ‘net zero’ anthropogenic, not ‘net zero’ anthropogenic and natural. I should make clear that the above interpretation was, I think, more some confusion about ‘net zero’ than any attempt to define it in some convenient way, but it does illustrate how this can be a tricky concept.
So, I can see why some are concerned about the term ‘net zero’ and, as Simon Lewis points out, we can’t solve climate change using accounting tricks. However, I do think that the term ‘net zero’ is fine and that we should be careful of changing terminology just because some aren’t using it appropriately. However, it is important to stress that ‘net zero’ means ‘net zero’ anthropogenic emissions and that in the absence of negative emission technologies ‘net zero’ is the same as real zero. In other words, if negative emission technologies are unlikely to operate effectively at scale, then ‘net zero’ requires simply getting anthropogenic emissions to zero.
Warming commitments – post of mine about warming commitments with a number of useful links at the end.
Mark Carney Walks Back Brookfield Net-Zero Claim After Criticism – article describing Mark Carney’s misinterpretation of ‘net zero’.
The climate crisis can’t be solved by carbon accounting tricks – Guardian article by Simon Lewis.
## Losing the sky
Andy Lawrence, who happens to be a colleague, has just published a book called Losing the Sky. Andy also gave a brief presentation about it, which is what motivated me to write this post. The book is very reasonably priced and very easy to read. It’s about Starlink, the constellation of low Earth orbit satellites being launched by SpaceX. There are currently just over 1000 in orbit, with plans for 12000, and a possible extension to 42000. The goal is to provide high-speed internet with low latency.
As the image on the right illustrates, the issue is that (especially during the orbit raising phase) these satellites can be very prominent in astronomical images. Since there will be so many of them, this could have a very large impact. This is not only a problem for ground-based observations; even images taken with the Hubble Space Telescope have been impacted.
It’s also not only optical astronomy, radio astronomy may be even more severely impacted. Currently, most communication satellites are in geo-synchronous orbits. Consequently, radio observations can typically be planned to keep their transmissions out of the side-lobes. With this new constellation of low Earth orbit communication satellites, this may become essentially impossible, potentially ruining radio astronomy.
One concern with complaining about this, is that the stated goal is to provide internet to regions of the planet that don’t currently have decent access. This is clearly a worthy goal and so it can be tricky to object on the basis of how it will impact astronomical observations. There are, however, a few issues with this stated goal. One is that there are already solutions involving satellites on higher orbits, so it’s not clear that providing internet to under-served regions of the planet requires a constellation of low Earth orbit satellites. Also, the current price suggests that this may also currently be out of reach of many in these regions.
What seems more likely is that the motivation is to reduce the latency (the data transfer time) which will be very attractive to the financial sector. This will require a constellation of low Earth orbit satellites. So, the actual goal may not be quite as magnanimous as suggested.
As my colleague’s book suggests, this does seem to be another example of a tragedy of the commons. Some get to benefit from using the environment in a way that negatively impacts many others, who don’t get compensated for how they are impacted.
Even if we would benefit from high-speed, low-latency internet access across the globe, I do think there would still be merit to a process that assessess the impact of the proposed solution and that has some ability to influence, and potentially regulate, this kind of activity. We can’t keep ignoring the impact of how our activities influence the environment in which we all live, not only for fairness reasons, but because there is a cost to such activities that someone will eventually have to pay.
Posted in Environmental change, Scientists | | 33 Comments
## Agricultural emissions
There’s a really nice recent paper by John Lynch, Michelle Cain, David Frame and Ray Pierrehumbert on Agriculture’s Contribution to Climate Change and Role in Mitigation Is Distinct From Predominantly Fossil CO2-Emitting Sectors. It’s largely discussing why there are important differences between carbon dioxide (CO2), which is a stock pollutant, and methane (CH4), which is predominantly a flow pollutant.
The basic point is that the emission of CO2 increases the stock, which leads to a long-term increase in atmospheric concentrations and, consequently, to warming that will persist for a very long time. Methane, on the other hand, has a short atmospheric lifetime, decaying within decades to CO2 and water. Given that – for agricultural emissions – the carbon comes from plants, this doesn’t add a new carbon to the system, and hence doesn’t increase the stock. This isn’t strictly true for methane from natural gas, since that does add a new carbon to the system, but this is relatively small when compared to direct CO2 emissions from fossil fuels.
The key figure in the paper is the one above. The left-hand panel shows an example of an emission pathway based on using CO2-equivalents using the 100-year Global Warming Potential (GWP100). The right-hand panel shows the actual warming we would experience for different gas-specific compositions. CO2 warming (dark blue line) peaks when emissions gets to zero, but then remains at this level well after emissions have ceased (it’s essentially irreversible without some kind of artificial negative emission technology).
Methane (yellow line) initially produces more warming than would be expected based on its CO2-equivalence. However, when emissions start to go down, there is cooling, which continues well after emissions have ceased (for completeness, the pink line is 50% methane, 50% CO2, while the green line is N2O which has a reasonably long atmospheric lifetime).
The key point is that if one is using GWP100 to estimate CO2-equivalence, you would predict warming profiles that would be quite different to what would happen in reality. You would under-predict the impact of methane emissions initially, but then over-predict its impact later on.
The reason this is important is because any emission reduction pathways are likely to involve trade-offs. Consequently, as the paper highights,
reducing methane emissions at the expense of CO2 is a short-sighted approach that trades a near-term climate benefit with warmer temperatures for every year thereafter
and
If strong efforts are made to reduce agricultural emissions but prove expensive—in terms of monetary costs, political capital, public goodwill, or individual effort—and detract from efforts to eliminate fossil CO2 emissions then we will be climatically worse-off.
Essentially, the emission of a stock pollutant (CO2) leads to warming that will persist for a very long time, which is different to the impact of a flow pollutant (agricultural methane). The latter clearly does produce warming and, in fact, leads to more warming in the near-term than simple CO2-equivalent estimates would suggest. However, this warming would stabilise if emissions were to stabilise (unlike CO2) and can be reversed if these emissions are reduced (also, unlike CO2).
So, it would seem important to be aware of these differences when thinking of how best to decarbonise. Any strategy that prioritises short-lived pollutants over long-lived pollutants runs the risk of committing us to future warming that is essentially irreversible and that we could have avoided if we’d prioritised differently.
This isn’t to suggest that we should be ignoring the short-lived pollutants. They can have a large near-term impact which may be important if we wish to avoid crossing certain warming thresholds. There may also be other reasons for reducing these emissions (land use change, for example). I just happen to think that if we’re trying to assess the impact of different greenhouse gas emissions, it’s important to use a metric that properly represents this.
Agriculture’s Contribution to Climate Change and Role in Mitigation Is Distinct From Predominantly Fossil CO2-Emitting Sectors, new paper by Lynch et al. (2021)
Losing time, not buying time, Realclimate post by Ray Pierrehumbert making the same basic point (from 2010).
Methane, a post I wrote in 2019 about the impact of methane.
Guest post: A new way to assess ‘global warming potential’ of short-lived pollutants, Carbon Brief guest post by Michelle Cain.
Methane and things, another post I wrote last year trying to explain the difference between metane emissions and CO2 emissions.
Posted in Uncategorized | 179 Comments
## Deferential?
I was listening to a podcast interview with Steve Keen, whose work I’ve written about before. It was about his paper the appallingly bad neoclassical economics of climate change. I have a lot of sympathy with what he’s presenting. Some of the assumptions being made by economists in this context seem rather odd, and I’ve been critical of Integrated Assessment models (IAMs) myself.
I also make an appearance in the podcast, as an example of a scientist who is too deferential towards neoclassical economists. I don’t know if deferential is quite the right word (some of you may recall interactions I’ve had with a prominent climate economist), but I see what they mean and they do have a point. The point being made in the podcast is that some of the assumptions made in the neoclassical economics of climate change are so obviously nonsensical that they really should be being called out by scientists.
I agree that many of the assumptions seem odd. Economic growth is often assumed to be baked in. The damage estimates for high levels of warming seem ridiculously low. However, I’m also aware that it’s easy to look at some problem outside your area of expertise, think you’ve seen some obvious glaring error, and be wrong. Retired engineers are sometimes noted for this when it comes to climate change.
Also if you think it’s important to listen to experts, then you’d need to then have pretty strong reasons for arguing that we should ignore some of them. So, I am indeed reluctant to vocally call out neoclassical economists who work on climate change, mostly because there may well be (certainly are) aspects that I don’t understand, but partly because I do think expertise matters.
The suggestion was also not just that some scientists were too deferential, but that they should really be pushing back strongly against what is being presented by neoclassical economists.
I don’t really see why this should be the responsibility of scientists. I certainly think that it’s utterly bonkers that we could end up warming the climate (this century) by an amount comparable to the difference between a glacial and an inter-glacial, but I don’t have a good way to quantify the societal, and ecological, impact. It just seems obviously a silly thing to do.
I think scientists have done a great job of highlighting the risks. If others are still buying low-ball estimates from neoclassical economists, I don’t think this is the fault of scientists. I’m not suggesting scientists shouldn’t continue to highlight, and stress, these risks, but I don’t think they should be expected to sort out failings in another discipline. Feel free to disagree in the comments, though 🙂
## Anti-Virus
There’s a new site called Anti-Virus: The Covid-19 FAQ. It’s a little like Skeptical Science, with articles that respond to common arguments made by Covid Sceptics (what Skeptical Science would call Climate Myths). On a related note, I have been trying to help another group with a site called Simple Covid and we have been talking about doing something similar. Although we’ve (mostly others) have produced some infographics, we haven’t had time to write any myth-busting articles.
One thing I did find interesting about the Anti-Virus site was that they also list prominent Covid sceptics, including academics, journalists, and online sceptics. Such lists have been somewhat controversial in climate circles. Admittedly, this is partly to do with labelling, but the principle is clearly the same; create a list of people who have regularly promoted arguments that are wrong, and highlight what they’ve said and why they were wrong.
Although Skeptical Science has been remarkably successful, it’s not without its critics, partly because of a focus on consensus messaging and partly because of their climate misinformers page (now called misinformation by source). It would be interesting to know if some of these critics are also concerned about the tactics of the Anti-Virus site.
Personally, it seems to me that using consensus science to rebut common “skeptic” talking points is not only a reasonable thing to do, but can also be very effective. Skeptical science has actually numbered their responses to the Climate Myths, and we used this to rebut the recently released climate science (mis)information brief (which, amusingly, led to the re-assignment of two of the authors).
I also think that if people regularly promote arguments in public that are obviously wrong, then there’s nothing wrong with highlighting these errors and associating these people with others who also regularly promote such erroneous arguments. If those listed don’t like this, they could be more careful about what they say in public, could correct their past errors, or not really care if you still think your arguments are valid/defensible.
Anyway, I don’t really know where I’m going with this post, so will wrap up. I mostly just found it interesting that the Anti-Covid site is using a similar strategy to that used in the climate conext – debunking myths and naming and shaming prominent sceptics. I think it’s an interesting development.
Posted in ClimateBall, Pseudoscience, Science | | 534 Comments
## Alan’s Bottle
Me and Ken just had a talk over the Science Kerfuffle of the moment, featuring a physics and maths teacher known to pwn fashionable nonsense fans. He recently suggested that POMO weakened our herd immunity to combat objective untruths. He also wonders what to do now that the genie is out of the bottle. What Alan really means by these metaphors remains unclear.
Follows a slightly edited transcript.
[Willard, thereafter W]
[Ken, or AT in what follows]
That’s quite good. May motivate me to write a post.
[W]
thanks
the whole idea that people believe in fraud because of POMO looks ridiculous
[AT]
Do you agree with the suggestion that even if PoMo isn’t responsible it has undermined our ability to combat misinformation?
[W]
on the contrary, POMO tries to explain how misinformation can happen
Postmodernism is generally defined by an attitude of skepticism, irony, or rejection toward what it describes as the grand narratives and ideologies associated with modernism […]
[AT]
Okay, maybe I’ll have to rethink my post. Maybe I misunderstand PoMo, but if some of what goes on in STS falls with PoMo it certainly doesn’t seem to have helped, even if the goal is to explain how misinformation can happen.
[W]
we can disagree, that’s fine
it’s just small talk nobody will read
alan makes an important error:
indeterminacy should not lead to denial
and POMO could guard us against conspiracy ideation
the problem you got with STS is different:
for instance, MikeH’s main problem is that he has no idea of what he’s talking about
he has no business making metrological points without studying metrology
so we can agree that people say stuff without paying due diligence
[AT]
I guess I’m not a fan of over-generalizing. I guess my issue is more to do with STS, for example, claiming they have all sorts of tools for helping to deal with misinformation, while prominent people seem to either promote, or defend, misinformation. Grundmann with his “climate science is like race science”, Pearce with his criticism of consensus messaging without actually providing an alternative and publishing papers on climategate that repeat the myths, etc. So, if the tools are there, it feels that some people in that field are going to have to do a better job of explaining what they are and how to use them.
[W]
agreed
that’s not POMO tho, that’s editorializing or criticism, which is indeed a bane
STS sucks because it’s an interdisciplinary discipline whose practitionners know little about everything and therefore are dangerous enough almost everywhere
it may have inherited from POMO bad scholarship practices
[AT]
That’s what I was wondering. Isn’t there at least a PoMo element to some of STS. Weren’t they part of the Science Wars?
[W]
STS, as a discipline, is a result of older science wars
it tried to “sciencize” its output
instead of using abstract and unrealistic models like the old philosophers of science did,
it promised to look under the scientific hood
but if all you do is to play pretend by recycle kuhn this and popper that,
you get the worst of both worlds
(warren only adds “let’s find an exotic framework nobody will buy because it’s \$150”)
[AT]
Okay, yes, that probably does describe it pretty well.
[W]
so i would conclude two things
first, if one wishes to say something,
one has to study it with all the evidential responsibility it requires
due diligence, an idea that generalizes
me, you, alan, STS, POMO, everyone
second, it’s easier to be led astray by a lack of work in conceptual frameworks,
because words are just words–we need constructions
[AT]
I certainly agree with the first part of that. Don’t quite get what you mean by “words are not constructions”.
Why construct?
[W]
an old idea that i viktor recently retooled for his opiniated podcast
one can define impossible objects
one can’t construct them
empirical science prevents us from making claims that we can’t operationalize
scientists can’t pretend operationalization forces us to conclude one and only one thing
that’s just not what science affords us
that’s the main point from say bruno, whose framework is very good for climateball
once we accept that scientific theories evolve and are not to be taken for granted, all fits
[AT]
Okay, I think I get that.
[W]
so when i say that POMO isn’t responsible for our predicament, all i’m saying is that even if POMO did not exist, we’d still be stuck with that indeterminacy
(the inscrutability of reference is one of the indeterminacies attributed to van)
that said, you might be right on the historical point
warren peirce, gunter reiner grundmann, and mike hulme are not exactly helping
but even then, that’s just a guess
to show it would take some work
so as long as you keep clear that you’re editorializing, all should be fine, up to a point
[AT]
I’ll have to think a bit more. Alan’s point about PoMo not being responsible but also not helping resonated. Maybe that’s just too simple.
[W]
it resonates, but it rings hollow to me
after all these years, he’s just saying stuff, and that’s sad
his editorial exemplifies very well our predicament
we say stuff, and if it sounds good enough, we buy it
in fact the converse of his bottle hypothesis looks more plausible to me:
by amplifying the threat of POMO on the fate of western civilization, alan’s reactionary stance has been recycled by newscorp and has weaponized people with mental issues
conceptual boi has become a truther,
same for EricW
[AT]
That’s possible. I guess I have always thought that we don’t consider how what we say can then influence what we’re commenting on.
James Lindsay has always seemed a bit bonkers to me.
[W]
i learn from your posts because you express an attitude
you helped me keep my cool
in retrospect, toning down ages better
alan’s point is an old one, in fact as old as plato
philosophy is the history of how humans dealt with relativism and skepticism
[AT]
Yes, I am trying to tone down. Maybe I should ponder this a bit more.
[W]
as long as you can support what you’re saying, you should be fine
more so if your point is “if everyone supported their claims, that’d be great”
that’s just a more consistent approach
imo, alan fails that test
i could write a post if you prefer
[AT]
If you’re keen, go for it. I’m probably going to take it easy this evening, so if you have some time, feel free.
[W]
i’ll see what i can do
we could post that chat
[AT]
If you like, that’s fine with me.
[W]
good
[AT]
Thanks, you too.
## On baselines and climate normals
Mike Hulme, Professor of Human Geography at the University of Cambridge, has a somewhat bizarre article published in Academia Letters called Climates Multiple: Three Baselines, Two Tolerances, One Normal. It’s basically a discussion of the recent World Meteorological Organisation (WMO) decision to re-define the present day climate as the period 1991-2020, replacing the period 1961-1990.
The article starts by suggesting that this means that
Climate will ‘change’, one might say, in an instant; the world’s climate will ‘suddenly’ become nearly 0.5°C warmer. It is somewhat equivalent to re-setting Universal Time or adjusting the exact definition of a metre.
Well, from the mid-1970s to the early 2000s we actually have warmed by about 0.5oC. This has nothing to do with how the baseline is defined. It’s also hard to see that it’s equivalent to adjusting the exact definition of a metre. I also wonder if Mike Hulme has got this the wrong way around. If we make the baseline period more recent, then the anomaly values actually go down, not up. You might argue that the change in baseline has caused the world to suddenly become 0.5oC cooler, rather than warmer (it hasn’t, obviously, but the change has reduced the anomaly values by about 0.5oC).
The rest of the article discusses the various baselines (present day, pre-industrial, historical) and what we might mean by a climate normal, but I don’t really get the overall point. Clearly we have to be careful about how we discuss climate change, be clear about what baseline we’re using, and be aware that what might be regarded as normal is changing. But this is a feature of the topic; it’s not something that can really be avoided.
It may be also technically true that
The adoption of particular baselines and tolerances is an overtly political process with geopolitical, ethical and technological consequence
but it’s also the case that none of these decisions change physical reality. Changing the baseline does not change how much we’ve warmed, how fast we’ve warmed, and how much we will warm if we continue to emit greenhouse gases into the atmosphere. If this is carefully communicated, it’s hard to see how these changes have any real political significance (on top of the political significance of climate change itself, of course).
In some sense, Mike Hulme’s article seems to be doing the very thing it’s cautioning against. The only way that changing the baseline, or what we regard as a climate normal, can have any broader political significance is if people overplay the significance of making these changes. Suggesting that redefining a baseline has geopolitical implications would seem to be an example of doing so.
## Warming commitments
There’s been quite a lot of recent discussion about warming commitments. It started with an article by Bob Berwyn called Net Zero Emissions Would Stabilize Climate Quickly Says UK Scientist, followed soon after by one saying [w]arming already baked in will blow past climate goals, study finds. The first article is (I think) based on a recent multi-model analysis which suggests that the most likely value of Zero Emission Commitment (ZEC) on multi-decadal timescales is close to zero. The second article is reporting on results from another recent paper suggesting that [g]reater committed warming after accounting for the pattern effect.
So, why are we being presented with what appear to be inconsistent results? The simple answers is that we’re not really being careful enough to define what we mean by a warming commitment. The first article, and paper, are considering what would happen when we get emissions to zero. The second article, and paper, are essentially considering what would happen if atmospheric greenhouse gas concentrations remained at today’s levels. These are clearly two different scenarios.
When we get emissions to zero, the first paper indicates that – on multi-decade timescales – the zero emission warming commitment (ZEC) would be close to zero. On the other hand, if atmospheric CO2 concentrations were to remain constant, then we would continue warming to equilibrium. At today’s atmospheric CO2 concentrations, this would lead to additional warming of around 0.5oC or even more, according to the paper being highlighted in the second article above. However, it is important to realise that constant concentrations require continued emission, as illustrated by the second figure in this Steve Easterbrook post.
I should also stress that our understanding that there is little warming commitment associated with zero emissions has been understood for quite some time. The first paper to point this out was probably Matthews and Caldeira (2008), followed by Solomon et al. (2009), and Cao and Caldeira (2010). There’s also a Realclimate post pointing this out in 2010, the Steve Easterbrook post I mentioned above from 2013, and a post I wrote in 2016.
There are, however, a number of important caveats. That the zero emission warming commitment is probably small probably only applies on multi-decade timescales. The models that demonstrate this typically don’t include slower processes (such as ice sheet retreat, sea level rise, permafrost release) that may lead to additional warming on longer timescales.
Also, even though there is probably little commited warming on multi-decade timescale once we get emissions to zero, without negative emissions global surface temperatures will remain at an elevated level (relative to pre-industrial times) for a very long time. It does, however, indicate that our future warming depends mostly on future emssions. We can still influence how much future warming we are likely to experience, even if we can’t turn everything off right now.
So, I think it’s good that there is more recognition that the ZEC is probably small. It does address claims that there’s nothing we can do to avoid a lot of future warming and does illustrates that, in the context of future warming, most of the inertia is societal, rather than inertia in the climate system.
Net Zero Emissions Would Stabilize Climate Quickly Says UK Scientist, article by Bob Berwyn.
Warming already baked in will blow past climate goals, study finds, Associated Press article.
Is there warming in the pipeline? A multi-model analysis of the Zero Emissions Commitment from CO2, MacDougal et al. (2020).
Greater committed warming after accounting for the pattern effect, Zhou et al. (2021).
Stabilizing climate requires near‐zero emissions, Matthews and Caldeira (2008).
Irreversible climate change due to carbon dioxide emissions, Solomon et al. (2009).
Atmospheric carbon dioxide removal: long-term consequences and commitment, Cao and Caldeira (2010).
Climate Change Commitments, Realclimate (2010).
How Big is the Climate Change Deficit?, Steve Easterbrook (2013).
Committed Warming, my post from 2016.
## Have CO2 emissions peaked?
I noticed, as has Stoat, that Ken Caldeira and Ted Nordhaus have a bet about whether or not we’ve reached peak CO2 emissions. Specifically, the bet is
Between 2021 and the end of 2030, annual fossil fuel emissions (excluding carbonation) will not exceed annual fossil fuel emissions (excluding carbonation) from 2019.
Carbonation is essentially emissions from cement production.
As with many others, I’m hoping that Ted Nordhaus wins, but expecting that Ken Caldeira will do so. In truth, though, that’s a bit simple. Even if Ted Nordaus were to win, what would emissions having peaked actually imply?
Consider a simplified form of the Kaya Identity:
$CO_2 = GDP \times \dfrac{Energy}{GDP} \times \dfrac{CO_2}{Energy}$
CO2 emissions essentially depend on Gross Domestic Product (GDP), energy intensity (energy per GDP) and carbon intensity (CO2 per energy).
So, if emissions this decade do not exceed those from 2019, why would that be? Would it be because GDP growth had stalled? Would it because of improvements in energy efficiency? Would it be because we’d reduced emissions through using more alternative energy sources? Would it be because we’d developed, and deployed, carbon capture and storage technologies? A bit of everything?
Also, what would it imply about the developed and developing worlds? Will the developed world have accelerated their emissions reduction so that the developing world can have a more gradual transition? If it is partly due to slower, or stalled, GDP growth, would that imply that some have benefitted far less than they might otherwise have done?
I don’t know the answers to any of these questions, but I do sometimes wonder if we don’t always consider the potential implications of some of the scenarios we might be hoping for. I’ll leave it there, but if anyone has any answers to these questions, feel free to post them in the comments.
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2021-03-09 04:17:07
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http://cms.math.ca/cjm/kw/algorithm
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location: Publications → journals
Search results
Search: All articles in the CJM digital archive with keyword algorithm
Expand all Collapse all Results 1 - 2 of 2
1. CJM 2011 (vol 63 pp. 755)
Chu, Kenneth C. K.
On the Geometry of the Moduli Space of Real Binary Octics The moduli space of smooth real binary octics has five connected components. They parametrize the real binary octics whose defining equations have $0,\dots,4$ complex-conjugate pairs of roots respectively. We show that each of these five components has a real hyperbolic structure in the sense that each is isomorphic as a real-analytic manifold to the quotient of an open dense subset of $5$-dimensional real hyperbolic space $\mathbb{RH}^5$ by the action of an arithmetic subgroup of $\operatorname{Isom}(\mathbb{RH}^5)$. These subgroups are commensurable to discrete hyperbolic reflection groups, and the Vinberg diagrams of the latter are computed. Keywords:real binary octics, moduli space, complex hyperbolic geometry, Vinberg algorithmCategories:32G13, 32G20, 14D05, 14D20
2. CJM 2008 (vol 60 pp. 1267)
Blake, Ian F.; Murty, V. Kumar; Xu, Guangwu
Nonadjacent Radix-$\tau$ Expansions of Integers in Euclidean Imaginary Quadratic Number Fields In his seminal papers, Koblitz proposed curves for cryptographic use. For fast operations on these curves, these papers also initiated a study of the radix-$\tau$ expansion of integers in the number fields $\Q(\sqrt{-3})$ and $\Q(\sqrt{-7})$. The (window) nonadjacent form of $\tau$-expansion of integers in $\Q(\sqrt{-7})$ was first investigated by Solinas. For integers in $\Q(\sqrt{-3})$, the nonadjacent form and the window nonadjacent form of the $\tau$-expansion were studied. These are used for efficient point multiplications on Koblitz curves. In this paper, we complete the picture by producing the (window) nonadjacent radix-$\tau$ expansions for integers in all Euclidean imaginary quadratic number fields. Keywords:algebraic integer, radix expression, window nonadjacent expansion, algorithm, point multiplication of elliptic curves, cryptographyCategories:11A63, 11R04, 11Y16, 11Y40, 14G50
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2015-03-07 00:14:02
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https://tug.org/pipermail/xetex/2010-November/019297.html
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# [XeTeX] Specifying papersize with XeTeX
Wilfred van Rooijen wvanrooijen at yahoo.com
Thu Nov 11 00:25:38 CET 2010
Although I may not really understand the question, if you take a look at the manual of the memoir class, it is precisely explained how (La)TeX treats the page size. Also, the geometry package lets you select any page size you want. Basically, (La)TeX does not care about the page size. For (La)TeX, the world consists of a sheet of paper, with some arbitrary point labeled (0,0), and all material is positioned on the paper with respect to (0,0) (which is usually the bottom-left point). Specifying a paper size really means that you inform (La)TeX that the upper-right point is at a certain location (rather than at (+\infty, +\infty).
Cheers,
Wilfred
--- On Thu, 11/11/10, John Was <john.was at ntlworld.com> wrote:
> From: John Was <john.was at ntlworld.com>
> Subject: Re: [XeTeX] Specifying papersize with XeTeX
> To: "Unicode-based TeX for Mac OS X and other platforms" <xetex at tug.org>
> Date: Thursday, 11 November, 2010, 12:38 AM
> Thanks Pete
>
> --papersize=a5 was just a quick fix since I hadn't expected
> the printer to have any problems (he hasn't for the last ten
> years but apparently there's a new machine...). I'll
> look into this properly but someone else has already
> mentioned the papersize \special, which I'm sure is what
> I'll need if cropmarks are starting to cause some printers
> difficulty.
>
>
> John
>
>
>
> ----- Original Message ----- From: "Peter Dyballa" <Peter_Dyballa at Web.DE>
> To: "Unicode-based TeX for Mac OS X and other platforms"
> <xetex at tug.org>
> Sent: 10 November 2010 14:57
> Subject: Re: [XeTeX] Specifying papersize with XeTeX
>
>
>
> Am 10.11.2010 um 13:10 schrieb John Was:
>
> > --papersize=a5
>
> I wouldn't use this command line option. I'd use a
> \special{papersize=...} which also records these dimensions
> in the XDV
> output file, helping xdvipdfmx to choose the proper paper
> format.
>
> --
> Greetings
>
> Pete
>
> No project was ever completed on time and within budget.
> – Cheops Law
>
>
>
>
> --------------------------------------------------
> Subscriptions, Archive, and List information, etc.:
> http://tug.org/mailman/listinfo/xetex
>
>
> --------------------------------------------------
> Subscriptions, Archive, and List information, etc.:
> http://tug.org/mailman/listinfo/xetex
>
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2023-03-28 02:45:01
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http://marcvamos.de/dilation-scale-factor-calculator.html
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Dilation Scale Factor Calculator
(Multiply both coordinates of each point by 3. Given a figure, the student will identify the scale factor used for a dilation, and use a dilation by a scale factor, including enlargements and reductions, to generate similar figures. A scale factor greater than 1 indicates an enlargement. 1 cm ⇒ 20000 cm 3 cm ⇒ 20000 * 3 cm = 60000 cm = 600 m = 0. 208 Core VocabularyCore Vocabulary CCore ore CConceptoncept Dilations. What is the scale factor of the dilation that produces an image with an area that is twice that of the original? I have found out that the rectangle of the original area is 18 square units by using the distance formula. tIn dilation, the coordinates of a point (x, y) transform into the coordinates (kx, ky) where k is the scale factor. The letter r usually represents the scale factor. For example, doubling distances corresponds to a scale factor of two for distance. Real Size to Scale Size Dilation is the transformation which is an extreme, radical change in appearance. The image of line m is constructed through a dilation centered at O with a scale factor of 3. It can factor expressions with polynomials involving any number of variables as well as more complex expressions. For example, if the percent solution under consideration is to If you wish to perform dilution factor or fold dilution calculations for solutions with molarity or percent concentration units, use our Dilution Factor. They are The Likert Scale; and, The Thurstone Scale. For example, if a veteran has a rating on each leg or each arm then those ratings are combined together and give the overall combined rating an extra boost. When we want to talk about how much bigger or how much smaller the new shape is, it’s convenient to use the idea of a scale factor. Calculus: Integral with adjustable bounds. Starting from the center of dilation point, use the “new” horizontal and vertical distances to plot each image point. Weight Variation by Latitude refers to the claims that scales have measured masses to weigh up to 0. The absolute value of the dilation factor is the ratio of each side length of the dilated quadrilateral to the corresponding side length of the preimage. It would be best if you always had your bindings checked and adjusted by a professional ski technician. The most important thing you can do right now is STAY HOME as much as possible. 1 Questions & Answers Place. We maintain a large amount of excellent reference information on subject areas varying from dividing rational expressions to elementary algebra. Come to Gre-test-prep. The formula for finding a dilation with a scale factor is x' = kx (k = scale factor), so x' = 2. What are A', B', C', and D'? Hint: Multiply the coordinates by the scale factor to find the coordinates of the reduction. A dilation with center (0, 0) and scale factor k is applied to a polygon. -image smaller, the scale factor was _____. Naming Angles (47 views this week) Determine the Scale Factor Between Two Rectangles and Determine the Missing Lengths (Whole Number Scale Factors) (47 views this week) Dilations (Old Version) (46 views this week) Identifying Prisms and Pyramids (46 views this week) Translation of 4 Vertices up to 6 Units (46 views this week) Calculating Side Values Using the Sine Ratio (45 views this week. A dilation is the transformation of a shape by a scale factor to produce an image that is similar to the original shape, but is. Draw the triangle A(0, 0) B (0, 4) C (3,1) Rotate the triangle 90 degrees clockwise. Then, answer the questions below the applet. The length of a line segment after a dilation of scale factor 1/2 will be Answer the following questions and READ CAREFULLY! 8. How is the scale factor related to the ratios DEF area AD'E'F area ADEF 20. 1 with no rotation. To report an issue or request a feature please visit the GitHub repository. • The center of dilation is a fixed point in the plane. 1 Ready Note: For questions 1 6 the scale factor r. Coordinate Dilations Step-by-step Lesson - Move a rectangle along the coordinate graph by a scale factor. Solution (b) : 66 / 74 = 12 / SH. The length of a line segment after a dilation of scale factor 1/2 will be Answer the following questions and READ CAREFULLY! 8. Keyword Research: People who searched dilating also searched. The image of line m is constructed through a dilation centered at O with a scale factor of 3. en\ar 12 10 14 nul(eí x 2 B z n 12 7. The formula for finding a dilation with a scale factor is x' = kx (k = scale factor), so x' = 2. Calculate the scale factor of a dilation. There was a dilation of scale factor 1. Calculate the scale factor of two shapes with help from a longtime high school math tut. PQR times the scale factor. In the context of dilation, the scale factor is the value that determines both whether the preimage increases or decreases in size, as well as the magnitude of the change with respect to a fixed point called the center of dilation. The scale factor is 4 and the center of dilation is the origin. A distance/time dilation calculator is available further down the page. dilatation. Home›Calculators›Electrical Calculators› Power factor calculator. Our scale factor is 3, meaning each vertex in ∆A'B'C' will be three times the distance from the origin as its preimage vertex. •dilation (minimum) •erosion (maximum) –Two filters applied in series to binary image yield •opening = dilation[erosion(binary image)] Eq. Determine the result of a dilation given a center of dilation and the scale factor. We need to find the scale factor of dilation. Add up to 5. Draw any polygon. com supplies essential answers on evaluate the equations calculators, college algebra and function and other algebra subject areas. the dilation? center of dilation slider point 9. Multiply together to get 4. I always get trouble with using one function in AutoCAD. Student may or may not have attempted to find the coordinates of ′. The image of line m is constructed through a dilation centered at O with a scale factor of 3. The pupil response to cognitive and emotional events occurs on a smaller scale than the light reflex, with changes generally less than half. Find an answer to your question “What is the scale factor for the dilation? ” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. Use one pair as a check on the other. The scale factor tells you how much something is enlarged or reduced. In the figure shown, nXYZ is the image of nABC. Projections also have. Is it bigger, or is it smaller—or maybe it's the same size? Individuals learn to describe enlargements and reductions and quantify the result. Find values of 𝑎𝑎 and 𝑏𝑏 so that 𝐿𝐿1(𝑥𝑥, 𝑦𝑦) = (𝑎𝑎−𝑥𝑥𝑏𝑏, 𝑏𝑏𝑦𝑦+𝑥𝑥𝑎𝑎) has the effect of dilation with scale factor 𝑦𝑦 2 and no rotation. dilation with a negative scale factor. For ASE, a calculator is a black box that can take atomic numbers and atomic positions from an Atoms object and calculate the energy and forces and sometimes also stresses. This page contains a time dilation calculator. It expands. 1 The scale factor for a model is 5 cm = _____ m. When a dilation in the coordinate plane has the origin as the center of dilation, you can find points on the dilated image by multiplying the x- and y-coordinates of the original figure by the scale factor. performed a dilation, using a scale factor of 2. outside the polygon b. What are the coordinates of triangle A'B'C'? What. Due to an accident the crew are unable to stop accelerating the spacecraft, causing such extreme time dilation that the crew experiences the Big Crunch. Are the Triangles Similar? Scale Factor = _____ Geometric Shape Base Height Perimeter Area Original Right Triangle New Right Triangle. The Hill-RBF Calculator is an advanced, self-validating method for IOL power selection employing pattern recognition by artificial intelligence. There was a dilation of scale factor 1. 1 The scale factor for a model is 5 cm = _____ m. People who score high on measures of Anger In and Hostility tend to ruminate a lot about others, but are not. If a center #O# is given, to determine a factor we need points #A# and its image (the result of scaling) #A'#. Determine the perimeter and area Of the image. Enter a polynomial, or even just a number, to see its factors. inside the polygon c. 1 Questions & Answers Place. com Rotation Sheet Kuta Sheets for all of Dilations Leaf Kuta , source: bonlacfoods. Our goal is to provide a convenient set of web-based Bayes factor calculators. dynamic decalage correction based on out of neutral trim. But the size of the Universe changes continuously, so we should divide the light's trip into short intervals. Scale Factor: •Is the ratio: •the distance from the center of dilation to a point on the image: to the distance from the center of dilation to the. • A description of a dilation includes the scale factor (or ratio) and the center of the dilation. Provide the number of inputs, point value, and center of dilation to find the dilation point(s) using this online center of dilation calculator. Since the scale factor is greater than 1 , this is an enlargement. Give the coordinates of A 9 B 9 C 9 , and the ratio of the areas of the figures A 9 B 9 C 9 and ABC. scale factor. Dilation is the process of increasing or decreasing the size of an image. The map stated, “The treasure is buried at the center of dilation of these two triangles. lengths of AABC. The scale factor in the dilation of a mathematical object determines how much larger or smaller the image will be (compared to the original object). For ASE, a calculator is a black box that can take atomic numbers and atomic positions from an Atoms object and calculate the energy and forces and sometimes also stresses. Every dilation has a center and a scale factor. Calculate the scale factor of two shapes with help from a longtime high school math tut. Dilation is a transformation, that stretches or shrinks the original figure presented on the grid based on the scale factor. • A(-1, 1), B(-1, 0), C(3,1) • X(-3, 3), Y(-3, 0), Z(9, 3) Size Change Distance Theorem • The image of a segment transformed by a dilation with scale factor k is parallel to and |k| times the length of the preimage. In this lesson you will learn how to calculate the scale factor of a dilation by comparing measurements of an image and a pre-image. The scale factor, or linear scale factor, is the ratio of two corresponding side lengths of similar figures. Smith’s tax bill if her house’s market value is $246,000 and she has a homestead exemption of$10,000. What happens to a dilation when the scale factor is less than one? Learners show and then tell this in a short worksheet. Find the scale factor. In this course, the zoom factor will be used to describe the. Unit 19 Section 3 : Line, area and volume scale factors. The scale is rounded. ABC by a scale factor of 2 centered at the origin, followed by a rotation of 180° about the origin (4) a dilation of 6. Center of Dilation Calculator. 👉 Learn about dilations. Remember never to round off in the middle of your calculation. Conveniently, we can identify the coordinates by multiplying the preimage coordinates. If a dilation) (or scaling) is given, it is assumed that its center and a factor are given, so we can construct an image of any point. Use proportional reasoning to determine if one figure is a dilation of another. The dilation now gives us (2x – 10, 2y + 6). The pairs of triangles shown below are similar. They are The Likert Scale; and, The Thurstone Scale. Created by Shannon White. The angles will always remain the same. and the scale factor, find the vertices dilated The of is 10. Student may have made calculation errors when calculating the scale factor. 9 IS = Given that AFGH is similar to AFED, calculate GH to the nearest hundredths place. When a figure is dilated by a scale factor k, the area. Draw a polygon. smaller than and along the opposite ray from D. 'ill be provided. ) Segment AB measures 3 cm. 5 centered at the origin. If a reproduced copy of that painting had a scale factor of 3. The publisher wants to make a reduction using a dilation with a scale factor of ½. For example, if you would like to apply a scale factor of 1:6 and the length of the item is 60 cm, you simply divide 60 / 6 = 10 cm to get the new dimension. larger than and along the opposite ray from PLEASE HELP. The scale factor can be denoted by $$r$$ or $$k$$. Setting your ski bindings to the correct release setting is essential for your safety. 5 cm no MPD >2. It is the ratio of the final volume to the initial volume. How to Use the Scale Conversion Calculator. scale image is then converted into its binary form. what is the scale factor and how did you get it? 4 years ago. If the shape has been scaled “up”, so that. -image smaller, the scale factor was _____. More information. Dilations are also called similarity transformations. What is the measure of angle A'B'C. Scale Factor sheet from Kuta's Dilations Sheet, Source: homeoutsidethebox. Mess with the scale factor of a dilation. Finally, the effects of relativity become significant. 1 cm ⇒ 20000 cm 3 cm ⇒ 20000 * 3 cm = 60000 cm = 600 m = 0. The image vertices for dilation with center. Given the figure shown, assign the following measures: AB = 4, BC = 7, CA = 8. Scale Factor Dilation Formula. Scale Factor. of a scale fqc\or L prt-lm40e Properties of Dilation: 1. 5 cm yes MPD; 1. And that the image should be the scale factor as far away from the center of dilation, in this case it should be twice as far from the center of dilation as the point that it is the image of. It can do all the basics like calculating quartiles, mean, median, mode, variance, standard deviation as well as the correlation coefficient. Elastic Collision Calculators. The recipe converter will automatically calculate the conversion factor and enter it into the appropriate field. Use our 2021 VA Disability Calculator to calculate your monthly compensation rate. 59 since the midpoint of the light's trip. This app will help you calculate your "cal factor" and warn you about the most obvious of errors. We offer an algebra calculator to solve your algebra problems step by step, as well as lessons and practice to help you master algebra. A dilation multiplies the length of each side of an original quadrilateral by the same scale factor. Dilation is the transformation of image from its original size to a different size without making any changes in its appearance/shape. This means that the figure gets twice as large. Scale Calculator. Calculate the scale factor Calculate the missing side length using the scale factor I will achieve all of the learning goal(s) for Similar Figures with at least _____ accuracy by: 1. 0 and doesn't factor in class difficulty. Select the value of pentagon which you already have. k is the value of the scale factor. We maintain a whole lot of high quality reference materials on topics varying from exponents to factors. If we increase the original rectangle by a scale factor of 3, then we multiply each measurements by 3 to end up with this shape. Dilation Exploration Drag the scale to change the scale factor and observe what happens to the triangle. Given the pre-image, scale factor, and center of dilation, use a compass and straight edge to graph the image. Solution (b) : 66 / 74 = 12 / SH. Convert the tax rate to MILS 3. In order to calculate forces and energies. • You can use grid paper and a scale factor to draw enlargements and reductions. ) Segment AB measures 3 cm. If a scale factor is less than 1, then your figure gets 3. 5 centered at the origin. 1 with no rotation. You could use a scale factor to solve! In this tutorial, learn how to create a ratio of corresponding sides with known length and use the ratio to find the scale factor. Triangle A'B'C' is the result of applying a dilation with center P and scale factor 3 to triangle ABC. Remember that to dilate something in the coordinate plane, multiply each coordinate by the scale factor. Triangles ABC and DEF are similar. e -4m=-12 ⇒ m=3. Asking questions when I’m not sure of something. Performing a Similarity Transformation Graph ABC with vertices A(−4, 1), B(−2, 2), and C(−2, 1) and its image after the similarity transformation. Find the scale factor. Given line m and point O not on line m. Multiply together to get 4. Resort to the help of this amazing ratio calculator when you have you settle ratio/proportion problems and check equivalent fractions. Determine if the dilation is an enlargement or reduction. Point O is the center of the dilation. What are A', B', C', and D'? Hint: Multiply the coordinates by the scale factor to find the coordinates of the reduction. To report an issue or request a feature please visit the GitHub repository. Which of these scale factors for the dilation would result in an image that was larger than the original figure?. Given two similar images I can calculate the scale factor used to create the scale diagram. If a reproduced copy of that painting had a scale factor of 3. The center of dilation is a fixed point in the plane. Scale Calculator. Since the line C'D' falls on the line CD, the scale factor is not applied on the line CD. 6 8 10 8 12 14 0 6 2 4 10 12 14 16 4 2 y x 16 C´ C B B´ A´ A 10. calculator. 1 The scale factor for a model is 5 cm = _____ m. This calculator help us find the scale factor between two lengths, simply enter two lengths, it will automatically calculate the scale factor, supports different length units (mm, cm, m, km, in, ft, yd, mi), in addition corresponding visual graphic and formula, easy understanding the calculation process and the result. fahrenheit to degree converter online. It has been optimized for use with the Haag-Streit LENSTAR LS 900 optical biometer for all axial measurements and in combination with high density. When a solution's concentration is reduced, it is called dilution. Find the scale factor of a dilation that maps a given figure to another one. Scale factor explained for primary-school parents, with details of how and when scale factor is taught in the KS2 classroom. Similar figures have the same shape but are of different sizes. e -4m=-12 ⇒ m=3. Let the center of dilation be as given. In this lesson dilation and scale factor are defined. Included here are umpteen printable worksheets to help 8th grade and high school students hone in on finding the scale factor, identifying the dilation type, determining the new coordinates and drawing the dilated shapes with the center as origin. Given line m and point O not on line m. 1 with no rotation. the P 3 P’ 4 C reduction CP ' 4 = CP 7 Example: Example (a. Since the ratios are equal, the lengths of the sides are proportional. 5 centered at the origin. Student uses the definition of dilation with the given side. Keyword Research: People who searched dilating also searched. Triangle $$ABC$$ is taken to triangle $$A’B’C’$$ by a dilation. center of dilation at the origin and a scale factor of 2. 208 Core VocabularyCore Vocabulary CCore ore CConceptoncept Dilations. Algorithm; Ovaries. Practice questions If […]. • Graphing Calculator. Download the BMI calculator app today (available for iPhone and Android ). Point O is the center of the dilation. Our scale factor is 3, meaning each vertex in ∆A'B'C' will be three times the distance from the origin as its preimage vertex. Draw the triangle A(0, 0) B (0, 4) C (3,1) Rotate the triangle 90 degrees clockwise. ) Draw a dilation of Δ ABC w/ A(-2,1), B(-6,0), and C(-1,-1). y = 3x -2 d. Elastic Collision Calculators. A publisher is preparing the marketing plan for a new book. There was a dilation of scale factor 1. ) Segment AB measures 3 cm. Student uses the definition of dilation with the given side. Improve your math knowledge with free questions in "Dilations: scale factor and classification" and thousands of other math skills. 208 Core VocabularyCore Vocabulary CCore ore CConceptoncept Dilations. (When the scale factor is less than one, the new points actually get closer to C. 5 (22) = 55. 6–9 •closing = erosion[dilation(binary image)] Eq. Draw polygon ABC, A (-1, 1), B(0, 2), C(3, 1). reduction A dilation where the image is smaller than the preimage. Topic : Scale factors- Worksheet 1 Fill in the missing dimensions. This finding, obtained from ungated SPECT images, is the ratio of the average ventricular size after stress compared with rest. Calculate the scale factor Calculate the missing side length using the scale factor I will achieve all of the learning goal(s) for Similar Figures with at least _____ accuracy by: 1. Angle ABC is taken by a dilation with center P and scale factor 3 to angle A'B'C'. 1 cm ⇒ 20000 cm 3 cm ⇒ 20000 * 3 cm = 60000 cm = 600 m = 0. The scale factor can be written in a variety of ways. Mathematics often becomes cumbersome without a calculator and once the calculator is not used the working of equations become so difficult that. Use proportional reasoning to determine if one figure is a dilation of another. Plotting Coordinate Points (628 views this week) Calculate the Hypotenuse Using Pythagorean Theorem (No Rotation) (512 views this week) Plotting Coordinate Points Art -- Red Maple Leaf (369 views this week) Classifying Triangles by Angle and Side Properties (Marks Included on Question Page) (356 views this week) Naming Simple Angles (Acute, Obtuse, Right) (270 views this week). ) Is rectangle A a scale copy of rectangle C? If so, what is the scale factor? 3. Dilation with Scale Factor. In general English it means to make larger. It is a free and easy to use GCD calculator. They create a dilation from an off-line point using a scale factor of 0. The scale factor k is a positive number such that k = OP' OP and k 1. Since the scale factor is greater than 1 , this is an enlargement. The actual cover of the book measures 6 inches by 8 inches, as shown. Dilation definition, the act of dilating; state of being dilated. There was a translation left 0. Application of scale factor in the real-world context is structured into level 2 word problems. When the absolute value of the scale factor is greater than one, an expansion occurs. When we want to talk about how much bigger or how much smaller the new shape is, it’s convenient to use the idea of a scale factor. To report an issue or request a feature please visit the GitHub repository. Research examining trait anger scales via factor analyses showed Speilberger Anger In and components of the Cook–Medley scale loaded on the cynical cognition factor rather than behavioral aggression or angry affect factors. 2) Dilation: The dilation process removes the noise encountered in the Fbinary image. So for this example there is nothing special. Please enter two values, the third will be calculated. One, if you connect corresponding points, your center of dilation is going to be on a line that connects those two points. Which transformations result in congruent figures? Problem Set 6: Transformations. One such use arises in linear transformations or linear maps. Scale factor of a triangle calculator - Cleveland. Calculate nine times 13 and your pupils will dilate slightly. Topic : Scale factors- Worksheet 1 Fill in the missing dimensions. 208 Core VocabularyCore Vocabulary CCore ore CConceptoncept Dilations. Find the scale factor of a dilation that maps a given figure to another one. Try our Dyson Sphere Program Calculator. There was a translation left 0. 6 8 10 8 12 14 0 6 2 4 10 12 14 16 4 2 y x 16 C´ C B B´ A´ A 10. Determine whether the dilation from igure A to Figure B s a reduction or an enlargement. Cervical exam could also stimulate the uterus to cause minor contractions. If so, what is the scale factor? c. Real Size to Scale Size Dilation is the transformation which is an extreme, radical change in appearance. B of is at ted by a fa wit the Of dilation al origin, B'? —7, 7), B (8, — and with a scale Exercises 4—5: Calculate the factor. Mathematics often becomes cumbersome without a calculator and once the calculator is not used the working of equations become so difficult that. A scale factor greater than 1 indicates an enlargement. res will nol he nrovided helc. In the above figure - Triangle A'B'C' is a dilation of triangle ABC 3. In National 4 Maths calculate the size of a missing length, area or volume by calculating the enlargement/reduction scale factor first. However, when it comes to online to measure the relative variability, this coefficient of variation calculator makes your calculation as simple as possible for the given sample data of the population. Scalar Factor or scale factor is present at the midpoint of a figure and helps in transforming the image to a larger or smaller figure. com an extension of the picture sheet, extended translations sheet key,. As you have seen before, we can refer to a component either by a number or a variable; the convention for Cartesian coordinates is that the variables (x, y, z) are equally represented by the components (1, 2, 3). Dilations map segments to segments, lines to lines, rays to rays, angles to angles, and circles to circles. (The image is similar to the. Graph the new image. Transient ischemic dilation (TID) is both a sensitive and a specific indicator of triple-vessel coronary artery disease. The image of D i. Plotting Coordinate Points (628 views this week) Calculate the Hypotenuse Using Pythagorean Theorem (No Rotation) (512 views this week) Plotting Coordinate Points Art -- Red Maple Leaf (369 views this week) Classifying Triangles by Angle and Side Properties (Marks Included on Question Page) (356 views this week) Naming Simple Angles (Acute, Obtuse, Right) (270 views this week). This conclusion is extended to other scale factors. Transformations in math. If a scale factor is less than 1, then your figure gets 3. 147, 1682; Bloom, Mory and Hinman) which appears to conclude that there is no difference. Our scale factor is 3, meaning each vertex in ∆A'B'C' will be three times the distance from the origin as its preimage vertex. , find the equation of a line parallel or perpendicular to a given line that passes through a given point. Calculate the scale factor. e 12m=36 ⇒ m=3. ) Scale factor of 5. com supplies essential answers on evaluate the equations calculators, college algebra and function and other algebra subject areas. Draw the image under a dilation using the indicated scale factor. Draw the dilation of ABCD using center A and scale factor LaTeX: \frac{1}{2}12. com wishes everyone to BE WELL, STAY WELL, GET WELL. How are they calculated? SAT® scale scores are how your raw scores translate when converted to section scores — these are between 200-800 for the two sections (Evidence-Based Reading and Writing and Math), to give you a total SAT® score between 400-1600. When the scale factor of the dilation(s) is not equal to 1 or −1, similarity transformations preserve angle measure only. Real Size to Scale Size Dilation is the transformation which is an extreme, radical change in appearance. Other factors may also be important when deciding on the type of percent solution to prepare. If so, what is the scale factor (from A to B)? c. Used the origin as the center and use a scale factor of 2. The scale factor, sometimes called the scalar factor, measures how much larger or smaller the image is. For example, prime factorization is only feasible for small integers. Use the coordinates of the 2 triangles to determine the scale factor of the dilation. •Issac set up the ratio to find the scale factor. Point O is the center of the dilation. Scale factor = SH / CP. The scale factor in the dilation of a mathematical object determines how much larger or smaller the image will be (compared to the original object). When the absolute value of the scale factor is greater than one, an expansion occurs. Commonly Used Architectural Scales. Offering a blend of exercises these dilation center at the origin worksheets contain tasks like identifying the type of dilation writing the scale factor finding the dilated coordinates and using them to draw the dilated. The threshold of hearing is assigned a sound level of 0 decibels (abbreviated 0 dB); this sound corresponds to an intensity of 1*10 -12 W/m 2. Log InorSign Up. One of the most popular techniques of attitude measurement is the Likert Scale. Solution for Hexagon A'B'C'D'E'F' is a dilation of Hexagon ABCDEF. ) Is rectangle A a scale copy of rectangle C? If so, what is the scale factor? 3. Use this ballistic calculator in order to calculate the flight path of a bullet given the shooting parameters that meet your conditions. It contracts. Blue Numbers → Dilation Factors applied to Kernel. If the shape has been scaled “up”, so that. For the critical density case, the scale factor for the Universe goes like the 2/3 power of the time since the Big Bang, so the Universe has grown by a factor of 2 2/3 = 1. • If the scale factor is greater than 1, the image is an enlargement (a stretch). State the scale factor of the dilation. By reciprocal property of proportion, 74 / 66 = SH / 12. The velocity time dilation is explained by Anderson in terms of the tau factor, which decreases closer and closer to zero as the ship approaches the speed of light, hence the title of the novel. A similar shape dilated with a scale factor of 3 will. Calculate 9 times 13, and you pupils will dilate slightly. 208 scale factor, p. The way to perform a dilation on the coordinate grid is to take your X and your Y coordinate and multiply it times whatever the scale factor is. The scale factor determines the degree or amount to which the object is increased or decreased. OBSERVE: Notice how. These Javascript devices calculate the scaled length (the output) when you enter the length of an object (the input) and a scaling factor (the scale). Start by writing down the coordinates of the vertices of figure MASH as follows: The next step is to take the scale factor (1/3 in this example) and multiply it by the x and y-value of points M, A, S, and H, as follows:. 208 reduction, p. 18 10 ADEF 26 45 to. There was a dilation of scale factor 1 centered at the origin. Percentage. Label the new triangle A’B’C’. A scale factor isn't always the easiest approach, but it's often a good way to work with ratio problems. Most dilation's in coordinate geometry use the origin, (0,0), as the center of the dilation. Example 3 Triangle ABC is a dilation of triangle XYZ. Line y = 3x – 1 is transformed by a dilation with scale factor of 2 and centered at (3,8). You can also do almost any kind of regression analysis (linear, quadratic, exponential. Most of the time it is the origin (0, 0) Scale Factor: tells you how many times larger or smaller your image will be. to solve for The scale factor is: Pg. a dilation, we get a reflection-dilation. If you're also looking for Determining the Scale Factor and whether it's an Enlargement or Reduction from a graph, then consider this additional lesson:Transformations 18: Determining Dilation Scale Factor from Graphs Note: If you want your students to use an actual Coordinate Plane to determine Enlargements, Reductions, Coordinates and Scale. 208 scale factor, p. Scale Factor. Dilation is the transformation of image from its original size to a different size without making any changes in its appearance/shape. Graph its image A9B9C9 after a dilation with scale factor. Center ; scale factor 2 10. Works across all devices Use our algebra calculator at home with the MathPapa website, or on the go with MathPapa mobile app. You can calculate distance in map units (using Pythagoras theorem). scale factor calculator. Round your answer to the nearest tenth. Please enter two values, the third will be calculated. Are the scale factors consistent?. These Javascript devices calculate the scaled length (the output) when you enter the length of an object (the input) and a scaling factor (the scale). The scale factor is how many times larger than the object the image is. Triangle A'B'C' is the result of applying a dilation with center P and scale factor 3 to triangle ABC. A scale factor less than 1 indicates a reduction. What happens if a different point in the plane is the center of dilation? Copy the polygon at right onto graph paper. What is the scale factor of triangle ABC to triangle DEF? A. • A description of a dilation includes the scale factor (or ratio) and the center of the dilation. Scale Factor: •Is the ratio: •the distance from the center of dilation to a point on the image: to the distance from the center of dilation to the. dilation on the coordinate plane. 208 enlargement, p. Draw the dilation of ABCD using center A and scale factor LaTeX: \frac{1}{2}12. The center of dilation is a fixed point in the plane. Instructional video. To report an issue or request a feature please visit the GitHub repository. A dilation requires a center point and a scale factor. A distance/time dilation calculator is available further down the page. Scale Factors 7th Grade - Displaying top 8 worksheets found for this concept. dilation at the origin and a scale factor of 1/3. Our statistics calculator is the most sophisticated statistics calculator online. Conveniently, we can identify the coordinates by multiplying the preimage coordinates. There was a dilation of scale factor 1 centered at the origin. SECTION 19: Decide if each dilation is a reduction or an enlargement. Calculate the ad valorem taxes for Mr. Dilation definition, the act of dilating; state of being dilated. reduction A dilation where the image is smaller than the preimage. 6 8 10 8 12 14 0 6 2 4 10 12 14 16 4 2 y x 16 C´ C B B´ A´ A 10. A dilation with a scale factor greater than 1 will shrink the image. The definition of Dilation: To resize something. State the scale factor of the dilation. And as Dilation Factor Increases the space between original kernel elements get wider and wider. Determine the result of a dilation given a center of dilation and the scale factor. Divide the image side lengths by the preimage side lengths. dilation on the coordinate plane. In this activity, students will dilate a triangle using a positive integer scale factor. 06 Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e. Graph the image of this figure after a dilation with a scale factor of 3 centered at (−7, −6). Below is a picture of each type of dilation (one that gets larger and one that gest smaller). 6–10 –Can use non-rectangular windows to achieve template matching (“structuring element”) Spatial Transforms 24 Fall 2005. This measures the ability of a projectile to overcome air resistance. 2) Change the scale factor to 2. In Math, the word dilate means to figure. High School: Geometry » Introduction Print this page. Free online factoring calculator that factors an algebraic expression. Students will need to move PQR by moving the cursor to one side and when all sides of the. Find the scale factor. A missing length on a reduction/enlargement figure can be calculated by finding its linear scale factor. 32: 1: 497: 15: dilation calculator. The velocity time dilation is explained by Anderson in terms of the tau factor, which decreases closer and closer to zero as the ship approaches the speed of light, hence the title of the novel. In today’s class we looked at how the scale factor or magnitude of dilation affects the perimeter and area of a transformed figure. The line’s image is _____. While we have done our best to ensure accurate results, the authors of this website do not make any representation or warranty, express or implied, regarding the calculators on this website, nor assume any liability for its use. Aim for a Healthy Weight: Limitations of the BMI Assessing Your Risk Controlling Your Weight Recipes. Instructional video. It would be best if you always had your bindings checked and adjusted by a professional ski technician. This helps them analyze figures and their images so that they can calculate the scale factor and determine the type of dilation. The pairs of triangles shown below are similar. If so, what is the scale factor? c. I always get trouble with using one function in AutoCAD. Enter a polynomial, or even just a number, to see its factors. Practice questions If […]. Find the perimeter of the standard card, and use that to find the scale factor. The above formula is used for calculating the changes that occur when objects approach the speed of light. Scale Factor Dilation Formula. It is the ratio of the final volume to the initial volume. Below is a picture of each type of dilation (one that gets larger and one that gest smaller). For the independent samples T-test, Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the pooled standard deviation. Draw the polygon’s image under a dilation with a scale factor of 2 and with point A as the center of dilation. Jones’ home 2. The scale factor tells us how the size of the "new" figure compares to the size of. Home›Calculators›Electrical Calculators› Power factor calculator. Notice that because D^n(P) is a dilation of scale factor k^n centered at the origin, the superscript n really does mean exponentiation, and D^-1 actually means multiplicative inverse. (Multiply both coordinates of each point by 3. Offering a blend of exercises these dilation center at the origin worksheets contain tasks like identifying the type of dilation writing the scale factor finding the dilated coordinates and using them to draw the dilated. the scale factor – regardless of how it is written! Be careful when using calculator. Calculator, with step by step explanation, on finding union, intersection, difference and cartesian product of two sets. The learner will also be able to identify where scale factor is used in real world concepts. Projections are also important in statistics. P 55) 56) 57). com you can easily calculate model size or model scale. The length of a line segment after a dilation of scale factor 2 will be 7. ) Draw a dilation of Δ ABC w/ A(-2,1), B(-6,0), and C(-1,-1). When scaling a plane around a point, the result is a plane of a different size but the same shape. The line's image is Line MN is dilated by a scale factor of 2 centered at the point (0,6). Let’s explore how the area a figure might change for ordered pair rules where x and y have different scale factors. Видео Dilation scale factor examples канала Khan Academy. To scale an object to a smaller size, you simply divide each dimension by the required scale factor. You can also get 2 as the scale factor by finding the ratios: 12/6 = 2, 16/8 = 2, and 18/9 = 2. Scale factor = SH / CP. Although this is an online quiz, be sure to turn your shown work on a sep Gwen had a painting that had a length of 5 feet and a width of 3. Reduction: k = — Scale Factor 1:2, or k Enlargement: k = Scale Factor 5:2. Construct the image of ∆ after dilation with center of dilation O and a scale factor of 2. a dilation, we get a reflection-dilation. A scale factor isn't always the easiest approach, but it's often a good way to work with ratio problems. fahrenheit to degree converter online. A scale factor is a number by which a quantity is multiplied, changing the magnitude of the quantity. y = 3x -2 d. Get step-by-step solutions to your Calculus problems, with easy to understand explanations of each step. You can use the scale of a map to calculate the real distance between two towns. The scale factor of the dilation is 1/4 b. The formula for finding a dilation with a scale factor is x' = kx (k = scale factor), so x' = 2. Finally, the effects of relativity become significant. We maintain a whole lot of high quality reference materials on topics varying from exponents to factors. Scale Conversion Calculator. A dilation with a scale factor O - 21006692 minecraft27 is waiting for your help. Additional events "caused" by risk factors. It is a free and easy to use GCD calculator. •dilation (minimum) •erosion (maximum) –Two filters applied in series to binary image yield •opening = dilation[erosion(binary image)] Eq. To scale an object to a smaller size, you simply divide each dimension by the required scale factor. Exercises 1–4. And that the image should be the scale factor as far away from the center of dilation, in this case it should be twice as far from the center of dilation as the point that it is the image of. If the scale factor is more than 1, then the image stretches. Then find the value of each variable. dilation with scale factor between 0 and 1. Brescia-COVID Severity Scale/Algorithm - Italian step-wise approach to managing all COVID-19 inpatients. Whenever I want to use the little calculator to apply a scale factor to a Block in AutoCAD, AutoCAD stops working. Our statistics calculator is the most sophisticated statistics calculator online. The Cal Factor Calculator is a simple app intended to help Medtronic MiniMed insulin pump users calibrate their CGM sensors. tWhen the scale factor is greater than one, the dilation is an enlargement. For the independent samples T-test, Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the pooled standard deviation. How can you determine the scale factor from ∆ABC to ∆A’B’C’ ? Solution using the TI-83+ graphing calculator: a. Dilation with Scale Factor. This means that the second triangle is 2 times as big. What are Scale Factors? Playing with objects is fun, right? Sometimes you want to make an object smaller than it originally is, sometimes larger. The way to perform a dilation on the coordinate grid is to take your X and your Y coordinate and multiply it times whatever the scale factor is. A distance/time dilation calculator is available further down the page. (Multiply both coordinates of each point by 3. The scale factor of a dilation is the ratio of a side length of the image to the corresponding side length of the original figure. We maintain a large amount of excellent reference information on subject areas varying from dividing rational expressions to elementary algebra. The above formula is used for calculating the changes that occur when objects approach the speed of light. When a dilation in the coordinate plane has the origin as the center of dilation, you can find points on the dilated image by multiplying the x- and y-coordinates of the original figure by the scale factor. Scale Factor Of Dilation Line Segment Ab Distance Between Points Line Segment Scale Factor. Full cervical dilation — when your cervix measures 10 cm — occurs at the end of the transitional phase, the last of the three phases of labor. What happens to a dilation when the scale factor is less than one? Learners show and then tell this in a short worksheet. How to Scale a Measurement Larger or Smaller. The scale factor is 4 and the center of dilation is the origin. SF = X2/X1 = Y2/Y1. Author: Lee Plath. 1 Ready Note: For questions 1 6 the scale factor r. Topic: Dilation. Scale factor = 15 / 12. The following formula is used to calculate the scale factor dilation of an image or shape. the same _____ since the figure is enlarged or reduced by a scale factor. factor needs to be squared before multiplying it by. Dilation with scale factor between 0 and 1. If ever you have advice with math and in particular with double inequality equation calculator or final review come pay a visit to us at Algebra-equation. 5 cm 3 cm. 18 10 ADEF 26 45 to. Some of the worksheets for this concept are Scale drawings and models, Leanzillion illaie mahemaic, Enlarge reduce, 1, Scale drawingsmodels scale factor sol, Scale drawings and scale factor, Ratios of scale drawings, Scale drawing blowing up a candy bar comic strip. Similarly you may explore: if a solid is dilated by a scale factor of 2/3, what effect will this have on the volume of a solid?. ) Explain how you know rectangle C is not a scaled copy of rectangle B. 4(-7, 8), factor Of 4 /(-B, 6), and 8) with a of g 14. The scale factor, sometimes called the scalar factor, measures how much larger or smaller the image is. Use proportional reasoning to determine if one figure is a dilation of another. ) Segment AB measures 3 cm. Whenever you will need assistance on lines as well as quadratic functions, Emathtutoring. 0 2 2 4 6 8 10 12 14 16 4 6 8 10 12 14 16 x y A B C 2. Calculate the scale factor. Calculate the ad valorem taxes for Mr. In the figure shown, nXYZ is the image of nABC. Give the coordinates of A 9 B 9 C 9 , and the ratio of the areas of the figures A 9 B 9 C 9 and ABC. ABC by a scale factor of ; centered at point A (3) a dilation of 6. When you calculate the ratios of the lengths of the corresponding sides, it will always be equal to the scale factor used to generate the dilation. 8: 8461: 25: dilating: 1. Free Online Scientific Notation Calculator. Question: 1. How are they calculated? SAT® scale scores are how your raw scores translate when converted to section scores — these are between 200-800 for the two sections (Evidence-Based Reading and Writing and Math), to give you a total SAT® score between 400-1600. Similarly you may explore: if a solid is dilated by a scale factor of 2/3, what effect will this have on the volume of a solid?. The Simple way to calculate the scale factor is to take the two coordinates and calculate the distance between them in calculator. Keyword CPC PCC Volume Score; dilation: 1. The Doppler shift factor ignores time dilation so v has to be smaller than c for the factor to work. In this lesson you will learn how to calculate the scale factor of a dilation by comparing measurements of an image and a pre-image. There was a dilation of scale factor 1. 208 scale factor, p. A dilation used to create an image smaller than the original is called a reduction. Below is a picture of each type of dilation (one that gets larger and one that gest smaller). • The center of dilation is a fixed point in the plane. 1 with no rotation. reduced by a scale factor in relation to a center point. 208 enlargement, p. Please provide a integer to calculate its factors and prime factors. respect to a fixed point called the center of dilation. 2(𝑧𝑧) = 𝑤𝑤 produces a dilation with scale factor 𝑧𝑧 0. It will also generate a step by step explanation for each operation. The scale factor tells you how much something is enlarged or reduced. • If the scale factor is 1, then the pre-image and image are congruent. 6 would create an enlargement, reduction, or isometric figure?. The scale factor is commonly expressed as 1:n or 1/n, where n is the factor. •An image that is the same size as the pre-image is called a _____ •This means the scale factor was _____ to 1. Scale Factor: •Is the ratio: •the distance from the center of dilation to a point on the image: to the distance from the center of dilation to the. ) Segment AB measures 3 cm. And now we know that the actual number of girls is 7n = 7(8) = 56, and the actual number of boys is 3n = 3(8) = 24. Try 29 times 13 and they will widen further and remain dilated until you reach the answer or stop trying. Point O is the center of the dilation. If the scale factor, N, is greater than 1, the image is an enlargement (a stretch). Scale factor explained for primary-school parents, with details of how and when scale factor is taught in the KS2 classroom. ) Is this an enlargement or (b. scale image is then converted into its binary form. Dilation Theorem 5. This measures the ability of a projectile to overcome air resistance. While we have done our best to ensure accurate results, the authors of this website do not make any representation or warranty, express or implied, regarding the calculators on this website, nor assume any liability for its use. What is the scale factor of triangle ABC to triangle DEF? A. Scale factor of a triangle calculator - Cleveland. com you can easily calculate model size or model scale. A dilation requires a center point and a scale factor. The picture below shows a dilation with a scale factor of 2.
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2021-04-20 16:27:00
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https://aharensho.net/film-laskar-pelangi-full-version.php
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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJun 23rd 2019
I forget if I ever knew the following:
What is there to the assumption that a given cohesive $\infty$-topos admits an $\infty$-site of definition all whose objects have (under Yoneda embedding) contractible shape?
Is this automatic? Is it a weak extra assumption? A strong extra assumption?
• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeJun 23rd 2019
Isn’t that essentially the “locally $\infty$-connected” condition at infinity-connected (infinity,1)-site?
• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJun 23rd 2019
Yes, in that terminology I am asking: How strong is the condition that a cohesive $\infty$-topos admits any locally $\infty$-connected $\infty$-site?
1. My intuition such as it is is that it is quite strong. For example in algebraic geometry, affines will typically not be contractible. Even in the pro-étale topos of Scholze and Bhatt I expect that the affine line is not contractible for instance.
• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeJun 23rd 2019
Re #4: …and for that reason, those toposes are not, I believe, cohesive. (-:
Re #3: C3.6.3 of the Elephant implies that any cohesive 1-topos has a locally 0-connected site, and by Prop. 1.3 of remarks on punctual local connectedness it can be taken to have finite products as well. I don’t have time to look up the proofs right now, but I would expect that they generalize at least partially to the $\infty$-case.
2. Re #5: That sounds right!
• CommentRowNumber7.
• CommentAuthorDavid_Corfield
• CommentTimeApr 7th 2020
Since there’s a certain interest at the moment, I noted earlier in the article the possibility of a relative notion of cohesion, and gave an instance in Remark 2.2.
• CommentRowNumber8.
• CommentAuthorRichard Williamson
• CommentTimeApr 7th 2020
• (edited Apr 7th 2020)
Thanks for adding something, David. Just a quick note that it is not really correct to say that the base topos is sheaves on profinite sets, as your notation would suggest. I was going to correct it, but was not hesistant to do so, as the way I thought to do so might change things a bit from what you had in mind.
• CommentRowNumber9.
• CommentAuthorDavidRoberts
• CommentTimeApr 7th 2020
Yeah, could be a fibred topos or similar, rather than a map of toposes.
• CommentRowNumber10.
• CommentAuthorDavid_Corfield
• CommentTimeApr 8th 2020
Re #8, I was just copying Urs from back here. What are you saying it should be?
• CommentRowNumber11.
• CommentAuthorDavid_Corfield
• CommentTimeApr 8th 2020
• (edited Apr 8th 2020)
Another case of relative cohesion we have is over $Sh_\infty\left(Sch_{\mathbb{Z}}\right)$ at differential algebraic K-theory. I’ll add that.
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2021-11-28 00:16:21
|
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https://chemistry.stackexchange.com/questions/107829/determining-mole-fraction-of-carbon-dioxide-knowing-kp-only
|
# Determining mole fraction of carbon dioxide knowing Kp only
I've been assigned the following problem at school:
Given the following reaction:
$$\ce{C (s) + CO2 (g) <=> 2 CO (g)}$$
Find the mole fraction of $$\ce{CO2}$$ provided that, in equilibrium, $$K_\mathrm{p} = 14.1$$.
Our teacher's solution is $$x(\ce{CO2}) = 0.324$$ (mole fraction).
I suspect that the problem as is provides insufficient information to arrive at the solution above. Assuming that the total pressure of the system is $$\pu{10 atm}$$, the resulting mole fraction is indeed $$0.324$$.
Is the problem truly incomplete or can it be solved without knowing the total pressure?
• You seem to be comfortable enough working out the answer when you're given a pressure. Why not try working it out without explicitly choosing a pressure? Use $p$ as a symbol instead of putting in something like 10 atm, and if you find that all the $p$'s cancel out in your answer, then it follows that you don't actually need to know the pressure. – orthocresol Jan 11 at 20:42
• I have attempted this. The final expression is: 14.1x = (1-x)^2*PT. The total pressure does not seem to cancel out. Here, x is the mole fraction. X seems to be a function of PT. The value of X changes with other values of PT. – Grego Jan 11 at 20:44
• Took me a while to understand what that equation meant, but yeah, you are correct! So, the question is indeed incomplete. [More common notation would be something like $p_\text{tot}$; PT suggests some pressure multiplied by temperature...] – orthocresol Jan 11 at 20:48
• You can format mathematical and chemical expressions on Chemistry.SE using MathJax; this post contains further details. – orthocresol Jan 11 at 20:50
• Ok! Thanks for the quick reply! My bad. I'll keep this in mind for the next question. – Grego Jan 11 at 20:51
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2019-12-08 16:58:31
|
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http://en.m.wikipedia.org/wiki/Proto-Indo-European_root
|
# Proto-Indo-European root
The roots of the reconstructed Proto-Indo-European language (PIE) are basic parts of words that carry a lexical meaning, so-called morphemes. PIE roots usually have verbal meaning like "eat" or "run". Roots never occur alone in the language. Complete inflected words like verbs, nouns or adjectives are formed by adding further morphemes to a root. Typically, a root plus a suffix forms a stem, and adding an ending forms a word.[1]
$\underbrace{\underbrace{\mathrm{root+suffix}}_{\mathrm{stem}} + \mathrm{ending}}_{\mathrm{word}}$
For example, *bʰéreti[2] "he carries" can be split into the root *bʰer- "to carry", the suffix *-e- "present tense" and the ending *-ti "third person singular".[3]
In its base form, a PIE root consists of a single vowel, preceded and followed by consonants. Except for a very few cases, the root is fully characterized by its consonants, while the vowel may alternate, a process called ablaut. Thus, the mentioned root *bʰer- can also appear as *bʰor-, with a long vowel as *bʰēr- or *bʰōr-, or even unsyllabic as *bʰr-, in different grammatical contexts.
## Phonotactics
Phonotactics describes the restrictions on the permissible combinations of phonemes (sounds).
### Basic root structure
The centre of a PIE root is the ablauting vowel (usually *e, perhaps sometimes *a[4] in its base form, the full grade). This vowel constitutes a sonority peak that is preceded and followed by a sequence of consonants with progressively decreasing sonority values. In other words, the sonority has to fall toward both edges of the root. The sonority hierarchy is as follows:[5]
1. *l *r *y *n
2. *w *m
3. plosives (sounds like *p *t * *k * or *; see Proto-Indo-European phonology for a complete table of PIE plosives)
This gives the following root structure (with P being any plosive and $\oslash$ an empty position):
$^* \begin{Bmatrix} P \\ \oslash \end{Bmatrix} \begin{Bmatrix} w \\ m \\ \oslash \end{Bmatrix} \begin{Bmatrix} l \\ r \\ y \\ n \\ \oslash \end{Bmatrix} e \begin{Bmatrix} l \\ r \\ y \\ n \\ \oslash \end{Bmatrix} \begin{Bmatrix} w \\ m \\ \oslash \end{Bmatrix} \begin{Bmatrix} P \\ \oslash \end{Bmatrix}-$
*w after a vowel is often written *u, and *y after a vowel is often written *i. Thus, *leiǵ- = *leyǵ- "to bind" and *dʰeu- = *dʰew- "to run" are allowed roots.
Other possible roots include *ped- "to tread", *dʰwes- "to breathe" and *wleikʷ- "to moisten". Forbidden are structures like **mter- (wrong order of phonemes: internal plosive) and **wmek- (two phonemes of the same group: unchanging sonority).
The remaining sounds, namely the laryngeals *h₁ *h₂ *h₃ and the sibilant *s, can occupy almost any place in the hierarchy.[5]*s is particularly common in initial position (see s-mobile).[6] Examples of such roots are *peth₂- "to fly", *treh₁w- "to nourish" and *streig- "to stroke".
Following the terminology of Sanskrit grammar, roots ending in laryngeals are referred to as seṭ-, all others as aniṭ-roots.
### Restrictions on the plosives
A root cannot contain two plain voiced plosives (**ged-), nor can it contain a voiced aspirate and a voiceless plosive (**tebʰ-), unless the latter occurs in a word-initial cluster after an *s (e.g. *stebʰ- "to stiffen").[6]
### Restrictions on the number of phonemes
The vowel has to be preceded and followed by at least one consonant each. The maximum number of consonants seems to be five (as in *strengʰ- "to twine").[6]
Early PIE scholars reconstructed a number of roots beginning or ending with a vowel.[7] The latter type always had a long vowel (*dʰē- "to put", *bʰwā- "to grow", *dō- "to give"), while this restriction did not hold for vowel-initial roots (*ed- "to eat", *aǵ- "to drive", *od- "to smell"). Laryngeal theory can explain this behaviour by reconstructing a laryngeal following the vowel (*dʰeh₁-, *bʰweh₂-, *deh₃-, resulting in a long vowel) or preceding it (*h₁ed-, *h₂eǵ-, *h₃ed-, resulting in a short vowel). These reconstructions obey the mentioned rules.[8]
### Roots without a full grade
Some roots have no central *e, an example being *bʰuH- "to grow, to become". Such roots can be seen as generalized zero grades of forms like **bʰweH-,[9] and thus follow the phonotactical rules.[10]
### Exceptions
Some roots like *pster- "to sneeze" or *pteh₂k- "to duck" do not appear to follow these rules.[5] This might be due to incomplete understanding of PIE phonotactics or to wrong reconstructions. *pster-, for example, might not have existed in PIE at all, if the Indo-European words usually traced back to it are onomatopoeias.[11]
Thorn clusters are sequences of a dental (*t *d *) plus a velar plosive (*k *g * etc.).[12] Their role in PIE phonotactics is unknown. Roots like *dʰgʷʰei- "to perish" apparently violate the phonotactical rules, but are quite common.
↑Jump back a section
## Lexical meaning
The meaning of a reconstructed root is conventionally that of a verb; the terms root and verbal root are almost synonymous in PIE grammar. This is because, apart from a limited number of so-called root nouns, PIE roots overwhelmingly participate in verbal inflection through well-established morphological and phonological mechanisms. Their meanings are not always directly reconstructible, due to semantic shifts that led to discrepancies in the meanings of reflexes in the attested daughter languages. Many nouns and adjectives are derived from verbal roots via suffixes and ablaut.
Nevertheless, some roots did exist that did not have a primary verbal derivation. Apart from the aforementioned root nouns, the most important of these were the so-called Caland roots, which had adjectival meaning. Such roots generally formed proterokinetic adjectives with the suffix *-u-, thematic adjectives in *-ró- and compounding stems in *-i-. They included at least *h₁rewdʰ- "red", *h₂erǵ- "white", *dʰewb- "deep" and *gʷreh₂- "heavy".[13]
↑Jump back a section
## Word formation
Fully inflected words are usually formed from a root plus a suffix plus an ending. The suffix is sometimes missing, which has been interpreted as a zero suffix.[14] Words with zero suffix are termed root verbs and root nouns. Beyond this basic structure, there is the nasal infix, a present tense marker, and reduplication, a sort of prefix with a number of grammatical and derivational functions.[15]
### Finite verbs
Verbal suffixes, including the zero suffix, convey grammatical information about tense and aspect, two grammatical categories that are not clearly distinguished. Present and aorist are universally recognised, while some of the other aspects remain controversial. Two of the four moods, the subjunctive and the optative, are also formed with suffixes, which sometimes results in forms with two consecutive suffixes: *bʰér-e-e-ti > *bʰérēti "he would carry", with the first *e being the present tense marker, and the second the subjunctive marker.[16] Reduplication can mark the present and the perfect.[15]
Verbal endings convey information about grammatical person, number and voice. The imperative mood has its own set of endings.[17]
Nouns are usually derived from roots or verb stems by suffixation or other means (see the morphology of the Proto-Indo-European noun for some examples). This can hold even for roots that are often translated as nouns: *ped-, for example, can mean "to tread" or "foot", depending on the ablaut grade and ending. Some nouns like *agʷn-o- "lamb" or *h₂ster- "star", however, are not derived from verbal roots.[18] In any case, the meaning of a noun is given by its stem, whether this is composed of a root plus a suffix or not. This leaves the ending, which conveys case and number.[19]
Adjectives are also derived by suffixation of (usually verbal) roots. An example is *ǵn̥h₁-tó-s "begotten, produced" from the root *ǵenh₁- "to beget, to produce". The endings are the same as with nouns.[20]
### Infinitives and participles
Infinitives are verbal nouns and, just like other nouns, are formed with suffixes. It is not clear whether any of the infinitive suffixes reconstructed from the daughter languages (*-dʰje-, *-tu-, *-ti-, among others) was actually used to express an infinitive in PIE.[21]
Participles are verbal adjectives formed with the suffixes *-ent- (active imperfective and aorist participle), *-wos- (perfect participle) and *-mh₁no- or *-m(e)no- (mediopassive participle), among others.[22]
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## Root extensions
Root extensions are additions of one or two sounds, often plosives, to the end of a root which do not seem to change its meaning. For *(s)teu- "to push, hit, thrust", we can reconstruct
• *(s)teu-k- > Ancient Greek τύκος (kos) "hammer"
• *(s)teu-g- > English stoke (Germanic k goes back to PIE *g.)
• *(s)teu-d- > Vedic tudáti "beats"
The source of these extensions is not known.[6]
↑Jump back a section
↑Jump back a section
## Notes
1. ^ Fortson (2004:76)
2. ^ The asterisk * indicates that this form is not directly attested, but has been reconstructed on the basis of other linguistic material.
3. ^ All examples of PIE roots are taken from Rix (2001) and Fortson (2004).
4. ^ The existence of *a as an ablauting vowel is disputed (see Indo-European ablaut: a-grade).
5. ^ a b c Rix (2001:5)
6. ^ a b c d Fortson (2004:70–73)
7. ^
8. ^
9. ^ Rix (2001:98–99)
10. ^ Jasanoff (2003:112)
11. ^
12. ^ Fortson (2004:59–60)
13. ^
14. ^ Fortson (2004:108)
15. ^ a b Rix (2001:14–21)
16. ^ Fortson (2004:81–83)
17. ^ Fortson (2004:83–85)
18. ^ Fortson (2004:116, 302)
19. ^ Fortson (2004:103)
20. ^ Fortson (2004:120–121)
21. ^ Fortson (2004:97)
22. ^ Fortson (2004:97–98)
↑Jump back a section
## References
↑Jump back a section
|
2013-06-19 07:27:21
|
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https://www.rdocumentation.org/packages/caret/versions/4.27/topics/filterVarImp
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# filterVarImp
0th
Percentile
##### Calculation of filter-based variable importance
Specific engines for variable importance on a model by model basis.
Keywords
models
##### Usage
filterVarImp(x, y, nonpara = FALSE, ...)
##### Arguments
x
A matrix or data frame of predictor data
y
A vector (numeric or factor) of outcomes)
nonpara
should nonparametric methods be used to assess the relationship between the features and response
...
options to pass to either lm or loess
##### Details
The importance of each predictor is evaluated individually using a filter'' approach.
For classification, ROC curve analysis is conducted on each predictor. For two class problems, a series of cutoffs is applied to the predictor data to predict the class. The sensitivity and specificity are computed for each cutoff and the ROC curve is computed. The trapezoidal rule is used to compute the area under the ROC curve. This area is used as the measure of variable importance. For multi--class outcomes, the problem is decomposed into all pair-wise problems and the area under the curve is calculated for each class pair (i.e class 1 vs. class 2, class 2 vs. class 3 etc.). For a specific class, the maximum area under the curve across the relevant pair--wise AUC's is used as the variable importance measure.
For regression, the relationship between each predictor and the outcome is evaluated. An argument, nonpara, is used to pick the model fitting technique. When nonpara = FALSE, a linear model is fit and the absolute value of the $t$--value for the slope of the predictor is used. Otherwise, a loess smoother is fit between the outcome and the predictor. The $R^2$ statistic is calculated for this model against the intercept only null model.
##### Value
• A data frame with variable importances. Column names depend on the problem type. For regression, the data frame contains one column: "Overall" for the importance values.
• filterVarImp
##### Examples
data(mdrr)
filterVarImp(mdrrDescr[, 1:5], mdrrClass)
data(BloodBrain)
filterVarImp(bbbDescr[, 1:5], logBBB, nonpara = FALSE)
apply(
bbbDescr[, 1:5],
2,
function(x, y) summary(lm(y~x))\$coefficients[2,3],
y = logBBB)
filterVarImp(bbbDescr[, 1:5], logBBB, nonpara = TRUE)
Documentation reproduced from package caret, version 4.27, License: GPL-2
### Community examples
Looks like there are no examples yet.
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2020-02-22 07:57:11
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http://vaguery.com/words/more-GP-benchmarks
|
Draft
# Looking for pattern-avoiding sets of numbers
Draft of 2019.04.07
May include: mathematical recreationsNumber Theory&c.
It’s seems remarkably easy to find problems and patterns in Number Theory that “feel” simple, but turn out to be tucked down low near the base of a big overbalanced stack of advanced and esoteric mathematics. In another piece, I recently wandered around over near Pell’s Equation, and discussed its use as a new “GP benchmark”.
An offhand passage from some other source—which I have now totally lost among my own overbalanced mental stacks—has made me think of a new one. Whoever it was that inspired this bit, either on Twitter or somewhere in a library book1 had been talking about taking integers and… probably multiplying them? And then looking at the resulting product (?) value and noticing that it didn’t contain a certain… digit?
¯\_(ツ)_/¯
As I write this, I’m starting to be more confident that it was something about digit-avoiding. Maybe “the product doesn’t have a 9 in it,” or something along those lines.
At any rate, that was enough to expand into some awful flower of Rather Quite Difficult problem-posing in my head. Here’s where I’ve ended up:
### Pattern-avoiding sets of integers
Suppose I give you a set $$N$$ containing integers (in base-10 notation, though that’s one of those parameters we can twiddle later if you want). Say, for example, I give you the set of integers $$0 < n < 1000$$. Here I’ve specified a range of values, but $$N$$ could be any set of positive integers (possibly including zero), and of any size.
I also specify a forbidden pattern, also in the form of a positive “integer”… though we won’t actually treat it quite that way in practice. Call this pattern $$p$$.
The score of the set $$N$$ is the sum of:
1. the number of items of $$N$$ which contain at least one contiguous occurrence of the digits of $$p$$, plus
2. the number of items in the set of pairwise sums of $$N$$ which contain at least one contiguous occurrence of the digits of $$p$$, plus
3. the number of items in the set of pairwise products of $$N$$ which contain at least one contiguous occurrence of the digits of $$p$$.
So for example, say I specify
$N = \{152, 231, 276, 427, 440, 706, 715, 741, 745, 756\} \\ p = 11$
Turns out there are no occurrences of the string pattern "11" in any of those ten numbers. Here though are the pairwise sums:
[383, 428, 579, 592, 858, 867, 893, 897, 908, 507, 658, 671, 937, 946, 972, 976, 987, 703, 716, 982, 991, 1017, 1021, 1032, 867, 1133, 1142, 1168, 1172, 1183, 1146, 1155, 1181, 1185, 1196, 1421, 1447, 1451, 1462, 1456, 1460, 1471, 1486, 1497, 1501]
If you look through the list, you’ll see 1133, 1142. 1168, 1172, 1183, 1146, 1155, 1181, 1185, 1196. That’s ten copies of “11” in the sums.
And here are the pairwise products:
[35112, 41952, 64904, 66880, 107312, 108680, 112632, 113240, 114912, 63756, 98637, 101640, 163086, 165165, 171171, 172095, 174636, 117852, 121440, 194856, 197340, 204516, 205620, 208656, 187880, 301462, 305305, 316407, 318115, 322812, 310640, 314600, 326040, 327800, 332640, 504790, 523146, 525970, 533736, 529815, 532675, 540540, 552045, 560196, 563220]
Here we have more digits, and more “tries”. But interestingly. there are only seven copies of "11" present, in 35112, 112632, 113240, 114912, 171171, 117852, and 318115.
So the score of the set {152, 231, 276, 427, 440, 706, 715, 741, 745, 756} is $$(0+10+7) = 17$$.
Compare that with this starting point:
$N = \{67, 101, 122, 370, 375, 384, 393, 445, 486, 838\} \\ p = 11$
Here, there are no copies of the pattern "11" in the values I give you, and also no copies in the set of pairwise sums or products. The score of this set on this pattern is zero.
## Some unsolved (by me) problems
Now those two little examples are just random sets of ten integers I found by sampling with some Ruby code (down at the bottom of this essay) and poking around. But when I tug on this little tidbit, a lot of interesting things feel like they’re connected. Here are the ones that have come to mind for me, so far.
### The largest pattern-avoiding subset
Suppose I give you some $$N$$ of integers, and ask you to remove items from it until the values remaining, and their pairwise sums, and their pairwise products all lack the bad pattern $$p$$. What is the largest subset of $$N$$ that avoids $$p$$?
For example, if I give you $$N$$ as the integers $$i, 1 \leq i \leq 999$$, and pattern $$p = 11$$, you’d obviously need to remove all values of the form “11*” and “*11” to avoid the simplest violations. That is, you have to pitch 110 and 111 and 119, and also 711 and 311 and certainly 111.
But then given the values you have remaining, you’ll almost certainly have pairs that sum or multiply to produce results containing “11”. And obviously you can remove every remaining value that participates in all those sums and products and be absolutely certain no violating values remain.
But that seems over-enthusiastic. It seems to me that if you remove certain “key” items, perhaps those that participate in multiple “violating” interactions, you could remove far fewer items to satisfy the specified constraints.
So what’s the largest subset we can retain?
How does one go about finding that subset? The “greedy” heuristic I’ve already spelled out—“Throw any value away that participates in any sum or product resulting in a violation”—is a bit heavy-handed. But as we step “away” from it, and try to be a little more thoughtful… one doesn’t immediately see what to look for. That is, what are the features of the numbers that we should pay attention to, besides “produce a sum or product”?
What does it even mean, mathematically or computationally, for a certain number to “cause” a pattern appear in a sum or product? Indeed, is this a property of any single number, or is it something to do with pairs of numbers? And if so, where in the pair does the “stuff” live that has a role in it?
And in my head, I’m hearing the word “hypergraph” bubbling up. That may not be the case for you… but it is for me.
### The hard-and-easy pattern problem
In the examples above, I used the avoided pattern “11”. What happens if I specify $$p = 9$$, or $$p=883$$, without changing the initial set of integers $$N$$?
When $$p=9$$, it certainly feels as if the problem has gotten “harder”. That is, there are more opportunities for the single digit “9” to appear in a given set $$N$$ of integers, and in their pairwise sums and products.
Similarly, when $$p=883$$, it feels pretty unlikely. There’s only one number below 1000 that contains that pattern, and not a lot in the pairwise sums, either.
But there are subtler reasons why the score of two patterns might differ. Consider the case of $$p=11$$ vs $$p=12$$. Here, at least for numbers of relatively few digits (say up to four), there’s actually a real difference in the number of “copies” of the pattern we can fit into any given integer. You can have two copies of "11" in a three-digit integer, but you can’t have two copies of "12"!
And on the easy-peasy end of the spectrum, if I specify $$N$$ as the integers between $$1$$ and $$1000$$, and say that $$p=12345678$$, well… I think we can agree that the score is easy to calculate. There will be no copies of that pattern of digits anywhere in $$N$$, nor in its pairwise sums and products. It just doesn’t fit.
Thus, for any set $$N$$ and pattern $$p$$, there will be some particular, measurable score. We can look and see, if nothing else.
So here are some more open questions:
• For a given set $$N$$ of integers, what are the worst-scoring patterns $$p$$? That is, which patterns produce the highest score?
• For a single set $$N$$, and two patterns $$p_1$$ and $$p_2$$, can one predict which of the patterns will have the better or worse score, without doing the explicit counting?
This latter one especially feels interesting. Obviously there are heuristics: if only one pattern is literally unable to fit in the set or the derived sets, it’s more likely that the other one will have a higher score. But not given.
Similarly, there are some odd-feeling things happening when I speak of pairwise sums vs pairwise products. Those are qualitatively different patches of number theory, algorithmically….
There are also questions here about algorithms. Which, for me, are kinda sorta the point.
### The self-avoiding set problem
Suppose I give you a set $$N$$ of integers, as before. But instead of specifying a particular $$p$$, I tell you to select any one of the elements of $$N$$ as $$p$$.
Obviously, there will already be one copy of $$p$$ already present, so the best possible score the set can receive for any one of its own elements will be $$1$$. But that’s only possible when there are also no pairwise sums or products that happen to “duplicate” $$p \in N$$ in their representations.
Are there any sets of integers $$N$$, with two or more elements, that have a score of $$\|N\|$$? That is, where the only occurrence of any element is in the set, not in the sets of sums or products? Sure! If I specify $$N=\{3, 4\}$$ then the pairwise sums and products are $$\{7\}$$ and $$\{12\}$$, respectively, and there are no copies of either $$p=3$$ or $$p=4$$ in those.
But sets of two items are boring.
What’s the largest set of integers we can construct such that the pairwise sums and products of all its elements avoid containing any (additional) copies of any element?
### The open-ended do-it-forever problem
You knew this was coming, surely.
In the play above, I’ve talked about taking set $$N$$, producing the set of pairwise sums (call this $$N_{+}$$) and pairwise products (call this $$N_{\times}$$). The score is defined as the number of occurrences of the pattern $$p$$ in all those.
Call that “all those” set of values, sums and products $$N' = (N \cup N_{+} \cup N_{\times})$$.
Now so far, I’ve been counting the contributions of $$N_{+}$$ and $$N_{\times}$$ separately, and under certain circumstances it’s entirely possible that a value might appear in two of the three sets; for example, what if $$0 \in N$$? When we add up the three contributions to get a score, I may therefore be “double counting”.
I’m good with that, but I want to spell it out before this next step.
If I define $$N' = (N \cup N_{+} \cup N_{\times})$$, we can obviously continue this process. That is I could produce $$N'' = (N' \cup N'_{+} \cup N'_{\times})$$, and so on.
Those sets $$N'$$ and $$N''$$ and so forth are getting very large very quickly. And their elements are also getting very large—with more digits—as well. At some point, there are going to be some very long strings of digits in play. It feels almost inevitable that for some pattern $$p$$ we will stumble across it in $$N'$$ (the “basic” problem outlined above), but if not then maybe in $$N''$$, or certainly somewhere not much farther along.
Are there sets $$N$$ and patterns $$p$$ for which we will never encounter $$p$$, anywhere in the infinite series of recursively-applied expansion?
Even if there aren’t such things, it seems as though there will (probably?) be a few “better” choices of $$N$$ and $$p$$, where the number of iterations we need to stumble over the first $$p$$ is large. What are those like?
### What do you mean “pairwise”?
Another problem-generating heuristic I’ve avoided so far should also be bubbling up in your mind, if you’re like me: Why did I say “pairwise” sums and “pairwise” products? What about sums and products of triples? Or addition and multiplication mapped onto all subsets of $$N$$?
Why not try those, Tozier?
Well, first, sure. OK, I will call you on that and happily heat up my laptop looking around. But bear in mind that the scores I’ve defined here still apply, and so yes there are probably some very cunning and subtle choices to be made for $$N$$ and $$p$$ that give wildly differing scores.
In fact, it feels as though this is probably the most interesting-but-difficult way to think about these problems. We’re quickly going to launch ourselves into the parts of Number Space where things need to be theoretical to be tractable. It’s hard for me to, for example, calculate the whole collection of subset-wise sums and products for a 1000-element set, for example.
Go ahead, work out how many subsets of a 100-element set there are. Got it? Now you see that there are $$2^{100} - 1$$ of those, at least if you avoid including an empty set. So by all means start your laptop working out all the sums and products over all of those.
I’m not even joking. While you’re waiting, I suspect something else will present itself. If not to you, then to your descendants.
For instance, given a set $$N=\{2,4,6, 20, 40, 60\}$$, and pattern $$p=7$$, what do you think this “super-score” of sums and products over all subsets of $$N$$ will be? Will it be large? Small? Compared to what?
And yet for $$N=\{1,2,3,10,20,30\}$$ and $$p=10$$ I think it might be a higher score. I could be wrong. Something interesting in that.
Now: How far can you push it?
## Why?
These are all perfectly reasonable mathematical explorations, but as I mentioned above, my interest is even more esoteric. I’m gathering examples of problems for which genetic programming might turn out to be useful. There are simple abstract data structures here, and there are well-known operators and groups involved in the Number Theory and algorithms people might bring to bear. So the big question for me is: Given an unsolved problem that’s this easy to specify, and a toolkit that includes all the parts and patterns that real hyoo-man mathematicians might invoke, what can GP come up with?
Can these be solved with arithmetic and a little conditional logic? With set theory? Compass-and-straightedge constructions? Gears and cogs? Are there approximations along the way, and are there insights to be gleaned from seeing the places where automated search finds a foothold?
Yes, of course. For that, though, we’ll all have to wait a little while and see.
## The Ruby code I used to poke around
def blocked?(number,pat_string)
number.to_s.include?(pat_string)
end
def remove_pattern(set,p)
pat = p.to_s
return set.reject {|n| blocked?(n,pat)}
end
def count_pattern(set,p)
pat = p.to_s
return set.find_all {|n| blocked?(n,pat)}.count
end
def pairwise_sums(set)
return set.combination(2).collect {|p| p[0]+p[1]}.uniq
end
def pairwise_products(set)
return set.combination(2).collect {|p| p[0]*p[1]}.uniq
end
def random_subset(set,size)
return set.sample(size)
end
range = 1000
s = (1..range)
p = 11
1000.times do |i|
size = 10
subset = s.to_a.sample(size)
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2022-05-22 06:58:21
|
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|
https://www.esaral.com/q/solve-this-following-42643
|
# Solve this following
Question:
If $2\left(\begin{array}{cc}3 & 4 \\ 5 & x\end{array}\right)+\left(\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right)=\left(\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right)$
A. $(x=-2, y=8)$
B. $(x=2, y=-8)$
C. $(x=3, y=-6)$
D. $(x=-3, y=6)$
Solution:
|
2023-02-01 13:29:10
|
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|
https://math.stackexchange.com/questions/477810/curve-stitch-primitive-calculation
|
Curve stitch primitive calculation
I want to calculate the intersection points of the following image:
Assume that the three points of the triangle could be located anywhere.
How would I do this?
• I think the formula for the quadratic spline can be adapted for my needs. Aug 28, 2013 at 22:15
• The image illustrates a parabola as the "envelope" of lines of the form $$x(a-t)+yt=t(a-t)$$ (where $a$ is the (constant) length of the "legs" of the figure, and $t$ is a parameter that varies from $0$ to $a$). The way envelopes work, the parabola doesn't contain points of intersection of the lines; the parabola is tangent to each line at some point (usually not the point of intersection with another line). Do you really want the points-of-intersection? Or do you want the points(-of-tangency) actually on the parabola?
– Blue
Aug 28, 2013 at 23:10
• I need the points of intersection. I already know how to generate the spline. Sep 4, 2013 at 0:55
1 Answer
Let's first solve the problem where the figures right-corner is at the origin, and its legs (of unit length) align with the axes.
The intercept-intercept form of the line equation gives this parameterization of the various lines: $$\frac{x}{t} + \frac{y}{1-t} = 1 \quad \text{or, in fraction-free form,} \quad x(1-t) + y t = t (1-t)$$ for $t$ a parameter between $0$ and $1$ (the extreme values being valid only in the fraction-free version).
Now, if we divide each leg into $n$ pieces (with $n+1$ equally-spaced points), then the intersection of the lines corresponding to $t=\frac{i}{n}$ and $t=\frac{j}{n}$, with $i, j \in \{0, 1, \dots, n\}$, is the point $$P_{ij} := \frac{1}{n^2}\large(\;ij\;,\;(n-i)(n-j)\;\large)$$
When the figure is located "anywhere", we need to apply a simple transformation. Note that, because the intersection points are defined via ratios of lengths of parallel segments, an affine transformation (that is, a linear transformation followed by a translation) will preserve the pattern.
Let's say that the resulting figure should have its (not-necessarily-right) corner at $C$ and its (not-necessarily-unit-length) legs ending at points $A$ and $B$.
We can write our original intersection points $P_{ij}$ as $$P_{ij} = \frac{ij}{n^2} u + \frac{(n-i)(n-j)}{n^2} v$$ where $u$ and $v$ are the unit vectors in the positive $x$ and $y$ directions.
Replacing $u$ with $A-C$ and $v$ with $B-C$ transforms the figure into the correct shape, though its corner remains at the origin; adding $C$ translates the figure into place. So, our transformed intersection points are given by
\begin{align} P^\prime_{ij} &= \frac{ij}{n^2} (A-C) + \frac{(n-i)(n-j)}{n^2} (B-C) + C \\[6pt] &= \frac{ij}{n^2} A \;+\; \frac{( n-i )( n - j )}{n^2} B \;+\; \frac{i (n-j) + j(n-i)}{n^2} C \end{align}
• As I said in my original post, the three points of the triangle formed by the two outer segments can be located anywhere, in any configuration. I don't think your formula works when that is the case. Sep 4, 2013 at 20:19
• Ah, right. Well, we just need to apply an appropriate transformation. I'll adjust my answer.
– Blue
Sep 4, 2013 at 20:37
• Sorry. How do I solve this type of equation? I am accustomed to using parametric formulas where there are two equations - one for the x value, one for the y value. Sep 4, 2013 at 23:14
• Give the triangle's vertices coordinates, $A(x_a,y_a)$, $B(x_b,y_b)$, $C(x_c,y_c)$. For brevity here, let $\alpha$, $\beta$, $\gamma$ be the multipliers on $A$, $B$, $C$. Then, writing $(x,y)$ for $P^\prime_{ij}$, the formula $$P^\prime_{ij} = \alpha \; A + \beta \; B + \gamma \; C$$ simply means \begin{align}x &= \alpha \; x_a + \beta \; x_b + \gamma \; x_c \\ y &= \alpha \; y_a + \beta \; y_b + \gamma \; y_c\end{align}
– Blue
Sep 4, 2013 at 23:40
|
2022-06-26 06:27:13
|
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|
https://150charles.com/equation-of-the-line-calculator/
|
Mortgages
# Online calculator: Equation of a line given two points
This online calculator finds the equation of a line given two points on that line, in slope-intercept and parametric forms
44
You can find an equation of a straight line given two points laying on that line. However, there exist different forms for a line equation. Here you can find two calculators for an equation of a line:
• first calculator finds the line equation in slope-intercept form, that is, $y=ax+b$ It also outputs slope and intercept parameters and displays the line on a graph.
• second calculator finds the line equation in parametric form, that is, $x=at+x_0\\y=bt+y_0$ It also outputs a direction vector and displays line and direction vector on a graph.
Also, the text and formulas below the calculators describe how to find the equation of a line from two points manually.
## How to find the equation of a line in slope-intercept form
Let’s find slope-intercept form of a line equation from the two known points $(x_0, y_0)$ and $(x_1, y_1)$. We need to find slope a and intercept b. For two known points we have two equations in respect to a and b $y_0=ax_0+b\\y_1=ax_1+b$
Let’s subtract the first from the second $y_1 - y_0=ax_1 - ax_0+b - b\\y_1 - y_0=ax_1 - ax_0\\y_1 - y_0=a(x_1 -x_0)$ And from there $a=\frac{y_1 - y_0}{x_1 -x_0}$
Note that b can be expressed like this $b=y-ax$ So, once we have a, it is easy to calculate b simply by plugging $x_0, y_0, a$ or $x_1, y_1, a$ to the expression above.
Finally, we use the calculated a and b to write the result as $y=ax+b$
### Equation of a vertical line
Note that in the case of a vertical line, the slope and the intercept are undefined because the line runs parallel to the y-axis. The line equation, in this case, becomes $x=x_1$
### Equation of a horizontal line
Read more: Mortgage Calculator | Bankrate
Note that in the case of a horizontal line, the slope is zero and the intercept is equal to the y-coordinate of points because the line runs parallel to the x-axis. The line equation, in this case, becomes $y=y_1$
### How to find the slope-intercept equation of a line example
Problem: Find the equation of a line in the slope-intercept form given points (-1, 1) and (2, 4) Solution:
1. Calculate the slope a: $a=\frac{y_1 - y_0}{x_1 -x_0} = \frac{4 - 1}{2 - (-1)} = \frac{3}{3} = 1$
2. Calculate the intercept b using coordinates of either point. Here we use the coordinates (-1, 1): $b=y_0 - a x_0 = 1 - 1\cdot(-1)=2$
3. Write the final line equation (we omit the slope, because it equals one): $y=x+2$
And here is how you should enter this problem into the calculator above: slope-intercept line equation example
## Parametric line equations
Let’s find out parametric form of a line equation from the two known points $(x_0, y_0)$ and $(x_1, y_1)$. We need to find components of the direction vector also known as displacement vector. $D=\begin{vmatrix}d_1\\d_2\end{vmatrix}=\begin{vmatrix}x_1-x_0\\y_1-y_0\end{vmatrix}$ This vector quantifies the distance and direction of an imaginary motion along a straight line from the first point to the second point.
Read more: 10/1 ARM vs. 30-Year Fixed | Bankrate
Once we have direction vector from $x_0, y_0$ to $x_1, y_1$, our parametric equations will be $x=d_1t+x_0\\y=d_2t+y_0$ Note that if $t = 0$, then $x = x_0, y = y_0$ and if $t = 1$, then $x = x_1, y = y_1$
### Equation of a vertical line
Note that in the case of a vertical line, the horizontal displacement is zero because the line runs parallel to the y-axis. The line equations, in this case, become $x=x_0\\y=d_2t+y_0$
### Equation of a horizontal line
Note that in the case of a horizontal line, the vertical displacement is zero because the line runs parallel to the x-axis. The line equations, in this case, become $x=d_1t+x_0\\y=y_0$
### How to find the parametric equation of a line example
Problem: Find the equation of a line in the parametric form given points (-1, 1) and (2, 4) Solution:
1. Calculate the displacement vector: $D=\begin{vmatrix}d_1\\d_2\end{vmatrix}=\begin{vmatrix}x_1-x_0\\y_1-y_0\end{vmatrix}=\begin{vmatrix}2-(-1)\\4-1\end{vmatrix}=\begin{vmatrix}3\\3\end{vmatrix}$
2. Write the final line equations: $x=3t-1\\y=3t+1$
0 ( 0 voted )
## 150 Charles
https://150charles.com
A remarkable site overlooking the expanse of the waterfront, 150 Charles Street is sited between the activity on the Hudson River and the history of the West Village
|
2022-11-27 18:01:52
|
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https://www.hepdata.net/record/ins1992937
|
• Browse all
Measurement of the production cross section for Z + b jets in proton-proton collisions at $\sqrt{s}$ = 13 TeV
The collaboration
CMS-SMP-20-015, 2021.
Abstract (data abstract)
The measurements of the cross section of the \PZ boson, decaying to dielectrons or dimuons, in association with at least one bottom quark jet are performed with proton-proton collision data at $\sqrt{s}=$ 13\TeV. The data sample corresponds to an integrated luminosity of 137.1\fbinv, collected by the CMS experiment at the LHC. The integrated cross sections of \PZ~$+\geq$~1~\cPqb~jet\xspace and \PZ~$+\geq$~2~\cPqb~jets\xspace are reported for the electron, muon, and combined channels with full Run 2 data. The measured integrated cross sections are 6.52 $\pm$ 0.04 (stat) $\pm$ 0.40 (syst) $\pm$ 0.14 (theo) pb for \PZ~$+\geq$~1~\cPqb~jet\xspace and 0.65 $\pm$ 0.03 (stat) $\pm$ 0.07 (syst) $\pm$ 0.02 (theo) pb for \PZ~$+\geq$~2~\cPqb~jets\xspace. The differential cross section distributions are measured as a function of various kinematic observables that are useful for precision tests of the perturbative quantum chromodynamics predictions. The ratios of integrated and differential cross sections of \PZ~$+\geq$~2~\cPqb~jets\xspace and \PZ~$+\geq$~1~\cPqb~jet\xspace processes are also determined. The value of the integrated cross section ratio measured in the combined channel is 0.100 $\pm$ 0.005 (stat) $\pm$ 0.007 (syst) $\pm$ 0.003 (theo). All of the measurements are compared with predictions from Monte Carlo simulations.
• #### Figure 2 (left)
Data from Figure 2 (left), located on page 15
10.17182/hepdata.115490.v1/t1
Differential cross section distribution as a function of Z transverse momentum for the Z + >= 1 b jet events
• #### Figure 2 (right)
Data from Figure 2 (right), located on page 15
10.17182/hepdata.115490.v1/t2
Normalized differential cross section distribution as a function of Z transverse momentum for the Z + >= 1 b jet...
• #### Figure 3 (left)
Data from Figure 3 (left), located on page 16
10.17182/hepdata.115490.v1/t3
Differential cross section distribution as a function of the leading b jet transverse momentum for the Z +>= 1 b...
• #### Figure 3 (right)
Data from Figure 3 (right), located on page 16
10.17182/hepdata.115490.v1/t4
Normalized differential cross section distribution as a function of the leading b jet transverse momentum for the Z + >=...
• #### Figure 4 (left)
Data from Figure 4 (left), located on page 16
10.17182/hepdata.115490.v1/t5
Differential cross section distribution as a function of the leading b jet absolute pseudorapidity for the Z + >= 1...
• #### Figure 4 (right)
Data from Figure 4 (right), located on page 16
10.17182/hepdata.115490.v1/t6
Normalized differential cross section distribution as a function of the leading b jet absolute pseudorapidity for the Z + >=...
• #### Figure 5 (left)
Data from Figure 5 (left), located on page 17
10.17182/hepdata.115490.v1/t7
Differential cross section distribution as a function of the leading b jet transverse momentum for Z +>= 1 b jet...
• #### Figure 5 (right)
Data from Figure 5 (right), located on page 17
10.17182/hepdata.115490.v1/t8
Normalized differential cross section distribution as a function of azimuthal difference between Z boson and the leading b jet for...
• #### Figure 6 (left)
Data from Figure 6 (left), located on page 18
10.17182/hepdata.115490.v1/t9
Differential cross section distribution as a function of the rapidity difference between Z boson and the leading b jet for...
• #### Figure 6 (right)
Data from Figure 6 (right), located on page 18
10.17182/hepdata.115490.v1/t10
Normalized differential cross section distribution as a function of the rapidity difference between Z boson and the leading b jet...
• #### Figure 7 (left)
Data from Figure 7 (left), located on page 19
10.17182/hepdata.115490.v1/t11
Differential cross section distribution as a function of the angular separation between the Z boson and the leading b jet...
• #### Figure 7 (right)
Data from Figure 7 (right), located on page 19
10.17182/hepdata.115490.v1/t12
Normalized differential cross section distribution as a function of the angular separation between the Z boson and the leading b...
• #### Figure 8 (left)
Data from Figure 8 (left), located on page 19
10.17182/hepdata.115490.v1/t13
Differential cross section distribution as a function of the leading b jet transverse momentum for the Z + >= 2...
• #### Figure 8 (right)
Data from Figure 8 (right), located on page 19
10.17182/hepdata.115490.v1/t14
Normalized differential cross section distribution as a function of the leading b jet transverse momentum for the Z + >=...
• #### Figure 9 (left)
Data from Figure 9 (left), located on page 20
10.17182/hepdata.115490.v1/t15
Differential cross section distribution as a function of the leading b jet absolute pseudorapidity
• #### Figure 9 (right)
Data from Figure 9 (right), located on page 20
10.17182/hepdata.115490.v1/t16
Normalized differential cross section distribution as a function of the leading b jet absolute pseudorapidity
• #### Figure 10 (left)
Data from Figure 10 (left), located on page 20
10.17182/hepdata.115490.v1/t17
Differential cross section distribution as a function of the subleading b jet transverse momentum for the Z + >= 2...
• #### Figure 10 (right)
Data from Figure 10 (right), located on page 20
10.17182/hepdata.115490.v1/t18
Normalized differential cross section distribution as a function of the subleading b jet transverse momentum for the Z + >=...
• #### Figure 11 (left)
Data from Figure 11 (left), located on page 21
10.17182/hepdata.115490.v1/t19
Differential cross section distribution as a function of the Z boson transverse momentum for the Z + >= 2 b...
• #### Figure 11 (right)
Data from Figure 11 (right), located on page 21
10.17182/hepdata.115490.v1/t20
Normalized differential cross section distribution as a function of the Z boson transverse momentum for the Z + >= 2...
• #### Figure 12 (left)
Data from Figure 12 (left), located on page 21
10.17182/hepdata.115490.v1/t21
Differential cross section as a function of the angular separation between two b jets for the Z + >= 2...
• #### Figure 12 (right)
Data from Figure 12 (right), located on page 21
10.17182/hepdata.115490.v1/t22
Normalized differential cross section as a function of the angular separtion between two b jets for the Z + >=...
• #### Figure 13 (left)
Data from Figure 13 (left), located on page 22
10.17182/hepdata.115490.v1/t23
Differential cross section as a function of the minimum angular separation between the Z boson and two b jets for...
• #### Figure 13 (right)
Data from Figure 13 (right), located on page 22
10.17182/hepdata.115490.v1/t24
Normalized differential cross section as a function of the minimum angular separation between the Z boson and two b jets...
• #### Figure 14 (left)
Data from Figure 14 (left), located on page 22
10.17182/hepdata.115490.v1/t25
Differential cross section as a function of the asymmetry of the Z + >= 2 b jets system
• #### Figure 14 (right)
Data from Figure 14 (right), located on page 22
10.17182/hepdata.115490.v1/t26
Normalized differential cross section as a function of the asymmetry of the Z + >= 2 b jets system
• #### Figure 15 (left)
Data from Figure 15 (left), located on page 23
10.17182/hepdata.115490.v1/t27
Differential cross section as a function of the invariant mass of two b jets for the Z + >= 2...
• #### Figure 15 (right)
Data from Figure 15 (right), located on page 23
10.17182/hepdata.115490.v1/t28
Normalized differential cross section as a function of the invariant mass of two b jets for the Z + >=...
• #### Figure 16 (left)
Data from Figure 16 (left), located on page 23
10.17182/hepdata.115490.v1/t29
Differential cross section as a function of the invariant mass of the Z boson and two b jets for the...
• #### Figure 16 (right)
Data from Figure 16 (right), located on page 23
10.17182/hepdata.115490.v1/t30
Normalized differential cross section as a function of invariant mass of the Z boson and two b jets for the...
• #### Figure 17 (left)
Data from Figure 17 (left), located on page 24
10.17182/hepdata.115490.v1/t31
Distributions of the cross section ratios as a function of the leading b jet transverse momentum
• #### Figure 17 (right)
Data from Figure 17 (right), located on page 24
10.17182/hepdata.115490.v1/t32
Distributions of the cross section ratios as a function of the leading b jet absolute pseudorapidity
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2022-01-17 00:32:43
|
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https://www.d3view.com/2006/09/102/
|
##### Modeling Friction in Contact
In contact-impact interactions, friction plays an important role in accurately capturing the sliding behavior. In LS-DYNA, the coulomb treatment of friction is used where a static and a dynamic friction can be defined which are used to determine the shear force while resisting penetration. By default, all contact definitions model a friction-less sliding which is a good start for initial models. The first step in activating friction in contacts is to use a non-zero static friction parameter, FS, in the *CONTACT_{OPTION} keyword. Optionally, one can define a dynamic friction, FD, which will only be used for a non-zero decay coefficient, DC. It is common to see a positive sliding energy when using non-zero friction parameters. It must be noted that non-zero friction parameters do not apply for FORCE_TRANSDUCER and TIED contacts which remain tied for the duration of the simulation. When TIED contacts are defined with failure parameters and are defined to be converted into penalty-type contacts after failure, then the non-zero friction parameters will be used to determine the shear contact forces. Figure 1 illustrates the nature of the penetration removal process when a node is detected as penetrating the closest master segment. When a penetrating node is first detected, by checking the sign of the projected normal distance to the closest master segment, a penalty force is first calculated based on the stiffness and the absolute penetration value. This force is then resolved in a local coordinate system embedded at the master element (contact point) to determine the normal and shear components. The sliding resistance is then computed using the friction parameters of the master segment and the normal force component as shown.
(Click image to enlarge)
Transitioning from Static to Dynamic Friction
By default, LS-DYNA considers only a static value (FS). In reality, the friction is dependent on relative velocity with which the parts are sliding and this friction is usually less than the static friction value. To model this behavior, two parameters, FD and DC. The transition from static to dynamic friction is modeled using an exponentially decaying function, $\mu = FD + (FS-FD)e^{-DC*\|v_{rel}|}$, that is based on the instantaneous relative velocity of the sliding node and the corresponding master segment. The transition from static to dynamic friction is as shown below.
(Click image to enlarge)
Part Based Friction
By default, the non-zero friction parameters defined in the *CONTACT_{OPTION} keyword are used for all segments. This may be acceptable when all parts involved are composed of similar material, but it may be inaccurate when dissimilar materials are defined to interact using SINGLE_SURFACE contact definitions. In such cases, LS-DYNA provides an option to define friction parameters at the component level using the *PART_CONTACT keyword. To let LS-DYNA use the values defined at the component level, FS must be set to -1. When FS=-1, a quick look-up is done prior to computing the shear force magnitude to determine the frictional parameters using the master segment’s part definition.
Part Pair Based Friction
While the part based friction parameter definition is an improvement compared to the global values, which is applied to all parts(segments), the concept of choosing the master segment friction values can still over/under predict the sliding resistance. To illustrate this, consider two dissimilar materials such as foam (F1) and steel (S1) that interact purely by contact treatment. Using component based definitions, we can input frictional values of 0.6 and 0.2 (using FS) for the respective materials using the *PART_CONTACT keyword. Now lets consider the case of S1 sliding on F1 which will cause LS-DYNA to look up the friction parameters of the master segment (F1 in this case) to give a value of 0.6. Next consider the case of F1 sliding on S1 which will result in a friction value of 0.2. As you can see, we come up with two different frictional parameters for the same PAIR of materials interacting with each other. This is easily overcome by using the *DEFINE_FRICTION keyword available in LS-DYNA versions 970 and later. The *DEFINE_FRICTION keyword allows us to define an unlimited number of interacting part pairs and their corresponding friction parameters. When using this option, the parameter FS must to set to -2, and the parameters defined in *DEFINE_FRICTION will override all values defined using the *PART_CONTACT keyword. Now using the earlier example, lets pick a pair of F1/S1 and define an average friction value of 0.4. With this definition, when either F1 slides on S1 or S1 slides on F1, instead of looking up the part based contact parameters, LS-DYNA looks up the pair definition and uses the average value of 0.4 which provides more accurate resistance.
Graphical viewing of frictional energy
The frictional sliding energy is dissipative and can be optionally written out to binary file using the option FRCENG in *CONTROL_CONTACT which requires the parameters SPR and MPR to be set to unity along with using the command line argument “s=interface_file”. Once these options are used, LS-DYNA outputs a binary file named “interface_file” which can be viewed using LS-PrePost. Among other variables, the frictional energy is output as the component “surface energy density” which can be fringed on the contact surface.
• Weldon says:
How can I read and view the frictional sliding energy (interface_file) using LS-PrePost?
• Suri Bala says:
When SPR=MPR=1 & s=interface file is requested, LS-DYNA outputs the interface file which can be read by LS-PREPOST directly as you would any D3PLOT file. Once open, you can view the history variables under the FCOMP button. The last paragraph in the above post describes this.
• Francesco Previtali says:
Dear Suri,
reading your article it seems to me that shear force in tied and tiebreak contacts is due exclusively to friction, which means F= FS*Fn where FS is the coefficient of friction and Fn is the force due to penetration normal to master segment.
In your presentation “Tie-Break Contacts in LS-DYNAâ€, at page 3 you wrote “There is no sliding allowed between the elements used in the tiebreak definitionsâ€, so I thought normal and tangential displacements of slave nodes were treated in the same way.
Now I am a bit confused: where am I wrong?
I am using tiebreak contacts in order to tie two parts of a mesh, one more coarse and the other finer (element dimension is half of the coarse ones), and I merged the coincident nodes. I used FS=0 and FD=0 in contact definition: might it be the cause of the problem?
Thank you
• Suri Bala says:
Francesco,
When using TIED or TIEBREAK contacts, it is not recommended to merge the nodes even if they are coincident. Coincident nodes are treated by slave and master nodes getting 100% of the forces.
There is also no sliding in TIED contacts. TIEBREAK allows sliding only after failure in which case the FS/FD are used when there is any sliding.
Regards,
Suri
• Francesco Previtali says:
Suri,
thank you for the kind answer.
I have one more question: having used a nodes_to_surface approach, I excluded coincident nodes from the slave node set. Is it still a problem?
Thank you again.
Kindest regards
Francesco
• Suri Bala says:
Francesco,
That should not be a problem.
Suri
• Paolo Capozzi says:
Dear Suri,
I’m trying to obtain the frictional energy plot but I can’t find the command line of LS-DYNA. I launch LS-DYNA through the Ansys product launcher and it just asks for the working directory and the keyword input file. How can I specify the s=interface_file without using the command line?
Thank you.
Best regards,
Paolo
• Suri Bala says:
Paolo,
I am sorry but I have never used Ansys product to run LS-DYNA.
Also, I don’t think there is any other way to output interface data besides using the ‘s’ option in the command line.
If you can locate the LSDYNA exe, it may be easier to simply it execute it your self.
Suri
• Magnus Bergh (FOI Sweden) says:
Dear Suri,
I’m trying to model friction in a 2D plane geometry. I’m interested in the dissipated energy (want to estimate frictional heating); is it possible to edit this in 2D? Seems like I need to set SPR and MPR on the generic CONTACT-card which is for 3D?
Best Regards,
Magnus Bergh
• Liliana Beldie says:
Dear Suri,
I have been trying for a while now to use *CONTACT_SLIDING_ONLY and/or *CONTACT_SLIDING_ONLY_PENALTY. In a small example the _penalty contact seems to be working ok, but in a few lager problems modelling brain skull it causes the models to blows up with negative volume in solid elements, although no deformations are present which could cause negative volumes, so I take it it is down to the contact. This happens even with using a simple linear material for both components (so it is not due to a viscous material). It has been suggested that this type of contact is not usable anymore in Dyna. Can you please let me know if this is the case and what other type of contact I can use instead to model the sliding sticking behaviour that I am after (or how can I make the sliding_penalty contact to work). Thank you, I appreciate your help.
Best regards,
Liliana
• Suri Bala says:
Liliana Beldie,
These are old contacts and I have not really worked with them much. It could be normals not aligned properly but I am not positive about that. Have you tried *CONTACT_NODES_TO_SURFACE_TIEBREAK with OPTION = 4 ?
Suri Bala
LSTC
• Liliana says:
Dear Suri,
Thank you for your reply. Do you mean using *CONTACT_AUTOMATIC_SURFACE_TO_SURFACE_TIEBREAK? (the *CONTACT_TIEBREAK_NODES_TO_SURFACE does not seem to have an option). Thanks.
Liliana
• Suri Bala says:
Dear Liliana
Yes.
Suri Bala
LSTC
• Mani says:
Dear Suri,
I am doing an bird strike analysis on an inclined wall. To perform this analysis, I have run two iterations to define the contact between the bird and the wall, in both iterations thickness of the wall remains same.
Iteration-1. Defined two contact definitions between bird and wall. CONTACT_ERODING_SURFACE_TO_SURFACE & CONTACT_AUTOMATIC_SURFACE_TO_SURFACE are used. The reason to use CONTACT_AUTOMATIC_SURFACE_TO_SURFACE is take care of contact between the wall and other parts of the assembly.
Iteration-2. Defined one contact definitions between bird and wall. CONTACT_ERODING_SURFACE_TO_SURFACE is used.
The results obtained from these two iterations are entirely different. In iteration-1, the bird skims over the wall without eroding the wall.
In iteration-2, the bird erodes the wall.
Could you please tell me, what is the reason for the differences in the results?
Thanks,
Mani
|
2019-07-23 22:59:18
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|
https://jp.mathworks.com/help/phased/ref/phased.freespace-system-object.html
|
Documentation
# phased.FreeSpace
Free space environment
## Description
The phased.FreeSpace System object™ models narrowband signal propagation from one point to another in a free-space environment. The object applies range-dependent time delay, gain and phase shift to the input signal. The object accounts for doppler shift when either the source or destination is moving. A free-space environment is a boundaryless medium with a speed of signal propagation independent of position and direction. The signal propagates along a straight line from source to destination. For example, you can use this object to model the propagation of a signal from a radar to a target and back to the radar.
For non-polarized signals, the FreeSpace System object lets you propagate signals from a single point to multiple points or from multiple points to a single point. Multiple-point to multiple-point propagation is not supported.
To compute the propagated signal in free space:
1. Define and set up your free space environment. See Construction.
2. Call step to propagate the signal through a free space environment according to the properties of phased.FreeSpace. The behavior of step is specific to each object in the toolbox.
When propagating a round trip signal in free-space, you can either use one FreeSpace System object to compute the two-way propagation delay or two separate FreeSpace System objects to compute one-way propagation delays in each direction. Due to filter distortion, the total round trip delay when you employ two-way propagation can differ from the delay when you use two one-way phased.FreeSpace System objects. It is more accurate to use a single two-way phased.FreeSpace System object. This option is set by the TwoWayPropagation property.
### Note
Starting in R2016b, instead of using the step method to perform the operation defined by the System object, you can call the object with arguments, as if it were a function. For example, y = step(obj,x) and y = obj(x) perform equivalent operations.
## Construction
H = phased.FreeSpace creates a free space environment System object, H.
H = phased.FreeSpace(Name,Value) creates a free space environment object, H, with each specified property Name set to the specified Value. You can specify additional name-value pair arguments in any order as (Name1,Value1,...,NameN,ValueN).
## Properties
PropagationSpeed Signal propagation speed Specify signal wave propagation speed in free space as a real positive scalar. Units are meters per second. Default: Speed of light OperatingFrequency Signal carrier frequency A scalar containing the carrier frequency of the narrowband signal. Units are hertz. Default: 3e8 TwoWayPropagation Perform two-way propagation Set this property to true to perform round-trip propagation between the origin and destination that you specify in the step command. Set this property to false to perform one-way propagation from the origin to the destination. Default: false SampleRate Sample rate A scalar containing the sample rate. Units of sample rate are hertz. The algorithm uses this value to determine the propagation delay in number of samples. Default: 1e6 MaximumDistanceSource Source of maximum distance value Source of maximum distance value, specified as 'Auto' or 'Property'. This choice selects how the maximum one-way propagation distance is determined. The maximum one-way propagation distance is used to allocate sufficient memory for delay computation. When you set this property to 'Auto, the System object automatically allocates memory. When you set this property to 'Property', you specify the maximum one-way propagation distance using the value of the MaximumDistance property. Default: 'Auto' MaximumDistance Maximum one-way propagation distance Maximum one-way propagation distance, specified as a real-valued positive scalar. Units are meters. This property applies when you set the MaximumDistanceSource property to 'Property'. Any signal that propagates more than the maximum one-way distance is ignored. The maximum distance should be greater than or equal to the largest position-to-position distance. Default: 10000 MaximumNumInputSamplesSource Source of maximum number of samples. The source of the maximum number of samples in the input signal, specified as 'Auto' or 'Property'. When you set this property to 'Auto', the propagation model automatically allocates enough memory to buffer the input signal. When you set this property to 'Property', specify the maximum number of samples in the input signal using the MaximumNumInputSamples property. Any input signal longer than that value is truncated. This property applies when you set the MaximumDistanceSource property to 'Property'. To use this object with variable-size input signals in a MATLAB® Function Block in Simulink®, set the MaximumNumInputSamplesSource property to 'Property' and set a value for the MaximumNumInputSamples property. Default: 'Auto' MaximumNumInputSamples Maximum number of input signal samples. Maximum number of samples in the input signal, specified as a positive integer. This property limits the size of the input signal. Any input signal longer than this value is truncated. The input signal is the first argument to the step method. The number of samples is the number of rows in the input. This property applies when you set the MaximumNumInputSamplesSource property to 'Property'. Default: 100
## Methods
reset Reset internal states of propagation channel step Propagate signal from one location to another
Common to All System Objects
release
Allow System object property value changes
## Examples
expand all
Calculate the amplitude of a signal propagating in free-space from a radar at (1000,0,0) to a target at (300,200,50). Assume both the radar and the target are stationary. The sample rate is 8000 Hz while the operating frequency of the radar is 300 MHz. Transmit five samples of a unit amplitude signal. The signal propagation speed takes the default value of the speed of light. Examine the amplitude of the signal at the target.
fs = 8e3;
fop = 3e8;
henv = phased.FreeSpace('SampleRate',fs,...
'OperatingFrequency',fop);
pos1 = [1000;0;0];
pos2 = [300;200;50];
vel1 = [0;0;0];
vel2 = [0;0;0];
Compute the received signal at the target.
x = ones(5,1);
y = step(henv,x,...
pos1,...
pos2,...
vel1,...
vel2);
disp(y)
1.0e-03 *
0.0126 - 0.1061i
0.0129 - 0.1082i
0.0129 - 0.1082i
0.0129 - 0.1082i
0.0129 - 0.1082i
The first sample is zero because the signal has not yet reached the target.
Manually compute the loss using the formula
$L=\left(4\pi R/\lambda {\right)}^{2}$
R = sqrt( (pos1-pos2)'*(pos1-pos2));
lambda = physconst('Lightspeed')/fop;
L = (4*pi*R/lambda)^2
L = 8.4205e+07
Because the transmitted amplitude is unity, the square of the signal at the target equals the inverse of the loss.
disp(1/abs(y(2))^2)
8.4205e+07
Calculate the result of propagating a signal in free space from a radar at (1000,0,0) to a target at (300,200,50). Assume the radar moves at 10 m/s along the x-axis, while the target moves at 15 m/s along the y-axis. The sample rate is 8000 Hz while the operating frequency of the radar is 300 MHz. The signal propagation speed takes the default value of the speed of light. Transmit five samples of a unit amplitude signal and examine the amplitude of the signal at the target.
fs = 8000;
fop = 3e8;
sProp = phased.FreeSpace('SampleRate',fs,...
'OperatingFrequency',fop);
pos1 = [1000;0;0];
pos2 = [300;200;50];
vel1 = [10;0;0];
vel2 = [0;15;0];
y = step(sProp,ones(5,1),...
pos1,...
pos2,...
vel1,...
vel2);
disp(y)
1.0e-03 *
0.0126 - 0.1061i
0.0117 - 0.1083i
0.0105 - 0.1085i
0.0094 - 0.1086i
0.0082 - 0.1087i
Because the transmitted amplitude is unity, the square of the signal at the target equals the inverse of the loss.
disp(1/abs(y(2))^2)
8.4206e+07
expand all
## References
[1] Proakis, J. Digital Communications. New York: McGraw-Hill, 2001.
[2] Skolnik, M. Introduction to Radar Systems, 3rd Ed. New York: McGraw-Hill, 2001.
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2019-12-06 18:08:48
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|
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-and-chemical-reactivity-9th-edition/chapter-2-atoms-molecules-and-ions-study-questions-page-95a/18
|
## Chemistry and Chemical Reactivity (9th Edition)
All but $_{18}^{9}X$ are isotopes of X.
Since the element has an atomic number of 9, all species with an atomic number of 9 are X's isotopes, so all but $_{18}^{9}X$ are isotopes of X.
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2018-06-18 04:32:52
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http://openstudy.com/updates/523c7b0be4b0fbf3cc7b976c
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## iiamentertainment Group Title Trig. 10 months ago 10 months ago
1. iiamentertainment Group Title
2. myininaya Group Title
So -180 degrees means you are going to go a half of the circle clockwise |dw:1379695755546:dw|
3. myininaya Group Title
Now tan(x)=sin(x)/cos(x) sin(x) and cos(x) are given to you in that picture for x=-180 degrees
4. iiamentertainment Group Title
so my answer would -1 correct
5. iiamentertainment Group Title
For number 2 its 4^2 + 3^2 = c^2 which equals 25 so c^2 = 5
6. myininaya Group Title
Well no. Because tan(-180 deg)=sin(-180 degrees)/cos(-180 degrees) You are only giving the me what cos(-180 degrees) equals.
7. iiamentertainment Group Title
well then im confused
8. myininaya Group Title
Let me show you a picture.
9. myininaya Group Title
|dw:1379696063163:dw| cos(theta)=adj/hyp=x/1=x So the x coordinate is cos(theta) sin(theta)=opp/hyp=y/1=y So the y coordinate is sin(theta) tan(theta)=opp/adj=y/x Your coordinates are given to you when you do that -180 degree rotation.
10. myininaya Group Title
Just put them in place of y and x.
11. myininaya Group Title
|dw:1379696195859:dw|
12. myininaya Group Title
tan(-180 deg)=?
13. iiamentertainment Group Title
(-1,0)
14. myininaya Group Title
well plug those numbers remember tan(theta)=y/x
15. iiamentertainment Group Title
im sorry im so lost
16. myininaya Group Title
what is y and what is x?
17. myininaya Group Title
Remember your coordinates are (x,y) which means that x is -1 and y is 0. Just replace your x with -1 and replace your y with 0.
18. myininaya Group Title
$\tan(-180^o) =\frac{\text{the y coordinate after rotating} -180^o}{\text{ the x coordinate after rotating} -180^o}$
19. myininaya Group Title
Replace that y coordinate with 0 Replace the x coordinate with -1 You know to do this since tan(theta)=y/x and your coordinate is (-1,0)
20. myininaya Group Title
So you should be able to do the last one. I will give you a hint use the same formula from above tan(theta)=y/x
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2014-07-23 07:53:52
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https://discourse.pymc.io/t/using-pm-data-to-predict-on-two-inputs-for-sample-posterior-predictive-why-is-there-no-change-in-the-results/7447
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# Using pm.Data to predict on two inputs for sample_posterior_predictive; why is there no change in the results?
Hi,
I’m having some issues understanding how to add new unseen data into my model, to see how it would effect my posterior.
I’ve looked at these posts (here, here but I still have not been able to understand what I’m doing wrong.
Hopefully this explanation will be enough to understand what the issue is.
Below I have created some data:
variation A ~ N(10, 3) and B ~ N(20, 3)
After I sample the first time, I wanted to see how the posterior would look when I add some new unseen data pm.set_data({"vid": [1] * 3, "val": [10000] * 3}). Basically I’m saying, what if we added 3 observations for variation B but with the observed ‘vals’ of 10000. Which, knowing out data generating process, should create quite quite different results.
However, when I look at the ‘predictions’ of my idata (image below) I cannot see any of the updated values? I can see there are 3 observations added but not any values associated with them. I would expect the values of the posterior to be much higher than the mean values of 10 and 20, but it just looks like the predictions are exactly the same.
My hypothesis is because I’m using another pm.Data class for the indexes in vid to separate the two variations. However I’m not sure how this would effect the outcome.
Thanks for taking a look at this and hopefully this is just a simple error on my side.
import arviz as az
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
import pandas as pd
import pymc3 as pm
import theano.tensor as tt
print(f"Running on PyMC3 v{pm.__version__}")
az.style.use("arviz-darkgrid")
pal = sns.color_palette("Set2")
pd.set_option('display.float_format', lambda x: '%.3f' % x)
#%%
def generate_data(n, m1, m2, sd):
variation = np.random.binomial(1, .5, n)
name = np.where(variation == 0, 'A', 'B')
val = np.where(variation == 0, np.random.normal(m1, sd, n), np.random.normal(m2, sd, n))
return {
'variation': variation,
'name': name,
'val': val
}
df = pd.DataFrame(generate_data(1000, 10, 20, 3))
#%%
var_idx, variations = pd.factorize(df['name'])
coords = {'variations': variations}
with pm.Model(coords=coords) as model:
var_id = pm.intX(pm.Data("vid", var_idx))
val = pm.intX(pm.Data('val', df["val"]))
mu = pm.Normal('mu', 15, 3, dims='variations')
sd = pm.Exponential('sd', 3)
mu_var = mu[var_id]
y = pm.Normal("y", mu=mu_var, sd=sd, observed=val)
trace = pm.sample(draws=2000, tune=1000, chains=2, target_accept=.90)
posterior_predictive = pm.sample_posterior_predictive(trace)
prior = pm.sample_prior_predictive()
idata_pymc3 = az.from_pymc3(
trace,
prior=prior,
posterior_predictive=posterior_predictive,
)
az.plot_trace(idata_pymc3);
az.summary(idata_pymc3)
with model:
# Switch out the observations and use sample_posterior_predictive to predict
pm.set_data({"vid": [1] * 3, "val": [10000] * 3})
posterior_predictive = pm.sample_posterior_predictive(trace, random_seed=1075)
az.from_pymc3_predictions(
posterior_predictive,
idata_orig=idata_pymc3,
inplace=True,
)
idata_pymc3
It looks like you are confusing the posterior predictive with the posterior. val are the observations, so changing them has no effect on posterior predictive sampling. sample_posterior_predictive uses the samples from the posterior p(\theta|y) to sample from the posterior predictive p(\tilde{y}|y) = \int p(\tilde{y}|\theta) p(\theta|y) d\theta. Changing var_idx does have an effect though. It defines which of the mus in mu are to be used. By setting it to 1, all 3 “predictions” will use mu[1] to draw values from y.
Both posts you linked are regressions, where the mean of y is defined by a beta * X kind of operation, so changing the values of X allows to interpolate or extrapolate and generate predictions from the posterior, but note that none of them are modifying the observed variables. If you want to condition on different observations you need to pm.sample again to find the new posterior.
1 Like
Hi @OriolAbril,
Thank you for a great explanation! I see where I made the confusion, and the inclusion of the formulas for posterior and posterior predictive were quite helpful.
Just as follow up, you mentioned If you want to condition on different observations you need to pm.sample again to find the new posterior. Does that mean if I pass in new observed variables using the same model the sampling starts over as if there was no previous sampling, or the new observed variables are simply updating the previous values from the first sample?
Thank you again.
The sampler would start over. Therefore you should include both old and new observations. There is no simple solution to do incremental updating in mcmc, but there are some less than perfect options such as using kennel density estimation as described in here: Updating priors — PyMC3 3.10.0 documentation
If speed is not an issue it’s better to just fit the model with the larger dataset.
2 Likes
Ah this is quite helpful. Thanks @ricardoV94 !
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2022-07-02 16:39:31
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http://www.geogebratube.org/student/m11531
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CCGPS CA 3.5.3 Example 2
At approximately what point does the value of $f(x)$ exceed the value of $g(x)$ if $f(x) = 2(4)^{\frac{x}{20}}$ and $g(x) = 0.5x$? Justify your answer with a graph.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
1. Make a general observation.
2. Create a table of values.
3. Graph both functions on the same coordinate plane.
4. Identify the approximate point where $f(x)$ is greater than $g(x)$.
Created with GeoGebraShared by Walch Education
View as Java Applet
View as HTML5 Applet
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2013-05-23 04:53:46
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https://blogs.mathworks.com/steve/2012/06/02/making-an-html-table-of-pixel-values-with-colored-cells/
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# Making an HTML table of pixel values with colored cells
Today's post shows you how to make a table with image colors and pixels appear when you publish your MATLAB scripts to HTML using the publish function. The result looks like this:
rgb = imread('peppers.png');
disp(im2html(rgb(88:92,200:204,:)))
R:80G:50B:77 R:78G:47B:74 R:76G:47B:71 R:95G:65B:73 R:158G:129B:117 R:76G:47B:77 R:75G:46B:71 R:91G:63B:72 R:158G:130B:119 R:192G:165B:141 R:77G:45B:70 R:82G:51B:63 R:148G:120B:114 R:192G:166B:146 R:203G:176B:153 R:75G:43B:65 R:126G: 95B: 97 R:186G:160B:145 R:197G:173B:154 R:208G:180B:160 R:100G: 70B: 72 R:174G:146B:135 R:193G:169B:151 R:198G:175B:158 R:211G:189B:170
I was inspired to do something like this when I saw Printing Variables to HTML Tables in Published Code (by Ned) on the File Exchange a while back. It also produces an HTML table with colored cells and superimposed values.
I was also thinking about the Pixel Region Tool in the Image Processing Toolbox. Here's a screen shot:
I wanted to go a bit further than Ned's original. I wanted to handle all the different kinds of image types (grayscale, truecolor, indexed with direct mapping, indexed with scaled mapping). I also wanted to replicate the feature of the Pixel Region Tool that automatically changed the color of the superimposed text depending on whether the underlying pixel was dark or light. (You can see that effect in the screen shot above.)
I've packaged all this in a function called im2html. You can download this function from the File Exchange.
Here are some examples showing how to use im2html with different types of images.
Display a table of values for a gray-scale image:
I = imread('pout.tif');
disp(im2html(I(125:134, 104:114)))
112 112 107 97 91 87 86 84 83 84 84 120 126 128 128 114 101 87 87 86 86 87 116 132 138 142 142 132 98 91 89 87 89 110 133 145 150 149 147 121 101 93 93 91 109 133 145 156 159 153 142 130 109 102 99 109 131 143 154 169 171 169 169 154 139 137 108 126 142 151 169 175 186 190 189 180 179 110 121 137 148 158 167 177 187 199 189 185 112 117 136 146 151 159 159 163 189 189 180 114 113 132 142 147 151 156 154 162 184 179
I = magic(10);
disp(im2html(I,[]))
92 99 1 8 15 67 74 51 58 40 98 80 7 14 16 73 55 57 64 41 4 81 88 20 22 54 56 63 70 47 85 87 19 21 3 60 62 69 71 28 86 93 25 2 9 61 68 75 52 34 17 24 76 83 90 42 49 26 33 65 23 5 82 89 91 48 30 32 39 66 79 6 13 95 97 29 31 38 45 72 10 12 94 96 78 35 37 44 46 53 11 18 100 77 84 36 43 50 27 59
Display a table of values from an indexed image:
[X,map] = imread('trees.tif');
disp(im2html(X(156:160,244:248),map))
<93>R:0.42G:0.68B:0.87 <93>R:0.42G:0.68B:0.87 <82>R:0.35G:0.65B:0.81 <77>R:0.39G:0.61B:0.81 <93>R:0.42G:0.68B:0.87 <82>R:0.35G:0.65B:0.81 <45>R:0.22G:0.45B:0.68 <50>R:0.42G:0.42B:0.52 <82>R:0.35G:0.65B:0.81 <82>R:0.35G:0.65B:0.81 <93>R:0.42G:0.68B:0.87 <50>R:0.42G:0.42B:0.52 <32>R:0.39G:0.29B:0.55 <44>R:0.52G:0.32B:0.52 <93>R:0.42G:0.68B:0.87 <93>R:0.42G:0.68B:0.87 <93>R:0.42G:0.68B:0.87 <44>R:0.52G:0.32B:0.52 <20>R:0.58G:0.13B:0.29 <27>R:0.45G:0.22B:0.42 <105>R:0.55G:0.74B:0.91 <93>R:0.42G:0.68B:0.87 <77>R:0.39G:0.61B:0.81 <44>R:0.52G:0.32B:0.52 <20>R:0.58G:0.13B:0.29
You can also capture the output of im2html as a string, or write it directly to a file.
s = im2html(magic(10),[]);
im2html(magic(10),[],'OutputFile','magic_table.html')
Give im2html a try. Comment here (or on the File Exchange page) if you find a good use for it, or if you have ideas about making it better.
Next time I'll go into some of the details about how im2html works, including the use of raw HTML in your publishable MATLAB scripts, as well as an obscure thing in the Image Processing Toolbox called imagemodel.
Published with MATLAB® 7.14
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2021-07-26 16:12:43
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=145&t=61501&p=233750
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## average rate vs instantaneous
$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$
Sarah Blake-2I
Posts: 153
Joined: Fri Aug 30, 2019 12:16 am
### average rate vs instantaneous
What is the difference between average rate and instantaneous rate in terms of the equations and calculations?
Posts: 103
Joined: Sat Aug 24, 2019 12:15 am
### Re: average rate vs instantaneous
The instantaneous rate of change is the average rate of change as the difference in time approaches 0. It is basically average rate of change over a really tiny time.
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2021-03-05 14:12:22
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http://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition/chapter-3-applications-of-differentiation-3-3-exercises-page-186/92
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# Chapter 3 - Applications of Differentiation - 3.3 Exercises: 92
False.
#### Work Step by Step
Let $h(x)=f(x)\times g(x)$ and $f'(x)gt;0; g'(x)gt;0$ over some interval $(a, b).$ Then $h'(x)=f'(x)g(x)+g'(x)f(x)$ which is not necessarily positive over the interval $(a, b)$ since we do not know the values of $f(x)$ and $g(x).$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2017-03-24 08:36:49
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https://dl.dropboxusercontent.com/s/8vhxsgyzwy2060r/syllabus.html
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# SSDS Syllabus
Seminar for the Study of Development Strategies
# General Information
The focus of the course is close reading and re-analysis of emerging research in the political economy of development, broadly construed. The focus is on well identified research whether based on experimental or observational data. It is intended for advanced graduate students (3rd - 4th year) that already have strong analytic skills. Auditors are welcome as long as they put in the work. Second time takers/auditors are also welcome.
The overall structure is that in most weeks an external speaker comes to discuss new or in-progress research. The speaker does not present the work however; instead they share their papers, data and code in advance with the class and a “replication team” has a week to put together a detailed discussion of the work. In other weeks we do something similar with work in progress of students in the class.
Note this course has an unusual format, meeting roughly once every two weeks over the course of a year. This course meets in room 711 of IAB building on Wednesdays from 4:10 - 6:00, generally followed by a dinner for a group of participants. It is led by Macartan Humphreys ([email protected]). If you want to see how this document was made, you can see the code here. Thanks to Jasper Cooper and Tara Slough who have done enormous work on the schedule and thinking through the structure and workflow of the class.
# Expectations
The reading loads are not especially heavy; typically the speaker will provide 1 or 2 readings that give a sense of their research agenda. You should read these carefully. You should also look at the data whether or not you are on the “rep” team. There is no point coming to the class unprepared. My thoughts on reading and discussanting are here http://www.macartan.nyc/teaching/how-to-read/ and here http://www.macartan.nyc/teaching/a-checklist-for-discussants/.
## Participation
The course will alternate between External and Internal weeks. During external weeks, guest speakers will present research to the class. Student research will be presented during Internal weeks.
### External weeks:
Guest speakers will be asked to share data in advance, and students are encouraged to replicate results and submit the results to robustness checks before each class.
• Every registered student will be expected to write a one-page response paper in advance of the talk each week. This is due into the class dropbox by midnight Monday of the day before. If you are presenting in a given week this is not required.
• A “rep” team of two students will be assigned a formal role as discussants and prepare oral and written commentary for the guest speaker.
Key elements of this are:
1. Be in touch with authors and be sure you have the data, papers, and all you need at least a week in advance
2. Make sure you can make sense of the data and run a basic replication.
3. When you have a feel of things jot down a brief pre-replication plan. What do you plan to look at? What do you expect to find? Archive this on dropbox.
4. Then there are two ways to expand the analysis;
• One is to check for robustness. How much do things depend on the particular models or measurements?
• The second is to go more deeply into the logic of the explanation. This might sometimes require assembling more data, constructing new tests and so on.
5. Meet me briefly on the Monday before class to go over your main material.
6. Generate a presentation that
• presents the paper in general
• uses experimentr (if it works; see below) to characterize the research design in abstract terms
• goes through the results and replication and
• goes through robustness and extensions
• does all this in rmarkdown so that speaker has content and code in a single file
7. Note that while we focus a lot on statistical replication and re-analysis there are many sides to a paper. Your presentation should not shy from discussing more fundamental conceptual or interpretational issues as appropriate.
### Internal weeks
During Internal weeks, student research will be presented.
• I strongly encourage participation from students returning from the field with main results in hand. The student will provide data and replication files to the class in advance but will not present his or her own research.
• Students that are not at that stage will be expected to provide an advanced draft of a research design by the end of the year. An advanced design means not only theory, hypothesis and identification strategy but also draft instruments and protocols and a dummy dataset and analysis.
• In internal weeks, two students will be assigned to present the research. The first will be assigned to act as the defender of the research and will prepare a presentation and defense of the research. The second student will serve as a devil's advocate, preparing a critique of the presented research.
Each student should expect to serve as a discussant for a guest speaker once per semester and to have his or her research presented once in the year and to act as both a defender and a devil's advocate for another student's research once in the year.
## Writing requirement
You will be expected to write a paper displaying original research to be presented during one of the internal weeks. These research papers will contain (i) a theoretical argument or motivation, (ii) an empirical test of that argument and (iii) a discussion of policy prescriptions resulting from the argument. A draft of this paper should be the paper used for your “internal” week; it does not have to have been written for this class specifically. However the final paper should however be the revised paper in light of the internal week discussions. Some thoughts on writing here http://www.macartan.nyc/teaching/on-writing/.
# The Speakers
## The Agenda
Our current speaker line up is as follows:
Date Speaker Provisional Topic
16-Sep Shira Mitchell Millennium Development Villages
23-Sep Rich Nielsen Violent Extremism
14-Oct Eli Berman Economics and Conflict
28-Oct Donald Green Vote-buying
4-Nov Pablo Querubin Accountability
18-Nov Leonard Wantchekon Deliberation
9-Dec Graeme Blair Nollywood or oil in the delta
3-Feb Jessica Gottlieb TBC
10-Feb Gwyneth McClendon TBC
24-Feb Thomas Fujiwara TBC
9-Mar Peter Bergman Education
23-Mar Daniel Hidalgo TBC
6-Apr Maarten Voors Health Systems Sierra Leone
13-Apr Jens Hainsmueller TBC
27-Apr Francesco Trebbi TBC
## The Rules
It is a very unusual thing for speakers to come and share data on unpublished work. It makes for terrific feedback and learning, but can also bring some risks to speakers. This cannot be thought of as a public presentation of research in the usual way and different rules apply. In particular:
• If a speaker requests that data not be shared outside the group, or perhaps even outside the replication team, this has to be adhered to strictly on pain of permanent ostracism.
• Any new findings from the analyses do not belong to the class or the students that engaged in the replication. You are working with the data for training purposes not for research purposes; you might see amazing patterns in the data but they don't belong to you.
• Any public commentary has to be bland at best. If you have to tweet or related after sessions, these should be of no cause for embarrassment for speakers.
# Workflow and Tools
We are going to be pretty hardcore about the workflow and using a set of very recent research tools to make sure all the work in the class is transparent and replicable.
The main tools that we will employ are:
• GitHub - for collaborating on code, publishing replications and raising issues
• Dropbox - for sharing data with one another
• R - for conducting statistical analysis and authoring documents in…
• Markdown - for authoring replications and pages on GitHub
## GitHub
GitHub will serve four main purposes:
1. Collaborating on code together
• Unlike Dropbox, GitHub allows for non-simultaneous editing of the same document, whether it is an R script, a .tex file or an .Rmd (Markdown) file.
• Each and every change is labeled, explained, and displayed in a simple interface. Reverting to previous versions or undoing certain changes is extremely easy. Three people can all make different changes to the same document on their own computers and then sync them whenever they want later.
• How it works: you make changes on your computer to a file, say an .R script. When you save, GitHub keeps a record of which changes you made. You label the changes with an explanation, and 'commit' them - but you haven't changed anything yet. To change the document on GitHub, you must 'push' or 'sync' your commits. To get your commits, others must 'pull' them from GitHub. The whole process becomes very easy and intuitive with a little familiarization.
• to push all of your commits and pull everyone else's in the desktop app, simply click the sync button.
2. Publishing replications as web pages
• When you submit and present replications you will write them in Markdown and compile them, then publish them to our GitHub page under your own subdirectory. This very page was created in R, using this file in 00_Admin.
• Publishing a page like this in GitHub is pretty easy.
• Firstly, create a new folder, for example in External Weeks, to host all the code for your replication.
• Secondly, write the publishable version of your code in a Markdown file in R, saving it as an .Rmd file. For example, mvp_replication.Rmd.
• Thirdly, compile that file into a file called readme.md using knitr in R (see Rmd_to_md.R for an example - feel free to add to this script).
• You're done: GitHub automatically converts any file called readme into a webpage. When you convert an .Rmd file to an .md file, you've told R to take the .Rmd, compile all the R code, and make a Markdown file out of it. In each subdirectory, GitHub reads the readme file and turns it into a webpage which everyone in the class can read and which you can use for the presentations.
3. Discussing and managing issues in the course using the 'issues' feature
• A range of issues will arise during our course. It could be anything from coding problems to trying to find a partner for a replication. You can post, label and assign issues here.
• All comments on issues can be formatted in Markdown!
4. Sharing code, functions, packages
• Everyone who contributes to the SSDS repository on GitHub can add code and other files to it. It can be a great incubator for new functions and other helpful general purpose tools.
To get started with GitHub, you will need two things:
1. a GitHub account
• you will use this to share code and make comments on the SSDS GitHub repository
2. the GitHub desktop app
• you will use this to label and sync any changes you make to one another's code
## Markdown
Please write all class reports in Markdown. Information on this here: http://rmarkdown.rstudio.com/. R markdown is fairly simple but has the advantage of letting you a) write $$\LaTeX$$ as needed b) integrate your R code directly c) compile to either a pdf, html or even word file. For transparency and error reduction b) is particularly important since we want to stay close to the data and set things up so that everyone in the class plus other presenters can follow your code and analysis.
To create a Markdown document in R:
• open Rstudio
• click File/New File/R Markdown…
• this creates an .Rmd file, containing both your code and text
• to compile the document, you can either use the knit() function in R (this is how we make .md files), or simply click the “Knit HTML / PDF / Word” button on the top panel of RStudio
## R
Analysis should be done in R. If you don't know R you should teach yourself. There are various online courses which you can take; have a look at http://tryr.codeschool.com/ and https://www.datacamp.com/. If you love your Stata or Excel and just cannot get on top of R, make sure you are on a team with someone who can so that final analyses can be implemented in R.
We will keep an updated list of packages that you will need in the the install_packages.R script. Run this script to get the new version of all
## Using Dropbox
We will keep all data on dropbox so that it can be sourced in from a single location. This is good practice and means that everything has to run off core data and not from individually customized files.
The easiest way to share data on Dropbox is to:
• put the data on your own Dropbox account
• get the link to the data document
• the link will have a “key” component (a random sequence of numbers and letters), and the filename (i.e. mydata.csv)
• see below for the use of the source_DropboxData() function, from the repmis package
Using this method we don't have to each store the data on our own computer, but can just temporarily use it in R. This avoids over-burdening our hard drives with large datasets.
## Using R, Markdown and Dropbox together
So for example here is some data:
rm(list = ls(all = TRUE))
library(repmis)
data <- source_DropboxData(key = "5zqvxaz6evtc16d",file = "dummydata.csv")
## Downloading data from: https://dl.dropboxusercontent.com/s/5zqvxaz6evtc16d/dummydata.csv
##
## SHA-1 hash of the downloaded data file is:
It looks like this
data
## ID Age Voted
## 1 1 21 1
## 2 2 NA 1
## 3 3 25 0
## 4 4 60 1
## 5 5 30 0
## 6 6 15 0
And here is some analysis:
# Age difference between voters and non voters:
t.test(Age ~ Voted, data = data)
##
## Welch Two Sample t-test
##
## data: Age by Voted
## t = -0.85866, df = 1.1034, p-value = 0.5372
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -221.2860 186.9526
## sample estimates:
## mean in group 0 mean in group 1
## 23.33333 40.50000
All claims in your text should come from the data. For example the average age is 30.2.
See here for an example of what a replication published to GitHub might look like.
## Using DeclareDesign to formally characterize the research designs
For each analysis we want to try to formally characterize the design and wring it through the alpha version (alpha as in struggling, not as in tough) of DeclareDesign. DeclareDesign is a package that I am working on with Graeme Blair, Jasper Cooper and Alex Coppock. It is designed to let you describe the core elements of a research design in an abstract way and then you get a set of outputs that provide information on the features of the design — bias, power, coverage — as well as objects useful for registration such as dummy data and mock analyses.
In the DeclareDesign framework, there are six core elements of a research design. You should be able to identify each of these for each replication:
1. The population. The set of units about which inferences are sought;
2. The potential outcomes function. The outcomes that each unit might exhibit depending on how the causal process being studied changes the world;
3. The sampling strategy. The strategy used to select units to include in the study sample;
4. The estimands. The specification of the things that we want to learn about the world, described in terms of potential outcomes;
5. The assignment function. The manner in which units are assigned to reveal one potential outcome or another;
6. The estimator function. The procedure for generating estimates of quantities we want to learn about.
In a replication, you will typically already have the data. The instructions below demonstrate how DeclareDesign can be used with pre-existing data.
To install the package, use devtools in combination with the access key. Please do not share the key during this alpha phase.
# Use this code here to install the DeclareDesign package
rm(list=ls())
devtools::install_github(repo = "egap/DeclareDesign",
auth_token = "7c4a0e3d05e33bd9bc15eae4a198a69f614e77ac"
)
We generate some example data using DeclareDesign DGP functions. You should already have data, so this step will not be necessary.
population_user <- declare_population(
individuals = list(
income = declare_variable()),
villages = list(
development_level = declare_variable(multinomial_probabilities = 1:5/sum(1:5))
),
group_sizes_per_level = list(
individuals = rep(1,1000),
villages = rep(5,200)
))
user_data <- draw_population(population = population_user)
save(user_data, file = "baseline_data.RData")
First, we load the baseline data created by the user, and then define a set of covariates that will be simulated to conduct power analysis and for simulated analyses.
load("baseline_data.RData")
kable(head(user_data), digits = 3)
villages_ID income individuals_ID development_level
1 1 -0.939 1 5
314 63 -0.042 2 4
636 128 0.829 3 5
681 137 -0.439 4 2
627 126 -0.314 5 5
692 139 -2.129 6 5
Second, we define the potential outcomes, which will be simulated based on the baseline covariate data.
potential_outcomes <- declare_potential_outcomes(
condition_names = c("Z0","Z1"),
outcome_formula = Y ~ .01 + 0*Z0 + .2*Z1 + .1*income
)
Then resample (bootstrap) from user data, respecting levels
population <- declare_population(
individuals = list(),
villages = list(),
N_per_level = c(500, 10),
data = user_data)
Fourth, we define one or more analyses we will run based on simulated data. This analysis will also be used for power analysis.
estimand <- declare_estimand(declare_ATE(), target = "population", label = "ATE")
Then we declare the design of the experiment, in this case a simple one without clusters or blocking.
assignment <- declare_assignment(potential_outcomes = potential_outcomes)
Then declare the estimator.
estimator <- declare_estimator(formula = Y ~ Z, estimates = difference_in_means, estimand = estimand)
Before finalizing the design, we conduct a power analysis to determine whether 500 units and 10 clusters (villages) are sufficient. To do this, we use the diagnose function.
The output of the diagnose() function is a summary of important statistical properties of the design, including the statistical power, bias, and frequentist coverage (among other uses, an indicator of whether the statistical power is calculated correctly). Here is the diagnosis summary for our simple experiment:
diagnosis <- diagnose(population = population, assignment = assignment,
estimator = estimator, potential_outcomes = potential_outcomes, sims = 1000)
kable(summary(diagnosis), digits = 3)
PATE sd(SATE) Power RMSE Bias Coverage
Y~Z1-Z0_diff_in_means_estimator 0.2 0 1 0.006 0 0.96
The information that diagnose outputs can be very useful for characterizing designs ex post.
The output has six important pieces of information. The first is the population average treatment effect, or PATE, the causal effect of the treatment on those in a finite population from which we have sampled. The sample average treatment effect, or SATE, is different: when we sample a particular set of units, the true average difference in potential outcomes might deviate from the PATE. In this example, we are treating the sample as the population, so there is no deviation of the SATE. Power in this simulation is defined as the probability of obtaining a statistically significant difference-in-means – this occurred in 100\% of the simulations. Reassuringly, the difference-in-means estimator does not exhibit any bias. Moreover, the coverage is very close to the theoretical target of 0.95, implying that the estimated confidence interval covers the true effect roughly 95\% of the time, as it should.
For more details on how to use DeclareDesign, visit the alpha version of the website here.
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2019-05-26 14:19:41
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https://rdrr.io/cran/HTSCluster/man/PoisMixClus.html
|
# PoisMixClus: Poisson mixture model estimation and model selection In HTSCluster: Clustering High-Throughput Transcriptome Sequencing (HTS) Data
## Description
These functions implement the EM and CEM algorithms for parameter estimation in a Poisson mixture model for clustering high throughput sequencing observations (e.g., genes) for a single number of clusters (PoisMixClus) or a sequence of cluster numbers (PoisMixClusWrapper). Parameters are initialized using a Small-EM strategy as described in Rau et al. (2011) or the splitting small-EM strategy described in Papastamoulis et al. (2014), and model selection is performed using the ICL criteria. Note that these functions implement the PMM-I and PMM-II models described in Rau et al. (2011).
## Usage
1 2 3 4 5 6 7 8 9 10 11 12 13 PoisMixClus(y, g, conds, norm = "TMM", init.type = "small-em", init.runs = 1, init.iter = 10, alg.type = "EM", cutoff = 10e-6, iter = 1000, fixed.lambda = NA, equal.proportions = FALSE, prev.labels = NA, prev.probaPost = NA, verbose = FALSE, interpretation = "sum", EM.verbose = FALSE, wrapper = FALSE, subset.index = NA) PoisMixClusWrapper(y, gmin = 1, gmax, conds, norm = "TMM", gmin.init.type = "small-em", init.runs = 1, init.iter = 10, split.init = TRUE, alg.type = "EM", cutoff = 10e-6, iter = 1000, fixed.lambda = NA, equal.proportions = FALSE, verbose = FALSE, interpretation = "sum", EM.verbose = FALSE, subset.index = NA)
## Arguments
y (n x q) matrix of observed counts for n observations and q variables g Number of clusters (a single value). If fixed.lambda contains a list of lambda values to be fixed, g corresponds to the number of clusters in addition to those fixed. gmin The minimum number of clusters in a sequence to be tested. In cases where clusters are included with a fixed value of lambda, gmin corresponds to the minimum number of clusters in addition to those that are fixed. gmax The maximum number of clusters in a sequence to be tested. In cases where clusters are included with a fixed value of lambda, gmax corresponds to the maximum number of clusters in addition to those that are fixed. conds Vector of length q defining the condition (treatment group) for each variable (column) in y norm The type of estimator to be used to normalize for differences in library size: (“TC” for total count, “UQ” for upper quantile, “Med” for median, “DESeq” for the normalization method in the DESeq package, and “TMM” for the TMM normalization method (Robinson and Oshlack, 2010). Can also be a vector (of length q) containing pre-estimated library size estimates for each sample. Note that if the user provides pre-calculated normalization factors, the package will make use of norm/sum(norm) as normalization factors. init.type Type of initialization strategy to be used (“small-em” for the Small-EM strategy described in Rau et al. (2011), and “kmeans” for a simple K-means initialization) gmin.init.type Type of initialization strategy to be used for the minimum number of clusters in a sequence (gmin): (“small-em” for the Small-EM strategy described in Rau et al. (2011), and “kmeans” for a simple K-means initialization) init.runs Number of runs to be used for the Small-EM strategy described in Rau et al. (2011), with a default value of 1 init.iter Number of iterations to be used within each run for the Small-EM strategry, with a default value of 10 split.init If TRUE, the splitting initialization strategy of Papastamoulis et al. (2014) will be used for cluster sizes (gmin+1, ..., gmax). If FALSE, the initialization strategy specified in gmin.init.type is used for all cluster sizes in the sequence. alg.type Algorithm to be used for parameter estimation (“EM” or “CEM”) cutoff Cutoff to declare algorithm convergence (in terms of differences in log likelihoods from one iteration to the next) iter Maximum number of iterations to be run for the chosen algorithm fixed.lambda If one (or more) clusters with fixed values of lambda is desired, a list containing vectors of length d (the number of conditions). specifying the fixed values of lambda for each fixed cluster. equal.proportions If TRUE, the cluster proportions are set to be equal for all clusters. Default is FALSE (unequal cluster proportions). prev.labels A vector of length n of cluster labels obtained from the previous run (g-1 clusters) to be used with the splitting small-EM strategy described in described in Papastamoulis et al. (2014). For other initialization strategies, this parameter takes the value NA prev.probaPost An n x (g-1) matrix of the conditional probabilities of each observation belonging to each of the g-1 clusters from the previous run, to be used with the splitting small-EM strategy of described in Papastamoulis et al. (2012). For other initialization strategies, this parameter takes the value NA verbose If TRUE, include verbose output interpretation If "sum", cluster behavior is interpreted with respect to overall gene expression level (sums per gene), otherwise for "mean", cluster behavior is interpreted with respect to mean gene expression (means per gene). EM.verbose If TRUE, more informative output is printed about the EM algorithm, including the number of iterations run and the difference between log-likelihoods at the last and penultimate iterations. subset.index Optional vector providing the indices of a subset of genes that should be used for the co-expression analysis (i.e., row indices of the data matrix y. wrapper TRUE if the PoisMixClus function is run from within the PoisMixClusWrapper main function, and FALSE otherwise. This mainly helps to avoid recalculating parameters several times that are used throughout the algorithm (e.g., library sizes, etc.)
## Details
Output of PoisMixClus is an S3 object of class HTSCluster, and output of PoisMixClusWrapper is an S3 object of class HTSClusterWrapper.
In a Poisson mixture model, the data y are assumed to come from g distinct subpopulations (clusters), each of which is modeled separately; the overall population is thus a mixture of these subpopulations. In the case of a Poisson mixture model with g components, the model may be written as
f(y;g,ψ_g) = ∏_{i=1}^n ∑_{k=1}^g π_k ∏_{j=1}^{d}∏_{l=1}^{r_j} P(y_{ijl} ; θ_k)
for i = 1, …, n observations in l = 1, …, r_j replicates of j = 1, …, d conditions (treatment groups), where P(\cdot) is the standard Poisson density, ψ_g = (π_1,…,π_{g-1}, θ^\prime), θ^\prime contains all of the parameters in θ_1,…,θ_g assumed to be distinct, and π = (π_1,…,π_g)^\prime are the mixing proportions such that π_k is in (0,1) for all k and ∑_k π_k = 1.
We consider the following parameterization for the mean θ = (mu_{ijlk}). We consider
μ_{ijlk} = w_i s_{jl} λ_{jk}
where w_i corresponds to the expression level of observation i, λ_k = (λ_{1k},…,λ_{dk}) corresponds to the clustering parameters that define the profiles of the genes in cluster k across all variables, and s_{jl} is the normalized library size (a fixed constant) for replicate l of condition j.
There are two approaches to estimating the parameters of a finite mixture model and obtaining a clustering of the data: the estimation approach (via the EM algorithm) and the clustering approach (via the CEM algorithm). Parameter initialization is done using a Small-EM strategy as described in Rau et al. (2011) via the emInit function. Model selection may be performed using the BIC or ICL criteria, or the slope heuristics.
## Value
lambda (d x g) matrix containing the estimate of \hat{λ} pi Vector of length g containing the estimate of \hat{π} labels Vector of length n containing the cluster assignments of the n observations probaPost Matrix containing the conditional probabilities of belonging to each cluster for all observations log.like Value of log likelihood BIC Value of BIC criterion ICL Value of ICL criterion alg.type Estimation algorithm used; matches the argument alg.type above) norm Library size normalization factors used conds Conditions specified by user iterations Number of iterations run logLikeDiff Difference in log-likelihood between the last and penultimate iterations of the algorithm subset.index If provided by the user, the indices of subset of genes used for co-expression analyses loglike.all Log likelihoods calculated for each of the fitted models for cluster sizes gmin, ..., gmax capushe Results of capushe model selection, an object of class "Capushe" ICL.all ICL values calculated for each of the fitted models for cluster sizes gmin, ..., gmax ICL.results Object of class HTSCluster giving the results from the model chosen via the ICL criterion BIC.results Object of class HTSCluster giving the results from the model chosen via the BIC DDSE.results Object of class HTSCluster giving the results from the model chosen via the DDSE slope heuristics criterion Djump.results Object of class HTSCluster giving the results from the model chosen via the Djump slope heuristics criterion all.results List of objects of class HTSCluster giving the results for all models for cluster sizes gmin, ..., gmax model.selection Type of criteria used for model selection, equal to NA for direct calls to PoisMixClus or "DDSE", "Djump", "BIC", or "ICL" for the respective selected models for calls to PoisMixClusWrapper
## Note
Note that the fixed.lambda argument is primarily intended to be used in the case when a single cluster is fixed to have equal clustering parameters lambda across all conditions (i.e., λ_{j1}=λ_{1}=1); this is particularly useful when identifying genes with non-differential expression across all conditions (see the HTSDiff R package for more details). Alternatively, this argument could be used to specify a cluster for which genes are only expressed in a single condition (e.g., λ_{11} = 1 and λ_{j1} = 0 for all j > 1). Other possibilities could be considered, but note that the fixed values of lambda must satisfy the constraint ∑_j λ_{jk}s_{j.} = 1 for all k imposed in the model; if this is not the case, a warning message will be printed.
## Author(s)
Andrea Rau <[email protected]>
## References
Anders, S. and Huber, W. (2010) Differential expression analysis for sequence count data. Genome Biology, 11(R106), 1-28.
Papastamoulis, P., Martin-Magniette, M.-L., and Maugis-Rabusseau, C. (2014). On the estimation of mixtures of Poisson regression models with large number of components. Computational Statistics and Data Analysis: 3rd special Issue on Advances in Mixture Models, DOI: 10.1016/j.csda.2014.07.005.
Rau, A., Maugis-Rabusseau, C., Martin-Magniette, M.-L., Celeux G. (2015). Co-expression analysis of high-throughput transcriptome sequencing data with Poisson mixture models. Bioinformatics, 31(9):1420-1427.
Rau, A., Celeux, G., Martin-Magniette, M.-L., Maugis-Rabusseau, C (2011). Clustering high-throughput sequencing data with Poisson mixture models. Inria Research Report 7786. Available at http://hal.inria.fr/inria-00638082.
probaPost for the calculation of the conditional probability of belonging to a cluster; PoisMixMean for the calculation of the per-cluster conditional mean of each observation; logLikePoisMixDiff for the calculation of the log likelihood of a Poisson mixture model; emInit and kmeanInit for the Small-EM parameter initialization strategy
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 51 52 set.seed(12345) ## Simulate data as shown in Rau et al. (2011) ## Library size setting "A", high cluster separation ## n = 200 observations simulate <- PoisMixSim(n = 200, libsize = "A", separation = "high") y <- simulate$y conds <- simulate$conditions ## Run the PMM model for g = 3 ## "TC" library size estimate, EM algorithm run <- PoisMixClus(y, g = 3, conds = conds, norm = "TC") ## Estimates of pi and lambda for the selected model pi.est <- run$pi lambda.est <- run$lambda ## Not run: PMM for 4 total clusters, with one fixed class ## "TC" library size estimate, EM algorithm ## ## run <- PoisMixClus(y, g = 3, norm = "TC", conds = conds, ## fixed.lambda = list(c(1,1,1))) ## ## ## Not run: PMM model for 4 clusters, with equal proportions ## "TC" library size estimate, EM algorithm ## ## run <- PoisMixClus(y, g = 4, norm = "TC", conds = conds, ## equal.proportions = TRUE) ## ## ## Not run: PMM model for g = 1, ..., 10 clusters, Split Small-EM init ## ## run1.10 <- PoisMixClusWrapper(y, gmin = 1, gmax = 10, conds = conds, ## norm = "TC") ## ## ## Not run: PMM model for g = 1, ..., 10 clusters, Small-EM init ## ## run1.10bis <- <- PoisMixClusWrapper(y, gmin = 1, gmax = 10, conds = conds, ## norm = "TC", split.init = FALSE) ## ## ## Not run: previous model equivalent to the following ## ## for(K in 1:10) { ## run <- PoisMixClus(y, g = K, conds = conds, norm = "TC") ## }
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2020-10-25 19:22:05
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https://stats.stackexchange.com/questions/228797/what-is-an-orthogonal-design
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# What is an orthogonal design?
I am a little bit confused about what an orthogonal design is and how it relates to the model matrix. It appears that there are many perspectives into the definition of orthogonal inside this post but none exactly helps me understand my problem. I am seeking an explanation with an experimental design context.
In Bailey 2008 pg 179, It introduces two factors G and F as orthogonal iff the subspaces $V_G \cap(V_F \cap V_G)^\bot$ and $V_F \cap(V_F \cap V_G)^\bot$ are both orthogonal to each other (or $V_G \cap V_{G \wedge F}^\bot$ and $V_F \cap V_{G \wedge F}^\bot$ is orthogonal)
A more intuitive theorem in the same book says (in terms of factors and levels) that
F and G on the same set are orthogonal to each other iff
• every F-class meets every G-class
• all these interesections have size proportional to the product of the sizes of the relevant F-class and G-class
However, the problem with these definitions is that it does help with 3 factors being orthogonal to each other or or when continuous covariates are included in your model. I was guessing that perhaps the model matrix when encoded can reveal something about orthogonality. However, my guess is that this is not true because It is obvious that when dummy coding the columns of the model matrix is not orthogonal to each other.
So my question is,
1. What is a general definition of an orthogonal design? Is there a more general definition of orthogonal design? Including continuous covariates
2. What are the advantages of orthogonal designs?
3. Does the model matrix reveal anything about orthogonality?
• I am not surprised that there is confusion about what is an Orthogonal Design. Orthogonal Designs were the subject of the 1979 book A.V.Geramita and Jennifer Seberry, Orthogonal Designs: Quadratic forms and Hadamard matrices, Marcel Dekker, New York - Basel, (1979), viii, 460 pages. This has now been republished as Jennifer Seberry, Orthogonal Designs, Springer Nature, 2017 Jan 3 '18 at 5:08
• The orthogonality of two factors in a design is a concept subtler than people would think of. A mathematical rigorous and intuitive definition (which has similar flavor to the definition you listed in your question) of orthogonality of design is given in Section 2.3 of The Coordinate-Free Approach to Linear Models. Though it might be quite abstract at the first reading, this is the best exposition of orthogonal design I have ever encountered. If I got time, I would like to summarize the author's idea therein. Jan 3 '18 at 5:25
1. Perhaps you haven't fully grasped the definition yet. The requirements for orthogonal designs are that the blocking is orthogonal and the treatment is orthogonal. This simply means that crossproducts total to zero, whether the blocking or treatment is continuous, pseudo-continuous, polytomous, or binary. As @whuber correctly points out, statisticians often call dot products cross products, and furthermore often assume blocking and treatment factors have mean 0. So any blocking factor or treatment factor "crossed" with any other will come out to 0.
2. Efficiency.
3. Absolutely. We would expect that cross products between any two columns of the design matrix will total out to zero.
• +1 In (2) it might perhaps be at least as important to list interpretability.
– whuber
Aug 8 '16 at 15:28
• @AdamO thanks for the reply. 1. I definitely haven't grasped the definition. Cross products of what exactly? 2. I am guessing that you mean dot product and not cross product (or do you really mean cross product). 3. Surely, one can make a design matrix (using dummy coding) such that dot product of two cols will not be zero. Aug 8 '16 at 16:33
• @tintinthong I neglected to mention that one should consider all factors to be mean-centered. That is an (unfortunately) common omission. You're right about dot product. Aug 8 '16 at 16:53
• tintin, in the regression context the term "cross product" often is used to refer to dot products: that is, a sum of squares or products. And yes, of course you can design a non-orthogonal experiment (almost surely a random design matrix will not be orthogonal): but that possibility seems to have no relevance to your questions.
– whuber
Aug 8 '16 at 16:53
• In addition to efficiency, if the design matrix is not orthogonal, there is no unique partition of the sums of squares. I think this is what @whuber means by interpretability: there would be different possible tests (w/ possibly different outcomes) & the data would not necessitate that 1 of them was the 'right' test. Aug 8 '16 at 16:58
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2021-10-16 17:37:32
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https://www.mathplanet.com/education/act/problems-21-40/32-what-is-the-value-of-cos-theta
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# 32. What is the value of cos theta?
$\begin{array}{lcl} The\;angle\;\theta\;lies\;in\;the\;second\\ quadrant\;and\;\sin \theta = \frac{4}{5}.\;What\;is\;the\\ value\;of\;\cos \theta ?\\ \\ (F)\; \frac{1}{5}\\ \\ (G)\; \frac{3}{5}\\ \\ (H)\; -\frac{3}{5}\\ \\ (J)\; -\frac{1}{5}\\ \\ (K)\; \frac{2}{5}\\ \end{array}$
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2023-02-04 11:51:11
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http://s116323.gridserver.com/3ave917f/match-the-following-fake-coin-problem-7439ad
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# match the following fake coin problem
One of the coins is fake. A Logic Brain Teaser: There are 12 gold coins. 6th Grade. The scale only tells you which side weighs more than the other side. Function Description. Have a look at the following video for another example: A mixture-type word problem (coins) One of the easiest of all the mixture word problems to understand is the coin problem since all students have some understanding of coins. We just launched a free weekly SC2 newsletter! Each man and son bought an apple, But when they returned ... A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocks over ... Let it be simple and as direct as possible. But we can do better than a factor of 2. Srabon got a prime!! Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. There are the two different variants of the puzzle given below. Include the coin: reduce the amount by coin value and use the sub problem solution (amount-v[i]). Russian Cyber Games 2020 - World Finals ! Equation: solution[coins+1][amount+1] = Coins game is a money game which introduces children to coinage in British, Australian, American and Euro currencies. If you knew the fake coin was lighter, then the solution would have an easy explanation. The most natural idea for solving this problem is to divide n coins into two piles of [n/2] coins each, leaving behind one extra coin if n is odd and then, compare the two piles and decrease the problem size by half. They get the following information: Both numbers ... Five puzzleFry ship’s pirates have obtained 100 gold coins and have to divide up the loot. Ragib: Yes. The colour of the coin does not match genuine coins. Put coins on each side, the real ones will balance each other out, the fake will make the scale go either way. Question: Match The Following: 1) Fake Coin Problem A) Shortest Hamiltonian Circuit 2) Floyd-Warshall Algorithm B) Class NHP 3) Traveling Salesman Problem C) Can Deal Negative Weight Edges 4) Graph Coloring Problem D) Divide And Conquer A) 1-D 2-B 3-A 4-C B) I-B 2-C 3-A 4-D C) 1-C 2-D 3-B 4-A D) 1-D … Weigh 9 against 10. ... Two fathers took their sons to a fruit stall. You see that the captains of the two teams participate in a coin toss wherein they pick one side of a coin each, that is head or tail. Tuesday, Thursday what are other two days staring with T? 4th Grade. Encourages children to partition amounts in different way and requires higher order thinking. Weigh coins 1,2,3,4 against coins 5,6,7,8. Fun Games for Kids ... MP1 - Make sense of problems and persevere in solving them. 1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8. It is known that a fake coin weighs either slightly less or slightly more than a real coin. But recently more and more coin dealers, precious metal sellers and attentive users of internet auction houses are finding items that are cause for alarm. Fariha’s mark was an even number. Complete the getWays function in the editor below. Okay? "We play turret d competitively", GSL Super Tournament #2 - RO16 Day 2 Preview, GSL Super Tournament #2: RO16 Day 1 Preview (2020), ASUS ROG Online: Showdown of Ultimate Destiny, Super Tournament 2 - RO16 concludes, RO8 bracket set. We have 12 coins. We know from fake £1 coins that forgers can achieve a good level of detail and colour match. Using a balance scale, how can you find the fake coin, and determine if it weighs less or more... :: Difficulty:2.6/4 1.2.3. It seems the number of forgeries that are offered in the US and Europe has increased sharply. etc. When you flip two coins at the same time — say, a penny and a nickel — you can get four possible outcomes: When you flip three coins at the same time — say, a penny, a nickel, and a … The umpire tosses the coin in the air. 2nd Grade. So first of all, let's check whether 8 and 9 and 10 and 11 can be indeed paid. If 7 and 8 do not balance, then the heavier coin is the counterfeit. The Problem: Start with n coins, all the same except for one fake coin which is lighter than the others. These are modern coins, so the fake coin is not necessarily lighter. I have seen examples of the very rare 1927–D Double Eagle offered that turned out to be fake … If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. The Treasury believes that around three per cent - amounting to a total of £45m - of pound coins are fake. You know that one is fake. For #1: imagine for a moment that all the coins are fake. Advertisement. If we took 0 coins from bag 0, 1 coin from bag 1, 2 coins from bag 2... we'd have $99\times100/2=4,950$ coins, and those 4,950 coins would weigh a total of 4,950 grams. We found another interesting puzzle for YOU-, Brain Development by Crazy Brain Teasers & Puzzles, Funny optical illusions to puzzle you and tease your brain, 1 to 50 Brain concentration level and focus on target Test, Five greedy pirates and gold coin distribution Puzzle. If the coins are objects you're handed, then you should be able to do that in a program quite easily. Here are the detailed conditions: 1) All 12 coins look identical. If we know that the counterfeit coin is necessarily lighter than the rest, then we can have a coin population is more than twice as large as the "heavier or lighter" problem for a given number of weighting: n=1 --> c = 3 vs 0. n=2 --> c = 9 vs 3. 1.) The one that loses a minute a day or the one that doesn’t work at all? Should be seven times. Most Analytical GOOGLE INTERVIEW Question Revealed. If they balance, 11 is light. He has to choose between three rooms. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. Assume for now that n is a power of 2, say 2 10 = 1024.This becomes an obvious binary search problem. Exclude the coin: solution for the same amount without considering that coin. So…. Weigh two coins at a time, so that's six times. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. Actually, max is 4. The orientation of the obverse and reverse designs is not in line. [STPL] Season 4 Cheerful Submission Thread. If the fake is lighter than normal, take the lighter of the 2 piles you just weighed. [PvT Build order] Beating Terrans with FE build, 2020 NFL and College Football Corona Season, Computer Build, Upgrade & Buying Resource Thread. Read TheYango's post to find out why. The coin was quickly panned by experts as a fake and was withdrawn from the auction sale. If they don’t balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake. If they balance, 11 is heavy. We are told that n − 1 of these coins are normal, that is, they have a head on one side and a tail in the other. It has three game modes. One side will be lighter, and one will be heavier. The problem of counterfeit coins is not a new phenomenon. There’s little … If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin). I need some assistance in solving the following problem: We are given a bag containing n unbiased coins. More Math Games to Play. You can see a full list of dates and designs of pound coins here. There are 10 stacks of 10 coins each. The problem is, we're only allowed the use of a marker (to make notes on the coins) and three uses of a balance scale. How is this done? 5th Grade. The currency defaults to British, but can be changed by clicking on the flags. Created for teachers, by … 1. Suppose we divide the coins into three piles, where at least two of them contain the same number of coins. 1) Fake Coin Problem – c) Can Deal Negative Weight Edges 2) Floyd Warshall algorithm – d) Divide and Conquer Floyd Warshall Algorithm – It is a dynamic programming algorithm which finds the shortest paths using recursive nature of problem. Find different combinations of coins that equal the same amounts of money teaching resources for 2014 National Curriculum Resources. It turned out ... A murderer is condemned to death. If it's balanced, the other coin from the imbalance is the fake. Weigh 1,2, and 5 against 3,6, and 9. Nabila got ... Robi is a very serious student. If you recall, for the "heavier or lighter" problem, c (n) = (3^n - 3)/2. The Mint told the investigators that the coin was “not a counterfeit”; instead it was likely to be one of the trial pieces despatched to shopkeepers to assist calibration processes during the pre-official launch. If they balance, 11 is light. The first activity, Sorting, helps children to recognise the different coins. The third weighing indicates whether it is heavy or light. The result of math class test came out. If the piles weigh the same, then the fake is heavier than normal. When the scales aren't balanced, one of the 2 you just put on is fake, try each against a real coin. The game is designed for 4-10 year olds. If they balance, 11 is heavy. TURF WARS: NA Team League by Rogues Gallery. Weigh 9 against 10. Which clock works best? Take all the coins off, and put 3 on each side from the set of 6 coins that were on the lighter side. By weighing 3 against a good coin the solution is easily arrived at. Take one of the coins from the imbalanced result and weigh it against a different coin. So our goal is to prove that any integer amount starting from 8 can be paid using coins of denominations 3 and 5. The recurrence relation for W (n): W (n)=W ( [n/2])+1 for n>1, W (1)=0. 1.2.1. @Sunyveil, "Wanna join my [combo] clan?" But now, say that bag 25 were the one with real coins … If the piles weigh the same, then the fake is heavier than normal. So it's possible there are better fake £2 coins out there, just good enough fakes that they haven't been reported. Paul, Sam and Dean are assigned the task of figuring out two numbers. Compare any 2 of those 3 on the scale. For example, if you have types of coins, and the value of each type is given as respectively, you can make change for units in three ways: , , and . Presume the worst case scenario, and don’t hope that you will pick the right coin on the first attempt. 1.2.2. Let us solve the classic “fake coin” puzzle using decision trees. 1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. 1.2. (1) On a fake 2 pound coin, the silver-coloured core isn’t quite flush with the gold-coloured outer ring. Is heavy or light of denominations 3 and 5 or light pirates are all... Four days there! That you have read and agree to the privacy policy and terms service... Coins is not in line and 5 against 3,6, and put 3 on each side from the set 6... “ 2 ” from 8 can be changed by clicking on the scale scale. Coins and a balance scale, one of the inner core the coin: solution for same... Piece has a lack of detail on the lighter side power of 2, say that bag 25 were one... Real coin strategy that detects the fake coin is the counterfeit we can do better than a real.. By coin value and use the sub problem solution ( amount-v [ i ].. Coins and a balance scale, one of which is lighter than,... Of money teaching resources for 2014 National Curriculum resources with calculations quickly containing two decimal points things. Given below: Start with n coins, no cutting coins, no cutting coins, all the coins,... Very strange number system weighs more than a real coin knew the fake is heavier the..., and 9 1 hour yet weighed ) is the fake coin six times paradigm. The obverse and reverse designs is not in line find different combinations of coins that were on the.... Detect the fake coin weighs less than the side with coins 5,6,7,8 are heavier than the others all 12 look. Scale also so we can do better than a real coin t.... Just put on is fake real coin detects the fake match the following fake coin problem heavier normal! First weighing ) the side with coins 1,2,3,4 problem, and the difference in weight is imperceptibly different there... Coins and a balance scale, one of the puzzle given below on! Denoting the number of times you must use the scale only tells you which side weighs more than the.! Take all the coins from the bag uniformly at random have read and agree to the privacy policy and of. Imbalance again, that 's six times 2 digit number Sakib: it! Coins, so the top coin is the fake is lighter than normal is. = 1024.This becomes an obvious binary search problem the rest the correct answer to this is Programming. There which Start with the letter ‘ t ‘ or slightly more than a factor of 2 say. Assigned the task of figuring out two numbers either 9 or 10 is light, that! Sign up '' you indicate that you will learn: how to create a brute force solution?! Detect the fake will make the scale 8 can be indeed paid would an. Have read and agree to the privacy policy and terms of service we divide the coins off, 5. [ 10 points ] you have read and agree to the privacy policy and terms of.. Difference in weight is imperceptible to your senses are better fake £2 coin often miss the dots... Dynamic Programming paradigm and not divide and conquer coin ” puzzle using decision trees, Thursday....... two fathers took their sons to a fruit stall by weighing 3 against a real...., match the following fake coin problem Wan na join my [ combo ] clan? less than the other side the are! Goal is to prove that any integer amount starting from 8 can be indeed paid 10 11! Off, and don ’ t balance, then you should be able to do that in a quite! Arrived at balances, we know coin 12 ( the only thing that distinguishes the fake coin is the.! List of dates and designs of pound coins here these are modern coins, no other weights, other! A time, so the fake coin from the imbalanced result and weigh against... Piles, where at least two of them as “ Tuesday, Thursday what are other days. American and Euro currencies that doesn ’ t balance, you know either. Again, that 's the fake coin, using only 3 times?. 'Re handed, then the fake coin bag uniformly at random assume n = 8 sub problem solution ( [! In weight is imperceptibly different the worst case scenario, and don ’ t balance, then the is. Heavier coin is the counterfeit or light order thinking so that 's the fake ”... 2 ) a counterfeit £2 piece has a lack of detail and colour match know that either 9 or is... Remaining one is fake should be seven times their sons to a fruit stall - make of. Worst case scenario, and put 3 on the Queen ’ s portrait is lighter than normal make of! Mp7 - look for and make use of structure if you knew the fake coin ” puzzle using trees... Coins here will be lighter, then the fake coin you have read agree. 6 coins on each side from the bag uniformly at random scale only tells you which side weighs than... Don ’ t balance, then the heavier coin is not in line because if 're! Other weights, no cutting coins, no cutting coins, so the coin... Find different combinations of coins that were on the flags side of the coin: solution the! Two different variants of the 2 piles you just weighed very strange number system ICCUP and proud of it,! Top coin is lighter than the others detail on the scale both sides that have... Approach works for finding the fake is lighter than the side with coins 5,6,7,8 are heavier than,. Is imperceptible to your senses American and Euro currencies that loses a minute a day or one... Amount starting from 8 can be paid using coins of denominations 3 and 5 three piles, where at two... Weighs either slightly less or slightly more than a factor of 2 of times you must the! Now that n is a money game which introduces children to coinage British. Amount-V [ i ] ) there is an imbalance again, that 's six times be seven times into. Is known that a fake coin was lighter, and don ’ balance! By comparing many coins at once,... Richie established a very strange number system is Dynamic paradigm... Of detail on the first attempt good coin the solution is obtained make use structure! Has heads on both sides different combinations of coins against a different coin were on the in! You knew the fake look for and make use of structure 's the fake was. Is that its weight is imperceptibly different also balances, we know coin 12 ( the only one yet! Difference in weight is imperceptibly different were the one with real coins that. Of which is fake, try each against a real coin by weighing 1 against 2 the solution is arrived. The task of figuring out two numbers finding the fake is lighter than normal, take the lighter of obverse. And Europe has increased sharply same amounts of money teaching resources for 2014 National Curriculum.! 8 can be indeed paid below, try each against a real coin “. Then you should be seven times days staring with t just weighed number system the number. Go either way considering that coin weighing also balances, we know from fake £1 coins that were the. When you flip a coin from the set of 6 coins on each side, the fake coin less! In the us and Europe has increased sharply generally get two possible outcomes heads! That were on the scale in order to always find the fake heavier... Sunyveil, Wan na join my [ combo ] clan? one side will be lighter, you. Weighing ) the side with coins 1,2,3,4 you flip a coin, using only 3 times scaling to a stall! Dates and designs of pound coins here number of ways to make change ‘ t.. Strategy that detects the fake coin ” puzzle using decision trees assume n = 8 only the twelve coins and.: //www.teamliquid.net/forum/viewmessage.php? topic_id=104154 & currentpage=316 # 6317 description of both the puzzles below, try to solve on own..., match the following fake coin problem na join my [ combo ] clan? ( ). Case scenario, and put 3 on each side, the fake to solve on own! Means that either 9 or 10 is light or 5,6,7,8 is heavy, so the fake will the! A brute force solution whether 8 and 9 imperceptible to your senses this year seriousness... Wan na join my [ combo ] clan?: heads or tails Curriculum resources we can do than! Coins game is a very strange number system be indeed paid coins.! We do not know whether the fake let 's try to solve on your own, assume n 8... All 12 coins look identical call “ 10 ” while looking at number “ 2 ” uniformly... Scale only tells you which side weighs more than a factor of 2 balance scale, one of which lighter. Amount-V [ i ] ) make sense of problems and persevere in solving them lighter side three,! Not balance, you can generally get two possible outcomes: heads tails. Only 3 times scaling agree to the privacy policy and terms of service coins into three piles where.: 1 ) all 12 coins look identical digit number Sakib: is it an odd first full... Strange number system “ 10 ” while looking at number “ 2 ” is known a. Of ways to make change got digits of a 2 digit number Sakib: is an... ( 2 ) a counterfeit £2 piece has a lack of detail on the lighter.... Better fake £2 coin often miss the fine dots around the perimeter of inner!
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2021-10-21 01:44:43
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https://testbook.com/question-answer/consider-a-coil-rotating-at-a-speed-of-n-rpm-in-th--5ff7fd83b4509f897072bbdf
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# Consider a coil rotating at a speed of N rpm in the field of P poles. As the coil moves past successive north the south poles, one complete cycle is generated. What is the frequency of the generated voltage?
This question was previously asked in
SSC JE EE Previous Paper 10 (Held on: 10 Dec 2020)
View all SSC JE EE Papers >
1. $$\frac{{PN}}{{60}}$$
2. $$\frac{{PN}}{{120}}$$
3. $$\frac{{120\;P}}{N}$$
4. $$\frac{{120\;f}}{P}$$
## Answer (Detailed Solution Below)
Option 2 : $$\frac{{PN}}{{120}}$$
Free
CT 1: Basic Concepts
18600
10 Questions 10 Marks 6 Mins
## Detailed Solution
Expression for frequency in a Generator:
• 3-phase supply with a frequency of ‘f’ is given to the 3-phase distributed winding the rotating magnetic field is set up by the windings.
• The speed of the rotating magnetic field is in synchronous with supply frequency and hence called synchronous speed ‘N’ (in rpm).
• 3-phase windings are wound for specific even number of poles ‘P’, 2/4/6/8…
• One cycle of AC current through the windings make the pole axes to move along one pair of poles (P/2).
• Hence fs cycle per second AC current will give rise the speed of rotating magnetic field as :
$$\frac{N}{{60}} = \frac{f}{{\frac{{\rm{P}}}{2}}}$$ rps.
• Rearrangement of the above equation will give the equation for synchronous speed in rpm as:
$$N = \frac{{120 \times {\rm{f}}}}{{\rm{P}}}$$ rpm.
• Frequency is expressed as $$\frac{{NP}}{{120}}$$ Hertz.
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2021-10-28 02:51:58
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http://math.stackexchange.com/questions/855789/solve-sum-i-1n-max-left-x-a-i-0-right-1
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# Solve: $\sum_{i=1}^n \max\left\{x-a_i,0 \right\}=1.$
Given $a_1,a_2,\ldots,a_n \in\mathbb{R}$. Solve the following equation on $\mathbb{R}$: $$\sum_{i=1}^n \max\left\{x-a_i,0 \right\}=1.$$
I am not sure that a closed-form solution exists, so iterative solutions are also welcome.
-
Without loss of generality, suppose that $a_1\leq a_2\leq\ldots\leq a_n$. We can distinguish several cases.
Case 1 Suppose that $x<a_1$. In this case, $$\sum_{i=1}^n\max\{x-a_i,0\}=0,$$ which can never equal $1$. We can conclude that any solution to the equation must involve $x\geq a_1$.
Case 2 At the other extreme, assume that $x>a_n$. In this case, $$\sum_{i=1}^n\max\{x-a_i,0\}=\sum_{i=1}^n(x-a_i)=n x-\sum_{i=1}^na_i,$$ and this equals $1$ if and only if $$x=\frac{1+\sum_{i=1}^n a_n}{n}.$$ If this value of this candidate solution satisfies $x>a_n$, the leading assumption for this case, then we just found a solution.
Case 3 Assume that $a_m\leq x\leq a_{m+1}$ for some $m\in\{1,\ldots,n-1\}$. Then, $$\sum_{i=1}^n\max\{x-a_i,0\}=\sum_{i=1}^m(x-a_i)+\sum_{i=m+1}^n0=mx-\sum_{i=1}^m a_i.$$ This is equal to $1$ if and only if $$x=\frac{1+\sum_{i=1}^m a_i}{m}.$$ If this value of the candidate solution satisfies also $a_m\leq x\leq a_{m+1}$, then we have found a solution.
By an exhaustive check of all cases (note that Case 3 consists of $n-1$ subcases), which can be automated by a computer program, we can find all solutions.
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2016-07-27 04:15:20
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https://www.semanticscholar.org/topic/Local-zeta-function/61973
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# Local zeta-function
Known as: Local, Local zeta function, Riemann hypothesis for curves over finite fields
Suppose that V is a non-singular n-dimensional projective algebraic variety over the field Fq with q elements. In number theory, the local zeta… Expand
Wikipedia
## Papers overview
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2014
2014
A p-adic field K is a finite extension of the p-adic numbers Qp. The ring of integers OK is the integral closure of the p-adic… Expand
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2005
2005
We give an explicit description of the poles of the Igusa local zeta function associated to a polynomial mapping g, in the case… Expand
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2005
2005
In this short note we compute for the polynomial $x^q -a$, $a\in K((\pi))$, its Igusa local zeta function and the corresponding… Expand
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2003
2003
Let K be a p−adic field, and ZΦ(s, f), s ∈ C, with Re(s) > 0, the Igusa local zeta function associated to f(x) = (f1(x), .., fl(x… Expand
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2002
2002
To a polynomial $f$ over a non-archimedean local field $K$ and a character $\chi$ of the group of units of the valuation ring of… Expand
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2001
2001
• 2001
• Corpus ID: 123128407
Abstract We give a very explicit formula for Igusa's local zeta function Z f ( s ) associated to a polynomial f in several… Expand
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1994
1994
We show the possibility of explicit calculation of the Fourier transforms of complex powers of relative invariants of some… Expand
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1993
1993
qui est meromorphe sur C. La conjecture de monodromie associe des valeurs propres de la monodromie (complexe) de l’hypersurface f… Expand
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Highly Cited
1991
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1991
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1985
1985
• 1985
• Corpus ID: 59066984
© Foundation Compositio Mathematica, 1985, tous droits réservés. L’accès aux archives de la revue « Compositio Mathematica… Expand
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2020-03-29 18:59:00
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https://socratic.org/questions/how-do-you-find-the-center-and-radius-of-the-circle-given-x-2-y-sqrt3-2-4x-25
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# How do you find the center and radius of the circle given x^2+(y-sqrt3)^2+4x=25?
Nov 18, 2016
The center is $\left(- 2 , \sqrt{3}\right)$ and the radius is $\sqrt{29}$
#### Explanation:
The standard form for the equation of a circle is:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center, and r is the radius.
Please notice that (y - sqrt(3))^2 is already in that form and we can see that $k = \sqrt{3}$.
Add ${h}^{2}$ to both sides of the given equation:
${x}^{2} + 4 x + {h}^{2} + {\left(y - \sqrt{3}\right)}^{2} = 25 + {h}^{2}$
When we expand the x square in the standard form we get:
${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$
We can find the value of h by setting the middle term from the standard form equal to the middle term of our equation:
$- 2 h x = 4 x$
$h = - 2$
This means that we can substitute ${\left(x - - 2\right)}^{2}$ for the terms on the left and 4 for ${h}^{2}$ on the right:
${\left(x - - 2\right)}^{2} + {\left(y - \sqrt{3}\right)}^{2} = 25 + 4$
Combine the right side:
${\left(x - - 2\right)}^{2} + {\left(y - \sqrt{3}\right)}^{2} = 29$
Write the 29 as ${\left(\sqrt{29}\right)}^{2}$
${\left(x - - 2\right)}^{2} + {\left(y - \sqrt{3}\right)}^{2} = {\left(\sqrt{29}\right)}^{2}$
In this form we can see the center and radius by observation:
The center is $\left(- 2 , \sqrt{3}\right)$ and the radius is $\sqrt{29}$
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2020-04-10 20:14:27
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https://wattsupwiththat.com/2022/09/02/greenhouse-efficiency/
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# Greenhouse Efficiency
Guest Post by Willis Eschenbach
Buoyed by equal parts of derision and praise for my last post, “Surface Radiation: Absorption And Emission“, I once again venture into the arena. I had an odd thought. The temperature has been generally rising over the period 2000-2021. I wondered if there was a way I could measure the efficiency of the greenhouse effect to see if the warming was due to increasing greenhouse gases (GHGs). If the GHGs were the cause, then the greenhouse effect would need to be more efficient in terms of warming the surface.
A Prologue: The earth is much warmer than the moon, which receives the same amount of solar energy per unit area. It’s generally accepted, including by me, that the warmth is from the very poorly named “greenhouse effect”, which has nothing to do with greenhouses.
Now, if you don’t think the “greenhouse effect” exists, this is NOT the thread for you. There are lots of places to make that argument. This isn’t one of them. We know the earth is warmer than expected. Nobody has ever come up with an explanation for that except the greenhouse effect.
If you are unclear about how the greenhouse effect works, the physical basis of it has nothing to do with CO2 or with the atmosphere at all. I explain this in my posts “People Living In Glass Planets“, and “The Steel Greenhouse“.
To reiterate: PLEASE do not post your opinions here on why the greenhouse effect isn’t real, or why there’s no such thing as downwelling radiation, or that scientists don’t understand the instruments that measure IR. The web is a very big universe. Somewhere out there is the perfect place to make those arguments.
This is not that place.
To return to the question at hand, which is the efficiency of the greenhouse effect, here’s the temperature change during the period of the CERES satellite data.
Figure 1. Surface temperature changes, CERES data. It is a conversion of the CERES surface upwelling longwave data to units of degrees Celsius using the Stefan-Boltzmann equation. It agrees well with e.g. the MSU lower tropical temperature, with a residual standard error of about a tenth of a degree C.
So the question, of course, is why did it warm over that period?
I thought, well, what the greenhouse effect does is to increase the surface temperature. The greenhouse effect starts with a certain amount of energy entering the climate system, and it ends with the surface being warmer and thus emitting more thermal radiation than would be expected if one were to look at say the moon, which gets the same energy from the sun as does the earth.
So … I figured that I could express the efficiency of the greenhouse effect by comparing the upwelling longwave radiation from the surface with the amount of solar energy entering the system. This measures the “end-to-end” efficiency of the entire system, including all feedbacks and interactions. I’ve chosen to express it as a “multiplier”—for every W/m2 of solar input, how many W/m2 of upwelling surface radiation do we get?
The amount of solar energy at the top of the atmosphere is about 340 watts per square meter (W/m2). However, about 100 W/m2 is reflected by the clouds and the surface. This means that the solar energy entering the system is on the order of 240 W/m2.
Upwelling longwave from the surface, on the other hand, is on the order of 400 W/m2. This means that the average greenhouse multiplier is approximately:
400 W/m2 / 240 W/m2 ≈ 1.66
In other words, for every watt per square meter of solar input, we get ~ 1.7 watts per square meter of upwelling surface radiation.
Now, we can run this calculation for each month, looking at the amount of thermal radiation emitted by the surface divided by the solar energy entering the system. Figure 2 shows that result. Remember that for increased greenhouse gases to be responsible for the warming, the greenhouse multiplier needs to increase.
Figure 2. Greenhouse multiplier. The multiplier is calculated as upwelling longwave surface radiation divided by incoming solar radiation (after albedo reflections). A multiplier of 2 would mean that the surface would be radiating two W/m2 of energy for each one W/m2 of solar energy actually entering the system. This shows that the greenhouse has increased the incoming solar radiation by about two-thirds, as measured at the surface.
Now, this is a most interesting finding. The efficiency of the planetary greenhouse has decreased slightly over the period—not significantly, but not increasing either.
In fact, the stability over the period is of interest in itself. Note that the standard deviation of the multiplier is 0.004 W/m2. Over the period, the end-to-end efficiency of the entire greenhouse system hardly varied at all. I’ve written before about the amazing stability of the system. This is another example.
So given the evidence above that the increase in upwelling surface radiation cannot be due to a change in greenhouse efficiency from increased CO2 or any other reason, what is the cause of the temperature increase? Here are the graphs of the two datasets that make up the greenhouse multiplier—the upwelling surface radiation, and the incoming solar radiation.
Figure 3. Upwelling surface thermal radiation (yellow, left panel), and incoming solar radiation after albedo reflections (red, right panel). Blue/black lines are LOWESS smooths of the data.
In Figure 3, we can see why the efficiency of the system hardly varied—the upwelling surface longwave was increasing pretty much in lockstep with the incoming solar energy actually entering the system.
Conclusions: We have observational evidence that the temperature increase from 2000-2021 was not due to an increase in greenhouse gases, or any increase in the efficiency of the greenhouse effect from any cause. The efficiency has been very stable over the period, with a standard deviation of 0.2% and no significant trend.
On the other hand, the change in incoming solar energy is both adequate to explain the increase in warming, and has the same shape as the change in surface radiation (blue LOWESS smooths in both panels in Figure 3). While there are undoubtedly other factors in play, the main cause of the warming is clearly the increase in the amount of solar energy after reflections from the clouds and the surface.
And once again, the clouds rule … go figure …
w.
Math Note: I tend to use “upwelling longwave surface radiation” and “temperature” interchangeably. Yes, I know that radiation varies as the fourth power of temperature, T4. However, the difference is trivial in the narrow range shown in e.g. Figure 3.
Figure 4 shows a comparison of the upwelling longwave shown in Figure 3 and the Stefan-Boltzmann derived temperature. Basically identical in form.
Figure 4. Temperature (yellow, left scale) and surface upwelling radiation (red, right scale)
Policy Note: I was 100% serious about asking people to refrain from commenting about things like how downwelling radiation doesn’t exist and the greenhouse effect isn’t real. Don’t make me tap the sign.
My Usual Request: Misunderstandings abound in communication. When commenting, PLEASE quote the exact words you are discussing, so we can all understand exactly who and what you are responding to.
Article Rating
Inline Feedbacks
Bill Powers
September 2, 2022 10:52 am
That sun. That pesky sun. Now who could have possibly imagined. Certainly not all the brilliant minds at the UN and the best bureaucrats and scientists they could buy with taxpayer money.
Call me a skeptic
Reply to Bill Powers
September 2, 2022 1:46 pm
They won’t listen, never will. Do the light bulb test. Put a small cloth towel over a light bulb and eureka. Less heat radiates out of the light bulb. You can pump excess CO2 into that room, still less heat radiates from the light bulb. After all climate fraudsters, it’s the sun stupid!
william Johnston
Reply to Call me a skeptic
September 2, 2022 4:29 pm
No money in that,tho’.
R_G
Reply to Bill Powers
September 5, 2022 5:29 pm
To be more precise; sun radiations modulated by clouds.
September 2, 2022 10:58 am
Yep, this is a great way to puncture the false arguments that alarmists use to support the hypothesis of CO2 caused warming, Willis. Thanks.
James Rouse
September 2, 2022 11:01 am
Excellent and clear article. Thank you.
Of course the pushback is going to be that what you have measured is a change in albedo of clouds (and Ice) which are feedbacks to temperature increases driven by CO2 radiative forcing.
Rud Istvan
Reply to James Rouse
September 2, 2022 11:18 am
None other than Dessler himself published a clear sky/all sky analysis that he claimed showed a positive cloud feedback, which NASA then touted officially and loudly. The problem was his r^2 was 0.02! Essentially random no correlation. So observationally there is NO cloud feedback despite what the IPCC and the fancy climate models say. That is one of three main reasons the models run hot. (The others are the parameterization attribution problem, and the now ARGO verified under modeling of ocean precipitation, which causes the modeled water vapor feedback to be too high, which is why the modeled tropical troposphere hot spot doesn’t exist in reality.)
Reply to Rud Istvan
September 2, 2022 11:36 am
Well put, Rud.
John Tillman
Reply to Rud Istvan
September 2, 2022 12:39 pm
There is negative cloud feedback, since they cause net cooling of about 5 degrees C with present cloud cover.
DMacKenzie
Reply to John Tillman
September 4, 2022 7:45 am
If cloud feedback was positive, Earth would have an atmosphere something like Venus only composed of superheated steam instead of CO2, since the dawn of time….
Reply to Rud Istvan
September 2, 2022 1:00 pm
The models run hot because that is what “management” wants predicted.
They run hot mainly from the exaggerated water vapor positive feedback not limited by increasing cloud cover.
But the reason does not matter — we get the predictions that the programmers are paid to predict — politics not science.
Apparently, that is not true of the Russian INM model that no one seems to care about.
There is insufficient knowledge of all causes of climate change to build a model that is accurate by design. But there is enough knowledge to be “in the ballpark” of observations (reality). if accuracy mattered. But accuracy does not matter. The models are used as science-like props to defend always wrong predictions of climate doom. They will never, as a group, be accurate. The average model represents the consensus on climate change. I’m surprised there are no sanctions to delete the Russian model from the average !
People with good scientific minds, like yourself, think models are intended to make accurate predictions. They are not. Not in modern climate “science” (politics). They are climate computer games intended to scare people. That’s why the CMIP6 models have a higher range than the CMIP 5 models. And CMIP7 models will likely have a higher range than CMIP6 models. This is very predictable if you think like a leftist –delete reason and accountability.
Last edited 1 month ago by Richard Greene
Reply to Rud Istvan
September 2, 2022 1:31 pm
Talking about Dessler is almost as bad as talking about Al Gore.
Two pretend scientists.
Ken
September 2, 2022 11:04 am
Great article. Elegant proof that albedo and solar irradience plays a major role in temperature. It’s intuitively obvious that CO2 has a negligible effect at 400 ppm compared to solar input.
Rud Istvan
September 2, 2022 11:10 am
WE, another nice post. I was curious about what the ‘settled science’ says the result should have been with the CO2 control knob changed over your CERES time interval.
Turns out did not even have to work it out myself. There is a ‘calculator’ available at scied.UCAR.edu. That is pretty official—UCAR is a major hub of settled climate science. I used it and the ‘official’ ECS (UCAR default) value of 3.0 (the calculator lets you input more or less ECS). The calculator runs in increments of 10 ppm CO2.
The CO2 in 2000 was 372 (annual average taking out seasonality). 370>>14.4C.
The CO2 in 2021 was 416. 420>>14.9C.
UCAR says you should have seen about a 0.5C increase in surface temperature from the GHE alone. You saw none (Fig 2). So much for UCAR’s settled climate science.
Reply to Rud Istvan
September 2, 2022 1:05 pm
Except UAH is up over +0.5 degrees C. from 2000 to 2020.
Johne Morton
Reply to Richard Greene
September 2, 2022 3:41 pm
From 2002 through July 2022, the trend is pretty small, ~0.2⁰ using the 12 month running mean (you can’t look at individual months). Even starting in the post-1998 La Niña dip, it’s less than 0.5⁰. Also, how do you explain the excellent correlation to solar activity?
Reply to Johne Morton
September 2, 2022 4:14 pm
Of course you can look at individual years — 2020 is about +0.5 degrees C warmer than 2000. These are not linear data so a linear trend line is not necessary. Did increased solar energy cause +0.5 degrees C. in 20 years. I’m not even close to being convinced the global warming in the past 20 years was caused by solar energy. And few others are convinced, for good reasons.
Maybe the CERES measure of solar activity is wrong.
There is a lot of calculating involved:
Clouds and the Earth’s Radiant Energy System (CERES) FluxByCldTyp Edition 4 Data Product in: Journal of Atmospheric and Oceanic Technology Volume 39 Issue 3 (2022) (ametsoc.org)
And correlation is not causation
Other CERES related publications in 2022:
Publications – CERES (nasa.gov)
Knowledge comes from study of a variety of sources, not just one WUWT article.
Richard M
Reply to Richard Greene
September 2, 2022 6:18 pm
Did you read Dubal/Vahrenholt 2021? Their data matches very closely to what Willis is reporting here.
Johne Morton
Reply to Richard Greene
September 3, 2022 2:49 pm
Well, I mentioned individual months, not years. Also, annual changes thanks to ENSO and other factors makes the year-to-year data noisy. Also, I’m not talking about a linear trend line, but the running mean. I could just as easily say that, thanks to the supposed increase in GHG concentrations and (non-existent) GHG efficiency, the temperature should never, ever go down. That would be silly, too.
Richard M
Reply to Richard Greene
September 2, 2022 6:16 pm
Willis explained why the temperature increase occurred. It was due to increased solar energy absorption. This was also reported in Dubal/Vahrenholt 2021. I’ve commented on this many times. They also used the CERES data.
Their work and now this work by Willis actually demonstrates the enhanced greenhouse effect, that is, that future warming is caused by CO2 increases that 1) increase DWIR warming the surface and 2) raising the effective radiation altitude, is shown to be false.
The reason increased DWIR does not lead to warming has to do with what I call boundary layer feedback. The reason the effective radiation altitude does not change is due to radiation exchange equilibrium in the atmosphere.
Jeff Alberts
Reply to Richard M
September 2, 2022 7:13 pm
There is no “the temperature”. This is all navel gazing.
Richard M
Reply to Jeff Alberts
September 2, 2022 8:55 pm
I think “the temperature” is fairly well defined and useful. Maybe a different name could be used, but that would probably lead to even more confusion.
Jeff Alberts
Reply to Richard M
September 8, 2022 6:43 pm
No it’s not well-defined or useful.
Reply to Richard M
September 2, 2022 10:15 pm
I never claimed Willie E, is misinterpreting CERES data. I do not believe his conclusion — that’s all. Nor do many climate scientists around the world.
Richard M
Reply to Richard Greene
September 3, 2022 7:53 am
Willis has been a luke-warmer for a long time. His admission that the data shows no greenhouse warming is a step forward. He should be congratulated for following the data. If you have another explanation feel free to express it.
Keep in mind that the very same lack of greenhouse effect increase was documented in Miskolczi 2010 using NOAA data since the 1940s.
The Dubal/Vahrenholt 2021 paper, which essentially shows the same result from CERES data, has been denied for the past year. Climate scientists are in denial likely for the same reasons you express. That would mean they have the physics wrong which in their minds is “inconceivable”.
?q=50&fit=crop&dpr=1.5
Marty Cornell
Reply to Richard Greene
September 2, 2022 7:34 pm
The UAH plot does not reflect attribution.
Reply to Marty Cornell
September 2, 2022 10:16 pm
UAH reflects a decent measurement methodology and I strongly doubt that solar energy was the cause of all warming from 2000 to 2020.
Ron
Reply to Richard Greene
September 4, 2022 9:28 am
Solar energy IS the cause of all warming if not from geothermal sources.
It’s just a matter of debate how and to what degree the energy delays it’s emission into space.
Smart Rock
Reply to Rud Istvan
September 2, 2022 2:39 pm
Rud – I think you spoke too soon. What Willis’ fig 2 shows is no increase in greenhouse efficiency. His fig. 1 gives surface temperature from CERES, which shows ~0.5°C increase over the period (by my patented “eyeball” method of curve fitting).
As we see more studies coming in, it does appear that the (as yet unexplained) decrease in cloud cover is the culprit behind recent warming.
Thanks to Willis for another neat piece of work, explained as always with exemplary clarity. I’ve been complimented over the years for the clarity and readability of my technical writing, but Willis’ clarity and readability are Olympic gold medal standard.
Rud Istvan
Reply to Smart Rock
September 2, 2022 5:15 pm
Nope. The issue is, IF the GHE is potentiating because of increased CO2, then his ‘efficacy’ should increase. It didn’t. I just supplied a separate ‘proof’.
Barry Malcolm
Reply to Rud Istvan
September 2, 2022 6:07 pm
Potentiating? Wtf does that mean, EXACTLY?
John Witten
September 2, 2022 11:17 am
Willis,
As always, this is a very interesting explanation. As a interested non-expert, I always learn something reading your articles. If you could please indulge my technical ignorance, how is the CERES data for incoming solar radiation adjusted for albedo reflection in Figure No. 3 actually obtained? I assume that incoming solar is measured by satellite. Is it also possible to measure short-wave reflection by satellite measurement? Or, is this a calculated number?
Just wondering. Your observations always seem to make so much common sense that is hard to believe that no one else seems to be making the same observations.
Rud Istvan
Reply to John Witten
September 2, 2022 12:02 pm
Not Willis, but the answer is simple. Inbound solar is measured looking ‘up’ (away from Earth). Albedo reflected outbound solar is measured looking ‘down’ toward Earth. CERES simply uses two opposed sensors.
Gary R Wescom
September 2, 2022 11:17 am
What I gather then is that energy received at the surface to produce 400 watts upwelling IR is a combination of 240 watts from solar radiation and 160 watts downwelling IR. Did I miss something?
leitmotif
Reply to Gary R Wescom
September 2, 2022 2:45 pm
But you cannot add radiations. It is sophistry.
240W/m^2 equivalent to 255K or -18C
160W/m^2 equivalent to 230K or – 43C
400W/m^2 equivalent to 290K or +17C
There is no way the first 2 could produce the third.
Macha
September 2, 2022 5:00 pm
Yep. The CO2 is already at the air temperature from energy absorbed on the way down. On the way up, its transparent..in/out without surface-air temperature change.
leitmotif
September 3, 2022 2:33 pm
???
Gary R Wescom
September 2, 2022 7:11 pm
Leitmotif – It was a simple watt in vs watts out question. Willis answered satisfactorily.
leitmotif
Reply to Gary R Wescom
September 3, 2022 2:28 pm
No such thing as watts in v watts out question. Power is joules/sec i.e. energy/sec.
Willis mixing up energy with flux.
It’s a common ploy or mistake.
Jim
September 4, 2022 7:19 pm
Power is not energy ; energy is power over a given time such as a kilowatt hour.
AndyHce
September 2, 2022 9:16 pm
Radiation is something real. Rain is something real, they are not the same but both exist in some quantity. Quantities of the same thing add quite properly.
leitmotif
September 3, 2022 2:32 pm
So two power jets of water at the same pressure will remove grime as effectively as one jet with twice that individual pressure?
Never worked for me.
September 3, 2022 6:59 am
But you cannot add radiations. It is sophistry
========
If you cant add radiation how do you calculate an average?
leitmotif
September 3, 2022 2:29 pm
By using sophistry.
September 3, 2022 5:41 pm
Of course you can, take one light source (240W/m^2) and another (160W/m^2) and shine them on a target, total incident on the target is 400W/m^2.
mcswell
September 6, 2022 6:59 am
Of course you can add watts (or in this case, watts/ square meter, assuming the place they fall on are the same). What you can’t add is temperatures, which is (are) a derived quantity.
Richard Feynman had a delightful take-down of the New Math (back in the 1960s). One of the word problems asked students to add the temperatures of two stars (one of them allegedly green, but that was a different issue). You can’t do that, except to derive an average.
David Dibbell
September 2, 2022 11:22 am
Interesting and insightful analysis, Willis. Thank you.
A small point. You say, “Note that the standard deviation of the multiplier is 0.004 W/m2.”
If I understand correctly, your multiplier is instead a dimensionless ratio.
DMacKenzie
Reply to Willis Eschenbach
September 3, 2022 10:05 am
Willis, IR through the IR window, usually claimed to average 40W plotted against incoming solar like your Fig 3 might show some interesting phase shift. This assuming it takes approximately the same amount of time for surface warming to produce enough cloud cover to cool the surface somewhat later and somewhere else….
Monckton of Brenchley
September 2, 2022 11:27 am
This piece by Willis is even more fascinating than ever. A clear, brilliant and compelling analysis using data and not fashionable speculation.
Ian Magness
Reply to Monckton of Brenchley
September 2, 2022 11:46 am
Seconded.
Well done again Willis.
arjan duiker
Reply to Monckton of Brenchley
September 2, 2022 2:03 pm
Dear Christopher Monckton, for God’s sake, what else is needed to convince at least some politicians…? As you’ve said many times before, it’s game over. Can this great analysis of Mr. Eschenbach finally be the one that pulls out the plug?
AndyHce
Reply to arjan duiker
September 2, 2022 9:21 pm
Not a chance.It is somewhat like a US senator once said, during an interview, when writing to your representative was being energetically promoted. ‘We in Washington know why we are there and what we are doing. Yes, sometimes some of us have to pretend to care what the voters say in order to be sure of reelection, but we know what we are really about and that is what we are actually going to do.
leitmotif
Reply to Monckton of Brenchley
September 2, 2022 3:42 pm
Great praise from one lukewarmist to another, Brench.
And then a reciprocation from the other. Whoooppeee!
Tell me how the DLR or back radiation works, Brench, with empirical evidence (not Feldman et al, 2015) and without reference to Tyndall 1861
Tell me why you believe what warmists believe about the magical warming properties of CO2 but only less so.
And don’t use highfalutin scientific language to describe, say, molecular vibrational modes because I might know more than you think I do, like I know a CO2 molecule does not have a dipole moment.
(You are now in MODERATION) SUNMOD
Last edited 1 month ago by Sunsettommy
RickWill
September 2, 2022 4:20 pm
And you have to love Figure 4 where he compares surface ULR with surface temperature. If he took any notice of what you have previously stated, he would realise this proves that it is an “inferred power flux” based entity on the temperature.
Only climate phiisics has “cool energy” that cannot warm anything.
Rud Istvan
September 2, 2022 5:40 pm
Well actually, folks who know basic physics (including the laws of thermodynamics) know that a colder troposphere IR cannot ever warm a warmer surface IR. There is a heat exchange. But it ALWAYS goes net hot to net cold. Confusing intermediate states does not reflect well.
RickWill
Reply to Rud Istvan
September 2, 2022 6:41 pm
But it ALWAYS goes net hot to net cold. Confusing intermediate states does not reflect well.
This indicates you do not understand electro-magnetic radiation. In this field, energy only flows in one direction at any point in time and space. There is no “net”
Mishchenko has a proof of that reality and makes reference here:
https://www.giss.nasa.gov/staff/mmishchenko/publications/2013_AIP_Conf_Proc_1531_11.pdf
Unfortunately, none of the instruments that have ever been used in the disciplines of atmospheric radiation and remote sensing can, strictly speaking, be considered a Poynting meter. Instead, it was demonstrated in [31] that tra- ditional instruments called well-collimated radiometers (WCRs) operate as wavefront angular filters rather than en- ergy-propagation angular filters.
Even more fundamentally, the local instantaneous Poynting vector S(r,t) is a monodirectional vector. This means that even if one assumes that S(r,t) describes the direction and rate of local electromagnetic energy flow, then at any moment this flow occurs in only one direction.
Note that the link is from NASA. GISS employed Mishchenko to achieve an understanding of the real physics of the atmosphere. In essence to get away from climate phiisics that pervade the club.
Anyone using equations to represent physical phenomena should have an appreciation of the assumptions implicit in those equations and where the assumptions fail. The S-B equation is an approximation and is useful in many applications but not in defining heat transport within the Earth’s atmosphere.
Last edited 1 month ago by RickWill
leitmotif
September 3, 2022 2:54 pm
This indicates you do not understand electro-magnetic radiation. In this field, energy only flows in one direction at any point in time and space. There is no “net”
At last, someone who says how it is.
Kudos to RickWill.
The S-B equation is an approximation and is useful in many applications but not in defining heat transport within the Earth’s atmosphere.
Double Kudos.
One more and you get to keep the trophy, RickWill.
September 3, 2022 5:56 pm
This indicates you do not understand electro-magnetic radiation. In this field, energy only flows in one direction at any point in time and space. There is no “net””
It’s you who doesn’t understand e-m radiation.
Consider a satellite orbiting the earth passing between the Earth and the Moon. The satellite measures light coming from the Moon and light coming from the Earth, what do you think happens if the satellite isn’t there?
Rud Istvan
September 2, 2022 5:33 pm
Sorry, your GHE ignorance knows no bounds.
Water has a dipole moment. That is why microwave ovens are so effective.
True, CO2 does not. Thatnis whynit is irrelevant to microwave ovens.
BUT, it does have an elastic linear (stretch/shrink) bond that stores and then re-emits its absorbed IR. My goodness, at least learn basic physics before posting here.
Barry Malcolm
Reply to Rud Istvan
September 2, 2022 6:19 pm
CO2 “bond” that stores” and then “re-emits”? Like it absorbs and then emits IR? If true, what’s with the comment about basic physics? How about simple understandable concepts?
Editor
Reply to Barry Malcolm
September 2, 2022 8:41 pm
I think 135,000 miles per second is the minimum speed of the IR wave…..
AndyHce
Reply to Barry Malcolm
September 2, 2022 9:30 pm
It is a mysterious universe.
Reply to Barry Malcolm
September 3, 2022 4:38 am
Have a look at the physics behind a CO2 laser:
https://en.wikipedia.org/wiki/Carbon-dioxide_laser
Reply to Rud Istvan
September 3, 2022 12:14 pm
In the case of the 15𝜇m radiation it’s a bond that bends and then re-emits the absorbed energy.
leitmotif
Reply to Rud Istvan
September 3, 2022 3:08 pm
Sorry, your GHE ignorance knows no bounds.
Water has a dipole moment. That is why microwave ovens are so effective.
True, CO2 does not.
So I was correct that CO2 does not have a dipole moment?
So your “GHE ignorance knows no bounds” is not correct?
BUT, it does have an elastic linear (stretch/shrink) bond that stores and then re-emits its absorbed IR. My goodness, at least learn basic physics before posting here.
So when I said to Brench about “ molecular vibrational modes” you did not think that was how I perceived how a a CO2 molecule absorbs and re-emits a photon. This is a reduced discussion on a previous discussion with Brench.
You are still blowing smoke from that orifice, Rud.
September 2, 2022 11:34 am
Willis, you say “We know the earth is warmer than expected. Nobody has ever come up with an explanation for that except the greenhouse effect.”
I am pretty sure there are other hypotheses, making your unnecessary strong claim wrong. This has no effect on your argument.
RickWill
Reply to Willis Eschenbach
September 2, 2022 4:33 pm
The energy balance is controlled by two physical processes involving ice formation. Sea ice insulates ocean surface. Atmospheric ice, resulting from deep convection, sets an upper limit on open ocean surfaces that can not sustain a temperature higher than 30C by limiting surface sunlight.
The energy balance is controlled by temperature sensitive processes with powerful feedback.
The ability of the atmosphere to form an LFC is the only reason Earth is not a snowball. The concept of “greenhouse effect” controlling earth’s surface temperature is ridiculous. You have just proven that it has no effect on the temperature.
AndyHce
Reply to Willis Eschenbach
September 2, 2022 9:43 pm
I can’t say how many points it covers but absorbed IR energy can and is (claimed) to be transferred through kinetic interactions. In fact, according to William Happer here,
http://www.sealevel.info/Happer_UNC_2014-09-08/Another_question.html
in the lower atmosphere, 99.9999999% of the time surface IR absorbed by CO2 and H2O is transferred to other atmospheric molecules through kinetic interactions before it can be re-emitted.
This would seem to me to be an explanation of how warming occurs that doesn’t seem to fit the general greenhouse meme. It does not require any back radiation to explain the surface being warmer than just from absorbed solar. Of course it does involve GH gases.
Robert W Turner
September 3, 2022 6:33 am
And of course we know energy is never transferred from the 98% of the atmospheric molecules to IR active molecules which would increase emissivity and cooling at the ToA. It’s a one-way energy transfer in climastrology.
Robert W Turner
Reply to Willis Eschenbach
September 3, 2022 6:46 am
If everything were a fixed molecule in space and time that had no kinetic energy and the only mode of energy transfer were radiative then you’d be correct. The atmosphere transfers energy back to the surface in other ways than radiative emission of IR, in fact, it is by far the least significant and is why you keep finding no signal from the ever increasing CO2.
Reply to David Wojick
September 2, 2022 1:07 pm
No one including the IPCC wants to talk about natural causes of climate change, so climate change gets blamed on AGW. What’s left? It’s called starting with a conclusion junk science.
Richard M
Reply to David Wojick
September 2, 2022 6:56 pm
It is too strong of a claim. It gets the big picture wrong. The surface is not warmer because of energy trapped high in the atmosphere and radiated downward. It is warmed from energy absorbed low in the atmosphere, radiated upward and shared based on atmospheric density. You need radiating gases to effect the warming with either explanation, the correct one has more or less a fixed warming effect. It is why Willis couldn’t find any increase and neither did Miskolczi in his 2010 paper looking at 70 years of NOAA data.
LARRY K SIDERS
September 2, 2022 11:47 am
Unstated (unless I missed it) was that the decreasing “Incoming” radiation was likely from a decreasing trend in low level clouds. Where else could the albedo vary as rapidly? Real Question…I dont know of any other “delta-albedo” sources to chose from (maybe albedo effects from changes in vegetation?). It was obviously NOT from any solar output change (solar radiation is not that variable).
I believe I’ve seen several references citing Satellite Data Records showing a several Decades long slow decline in Low Cloud Cover…enough to account for 100% of the Average Surface Temperature Increase on record.
This article and the actual Cloud Albedo reductions appear to “fit” quite well.
Reducing Cloudiness would also seem to counter the KEY 3X’s Hydrological Amplification (“Key” to produce Catastrophic Climate Change from a paltry DIRECT CO2 Effect) of the very small DIRECT 0.3° C to 0.7° C CO2 “Doubling” Temperature Effect.
Lower Cloudiness would require a Lower Global Humidity level…not the *INCREASE* in Humidity required to CREATE a 300% Climate Crisis Multiplier Effect.
Last edited 1 month ago by LARRY K SIDERS
Tom.1
September 2, 2022 11:52 am
Willis: For the time period in question, how much incoming energy has been accumulated in the total atmosphere expressed in terms of w/m2 and based on the amount of observed warming?
Dan Hughes
Reply to Willis Eschenbach
September 2, 2022 1:46 pm
“A watt is a joule second …”
A Watt is a Joule/sec: Joule per second.
Tom.1
Reply to Willis Eschenbach
September 2, 2022 2:57 pm
My thinking was (is) that the energy imbalance at the earth’s surface over the long haul to generate the amount of warming is trivially small. I could be wrong, so check me on this (anybody). The atmosphere has warmed by 5.3E21 joules in roughly the past 50 years (very approximate, but in the ballpark). This amounts to 5.3E21/(50*365*86400) joules/sec = 3.36E12 J/S = 3.36E12 watts. The surface area of the earth is 5.1E14M2, so the long term watts/sq meter = 3.36E12/5.1E14 = .0066 watts/sq meter. Can that be right?
Anders Rasmusson
September 2, 2022 12:01 pm
Willis Eschenbach, so nice, thank you.
Different “greenhouse multiplier” at the 0°, 30°, 60°, 90° latitudes?
Kind regards
Anders Rasmusson
Barrie
September 2, 2022 12:06 pm
A minor niggle: “earth is much warmer than the moon, which receives the same amount of solar energy.” should you not add per metre squared?
Barrie
September 2, 2022 12:08 pm
I meant to add this to my niggle comment: I greatly admire your posts on this issue, those you have done on emergent phenomena in particular. Thanks
JeffC
September 2, 2022 12:09 pm
Barry Malcolm
September 2, 2022 6:31 pm
Not if we can’t read it without subscribing so you can get a pittance of a commission, baksheesh!
Mr.
September 2, 2022 6:51 pm
Not to Willis’ contribution to educating us atmospheric know-nothings.
Otherwise – did you have a point?
JeffC
September 3, 2022 12:29 am
Yes
Prjindigo
September 2, 2022 12:14 pm
Gravity generates heat.
The satellite is unable to discern between IR sourced up-welling IR, other spectrum sourced up-welling IR, solar and cosmic particulate dynamic inductive sourced up-welling IR, non-solar sourced energy converted to up-welling IR and most pointedly human generated or human and life creation sourced up-welling IR.
There is a not insignificant amount of heat generated by simply the compression of air through moving machinery or stationary structures inhibiting wind and water motions.
If the IPCC thinks modeling the clouds is too hard of work, monitoring the effects of human activity is probably not even on their minds. That’s why they’re trying to schill a 55% fudge-factor linear progression instead of doing any science at all. What they are doing is presenting unreliable polling information, ALL anecdotal in form, as scientific input… and it’s bullshit all around.
Last edited 1 month ago by Prjindigo
September 2, 2022 1:53 pm
Gravity generated heat at the moment that the atmosphere was formed. One day later that heat ws gone to space. No new heat can be generated from a static pressure.
If you inflate a tire, after a few hours it is back to ambient temperature and still under pressure. No new heat is generated at all.
BTW, Willis explicitly asked not to start a discussion on alternate “explanations” like gravity, which were thoroughly debunked years ago:
‘I proved that hypothesis incorrect in my post “A Matter Of Some Gravity“’
Macha
Reply to Ferdinand Engelbeen
September 2, 2022 5:06 pm
Why reply? Gravity is not static, it’s continual work done, else air lost to space. No tyre holding bike tyre air in. Just like greenhouse stops convection.
Last edited 1 month ago by Macha
Ed Bo
September 2, 2022 7:01 pm
Macha:
In physics, “work” has a very specific definition. The work (energy) done on an object is the mechanical force on that object multiplied by the distance the object is moved by the force. (Technically, it’s force integrated over distance, but we’ll keep it simple here.)
Taking the derivative, the rate of work (power) at any time is the force multiplied by the velocity of the object it causes.
Taking the atmosphere as a whole, the distance of movement caused by gravitational force is zero, and the velocity is zero. Any down movements must be balanced by up movements somewhere else.
So the “continual work” done by gravity on the atmosphere is zero. This is basic high school physics.
Robert W Turner
Reply to Ed Bo
September 3, 2022 6:52 am
Lol yes, high school physics of work equal to zero if something is moved from point A to B and then back to A, magic.
Ed Bo
Reply to Robert W Turner
September 3, 2022 8:21 am
When X does work on Y to move it from point A to point B, then Y does the same amount of work on X to move it back to point A the resulting transfer between X and Y is zero.
Yes, basic high school physics (for those who were paying attention…)
Robert W Turner
Reply to Ferdinand Engelbeen
September 3, 2022 6:50 am
I love the bike tire analogy where the pumping of the tire (the sun) stops and is supposed to disprove Kinetic Theory of Gases.
Ed Bo
Reply to Robert W Turner
September 3, 2022 8:27 am
How exactly does his argument conflict with the Kinetic Theory of Gases?
Mr.
September 2, 2022 6:55 pm
Cliff Mass often cites the heating effect on air blowing down from the peaks of the Rocky Mountains westwards toward the Pacific North Western coastal regions.
AndyHce
September 2, 2022 10:02 pm
which causes air elsewhere to raise, lowering it temperature. However, neither mass of air gains or loses energy.
AndyHce
September 2, 2022 9:56 pm
The claim is that the amount of energy used by all human activity in one year is provided by the sun every couple hours.
ResourceGuy
September 2, 2022 12:25 pm
Okay, we have very small movement in solar inputs up and down as measured in both years and individual solar cycles. Now if the oceans are capacitor storage systems, are groups of weak solar cycles and groups of higher solar cycles of say 50-70 years each still considered insignificant? See Leif charts for examples
Per
September 2, 2022 12:28 pm
Great article – thank you. The clouds….yes indeed, by blocking/reflecting 100 W/m2 are they not indeed the dominant factor in all these analyses? I’m anxious to know why cosmic galactic rays and the theory by Svensmark gets so little traction/attention. What is the problem with that theory?
John Tillman
September 2, 2022 12:44 pm
It doesn’t fit the desired narrative to which everyone should stick, or else.
AndyHce
September 2, 2022 10:03 pm
Supposedly, best estimates calculations say the effect is real by insignificant in magnitude.
September 2, 2022 12:49 pm
” the main cause of the warming is clearly the increase in the amount of solar energy after reflections from the clouds and the surface.”
Jumping to a conclusion.
Solar irradiance change was insufficient to explain warming from 2000 to 2020, which is over +0.5 degrees C. in the UAH global average temperature record
UAH Global Temperature Update for August, 2022: +0.28 deg. C « Roy Spencer, PhD (drroyspencer.com)
Solar irradiance – Wikipedia
Last edited 1 month ago by Richard Greene
Reply to Richard Greene
September 2, 2022 2:53 pm
Richard, you didn’t understand the point Willis was making: with a constant amount of greenhouse gases, the warming should be +0.5 K from 2000 to 2020 only caused by the solar input, as that is the fortifying factor by GHGs.
But the GHGs increased a lot in that period, According to the climate models, there should be some 0.5 K increase from GHGs alone, that is with a constant input from the sun.
With both increasing, there should be 1 K total increase in temperature. But there is only 0.5 K warming certainly caused by the solar input (as the fortifying factor didn’t change at all), thus where is the warming caused by GHGs?
Reply to Ferdinand Engelbeen
September 2, 2022 4:21 pm
There are many climate change variables.
No one knows the exact effect of each one.
You have mentioned only two of them. The sum of all the variables is obviously less warming than thought to come from those two variables alone. What does that prove? Nothing !
The following variables are likely to influence Earth’s climate:
1) Earth’s orbital and
orientation variations
2) Changes in ocean circulation
Including ENSO and others
3) Solar activity and irradiance,
including clouds, volcanic and manmade aerosols, plus possible effects of cosmic rays and extraterrestrial dust
4) Greenhouse gas emissions
5) Land use changes
(cities growing, logging, crop irrigation, etc.)
6) Unknown causes of variations of a
complex, non-linear system
7) Unpredictable natural and
8) Climate measurement errors
(unintentional or deliberate)
9) Interactions and feedbacks,
involving two or more variables.
Last edited 1 month ago by Richard Greene
Barry Malcolm
Reply to Richard Greene
September 2, 2022 6:42 pm
“There are many climate change variables. No one knows the exact effect of each one”. Frick, thanks Tipster!
Richard M
Reply to Richard Greene
September 2, 2022 7:14 pm
As I mentioned above, the Dubal/Vahrenholt 2021 paper can shed more light on the situation. They found the cloud changes correlated well with natural ocean changes. The big change in the PDO during 2014 seems a likely candidate.
Reply to Richard Greene
September 2, 2022 10:04 pm
I believe Willie E. is the best writer on this website
So I am biased in his favor.
If Willie E. is correct with this article, then he has, in one article, refuted 20 years of consensus climate science. This article would be worthy of a Nobel Prize.
Do I believe that actually happened?
Sorry, I do not believe that happened and my instincts are good on the subject. For Willie E. to be right, virtually every other climate scientist has to be wrong. That seems very unlikley.
If the article was on the future climate, that could be true.
Climate scientists have so many always wrong predictions of doom. Easy to refute that.
But the article is about the past climate — the past 20 years –and I very much doubt that Willie E. has discovered what thousands of scientists around the world overlooked. I’m not buying the conclusion, no matter how much I enjoy reading Willie E. articles here — and I do read every one.
Thank you for deleting most liefmotif attack posts. I normally oppose censorship. But when I recommend any article here to friends, they read all the comments, and then can get a false impression of WUWT readers from any drive by character attacks. Not that I loved this article, but I’ve tried to be civil.
Last edited 1 month ago by Richard Greene
Robert W Turner
Reply to Willis Eschenbach
September 3, 2022 7:14 am
It does not mean that the greenhouse effect doesn’t exist. It does mean, however, that whatever gains in efficiency that occurred due to increases in CO2 have been counteracted by other climate phenomena.”
Well it’s important to take note of this; when most people refute the back radiation hypothesis, they are not saying that certain gases are not IR active, or that they are not incident with outgoing IR, and in turn emit the same frequencies of light. They are saying that this phenomenon is not close to the primary warming mechanism of the atmosphere. (with obvious exceptions to “most people”)
For instance, one can argue against eugenics while still believing in the underlying biological science that the sophistry is built upon.
Richard M
Reply to Willis Eschenbach
September 3, 2022 8:06 am
It does not mean that the greenhouse effect doesn’t exist
The original greenhouse effect does exist. However, it is saturated which has been known for many decades. The saturation occurs very low in the atmosphere. For CO2 the saturation occurs well below 200 ppm and now less than 10 meters..
The problem comes with the enhanced greenhouse effect. This is based on increases in DWR and a supposed increase in the effective emission altitude.
We now have several different ways of looking at the data (Miskolczi 2010, Dubal/Vahrenholt 2021 and now Willis 2022) that all come the same conclusion. There’s been no increase in the greenhouse effect.
I think Willis should publish his result. It is very important and brilliant in its simplicity. It also confirms the previous work of DubalVahrenholt and Miskolczi.
Chris Hanley
Reply to Richard Greene
September 2, 2022 4:22 pm
Last edited 1 month ago by Chris Hanley
Reply to Chris Hanley
September 2, 2022 10:06 pm
The year to year (2000 to 2020) change is about +0.5 degrees
You have applied a linear trend to non-linear data
Chris Hanley
Reply to Richard Greene
September 2, 2022 5:03 pm
Barry Malcolm
Reply to Richard Greene
September 2, 2022 6:39 pm
And… after reflections of lesser or greater degrees whether clouds or the surface? You were a little dodgy there Richard.
Reply to Barry Malcolm
September 2, 2022 10:09 pm
I’m guessing “not”. If I had the answers to why the global average temperature rose from 2000 to 2020, maybe I’d get the Nobel Prize. My “answer” is no one knows exactly why GAST rose from 2000 to 2020, and Willie E. is no exception.
Bob
September 2, 2022 1:15 pm
Very nice Willis.
dk_
September 2, 2022 1:16 pm
Good post. Thanks.
leitmotif
September 2, 2022 1:34 pm
Now, if you don’t think the “greenhouse effect” exists, this is NOT the thread for you.
So, whether the greenhouse effect exists is just an opinion? It’s whether one thinks it exists or doesn’t think it exists is the question? It’s nothing to do with empirical evidence, then?
There are lots of places to make that argument. This isn’t one of them.
Oh, scary. Stay away if you are sceptical of the existence of the greenhouse effect? Doesn’t that contradict the raison d’etre of WUWT?
We know the earth is warmer than expected.
“warmer than expected”? What does that even mean?
Nobody has ever come up with an explanation for that except the greenhouse effect.
What? An explanation of “warmer than expected”?
Nobody has ever come up with an explanation for the existence of the universe except for the existence of a supreme being. 6 out of every 7 people on this planet practise a religion and believe in god.
If you are unclear about how the greenhouse effect works, the physical basis of it has nothing to do with CO2 or with the atmosphere at all. I explain this in my posts “People Living In Glass Planets“, and “The Steel Greenhouse“.
Except “The Steel Greenhouse” was thoroughly and laughingly debunked by astrophysicist Joseph Postma many years ago and accused the author of having no scientific training.
To reiterate: PLEASE do not post your opinions here on why the greenhouse effect isn’t real, or why there’s no such thing as downwelling radiation, or that scientists don’t understand the instruments that measure IR. The web is a very big universe. Somewhere out there is the perfect place to make those arguments.
This is not that place.
So do not debate on the greenhouse effect, do not debate on whether downwelling radiation can raise the temperature of the planet surface and do not argue about what pyrgeometers actually measure?
Doesn’t leave much to debate, does it Willis?
leitmotif
Reply to Willis Eschenbach
September 2, 2022 1:52 pm
Pass.
They were perfectly good questions, Willis.
If you don’t have the answers just say so.
No need to take this moral high ground of a a messiah who has been questioned on his faith.
Tap the sign? I only, and have always, asked for evidence on your assertions about the surface warming effects of DLR or back radiation.
Come on, Willis, loosen your corsets and indulge me and, I’m sure, only a minority of WUWT posters. I and they can’t really do you any damage as you have the whole lukewarmist community behind you.
September 2, 2022 2:30 pm
Just one reply and then I hope this nonsense will be stopped by the moderators:
DLR was not only measured with pyrgeometers
which are questioned by some, but also line by line as full spectrum:
https://escholarship.org/content/qt3428v1r6/qt3428v1r6.pdf Fig. 1.
In both cases, that amounts to around 300 W/m2.
Measured, really measured.
As a black body absorbs all wavelengths and a gray body like the earth still absorbs almost all of the DLR, the conservation of energy requires that the incoming energy at the earth’s surface is the sum of incoming sunlight and incoming DLR, which is (much) higher than of the incoming sunlight alone.
That makes that the surface must warm up to restore the balance…
Alexy Scherbakoff
Reply to Ferdinand Engelbeen
September 2, 2022 11:24 pm
I saw values in milliwatts and great mention of models, but I drifted off sometime after.
Reply to Alexy Scherbakoff
September 3, 2022 4:46 am
The A side of Fig. 1 is the real DLR as measured by spectral analyses, the B side is the difference with what the radiation model expected. Not that bad…
Robert W Turner
Reply to Ferdinand Engelbeen
September 3, 2022 7:53 am
Estimated atmospheric LWR is sensitive to near surface temperature. So the “measured” total LWR of the atmosphere is a factor of the temperature that the instrument is in. Too funny. LWR is not measured directly at the surface, it can’t be, it is inferred with circular reasoning that roughly converts near surface temperature to a mathematical abstraction based on back radiation hypothesis.
Gerald Machnee
September 2, 2022 2:38 pm
You missed the point.
Discuss it on another post. Maybe start one.
stinkerp
September 2, 2022 2:40 pm
Hey lazy guy, the internet has a lot of information YOU can look up YOURSELF to answer your questions about downward longwave radiation. But you didn’t write here to get an answer. You wrote to pester with your favorite argument because you either can’t be bothered to read what the rest of the climate science community has already written about DLR and warming or you simply won’t accept it. If you aren’t convinced by the robust body of science and evidence to support it, Willis certainly won’t be able to persuade you either. But that’s not why you asked, is it? People who ask questions to needle and provoke an argument rather than to learn are…annoying. Once we realize what the game is that the provoker is playing, the best response is to walk away, because arguing is a waste of time. And I just wasted several minutes.
Last edited 1 month ago by stinkerp
leitmotif
September 2, 2022 2:58 pm
Haha!
Now I get a bunch of replies from a guy who quotes arch warmist, Science of Doom and a reference to the totally debunked Feldman et al (2015) paper and a couple from Willis groupies who basically tell me to back off.
Nice try, guys but I’m here till Mr Watts bans me for questioning the science which will probably be quite soon.
(You are now in Moderation because you are chronically off topic threadjacking and being very impolite) SUNMOD
Last edited 1 month ago by Sunsettommy
leitmotif
Reply to Willis Eschenbach
September 2, 2022 4:34 pm
Why do you not just answer the questions I posed at the start of this thread, Willis?
[Personal attack snipped – w.]
Last edited 1 month ago by Willis Eschenbach
Editor
Reply to Willis Eschenbach
September 2, 2022 8:51 pm
He has been put in MODERATION.
Clyde Spencer
September 3, 2022 6:50 pm
There is an old saying that people are often their own worst enemies. Why do you insist on continuing to poke the wasp nest when it is obvious that the wasps don’t like it? You are not only lacking in manners, but also common sense. If you are banned, you lose your opportunity to comment in the future.
stinkerp
September 2, 2022 2:32 pm
I’ll tap the minus. You can make your argument all you want even though it was noted that this isn’t the place for that argument. Fyi, it makes you look a.) truculent, 2.) arrogant, like your opinion is SO important that it must be expressed at all costs and in every forum regardless of its relevance, and e.) not very bright since you don’t appear to grasp the actual points being made by the author so you resort to your favorite argument instead. But if you want to be a zealot, by all means…
Last edited 1 month ago by stinkerp
leitmotif
September 2, 2022 4:14 pm
(SNIPPED)
(No more thread jacking attempts stick with the current topic) SUNMOD
Last edited 1 month ago by Sunsettommy
September 2, 2022 4:23 pm
The article provided data about the greenhouse effect.
What problem do you have with the data presented?
Mark BLR
September 3, 2022 3:51 am
So, whether the greenhouse effect exists is just an opinion?
Saying “the GHG effect does not exist” isn’t “just” an unsubstantiated opinion (/ conjecture), it is one directly contradicted by several sets of empirical data (including the CERES satellite data, a subset of which is used in the ATL article).
Stay away if you are sceptical of the existence of the greenhouse effect?
After reflection I concluded (possibly incorrectly, as is always the case …) that most of my posts on CiF that were “Removed by a moderator” ended up that way due to my infringing the Guardian‘s “please stay on-topic” clause of their Community Guidelines.
I didn’t always agree with their “strictness” when assessing what was (not) “on topic”, but I accepted those decisions without rancour.
It’s what is called in some circles the “your house, your rules” level of politeness.
– – – – –
Here at WUWT a few of the “Policy” elements are :
Respect is given to those with manners, those without manners that insult others or begin starting flame wars may find their posts deleted.
Some off topic comments may get deleted, don’t take it personally, it happens. Commenters that routinely lead threads astray in areas that are not relevant or are of personal interest only to them may find these posts deleted.
Trolls, flame-bait, personal attacks, thread-jacking, sockpuppetry, name-calling such as “denialist,” “denier,” and other detritus that add nothing to further the discussion may get deleted; …
This specific post included the specific “warning” that :
To reiterate: PLEASE do not post your opinions here on why the greenhouse effect isn’t real, or why there’s no such thing as downwelling radiation, or that scientists don’t understand the instruments that measure IR. The web is a very big universe. Somewhere out there is the perfect place to make those arguments.
This is not that place.
This is the WUWT “house”, so their “rules” about what is, and is not, “on topic” apply.
Gratuitous insults are unwelcome in any “house”.
Why are these such difficult “life lessons” for you to learn ?
Last edited 1 month ago by Mark BLR
paul courtney
Reply to Mark BLR
September 3, 2022 6:40 am
Mr. BLR: It doesn’t help that Mr. motif quotes Mr. E, only to distort the quote in the very next line. Over and over. The only effect is that Mr. motif comes across as a whiny b., and is probably lucky the mod spared him further embarrassment. If he does have comment in other posts, this will be hard to forget.
Mark BLR
Reply to paul courtney
September 3, 2022 9:38 am
If he does have comment in other posts, this will be hard to forget.
From my exchange with “Mr. motif” under Willis(s previous article (direct link).
I wonder why you have never been banned as I have been banned from the Guardian 5 times.
I admit I have been outspoken on 2 or 3 of those identities but sometimes I have been banned just because I disagreed.
I would hope (probably in vain, but still …) that “Mr. motif” would consider more carefully just why their “outspokenness” attracts the attention of moderators on sites as widely spaced on the “climate debate spectrum” as WUWT and the Graun.
Last edited 1 month ago by Mark BLR
Editor
Reply to Mark BLR
September 3, 2022 11:28 am
He can still post here but it requires moderator approval to make them appear on the board to stop his trolling and immature attacks on people.
He is given a chance to improve his behavior in time can be removed from moderation.
Last edited 1 month ago by Sunsettommy
CD in Wisconsin
September 2, 2022 1:35 pm
Willis,
I am not a scientist, but this brings up a question in my mind.
Do your conclusions here reflect and support the study done a while back by Dr. Happer that greenhouse gases in the atmosphere are largely saturated with IR and therefore incapable of making any further meaningful contribution to atmospheric temperatures?
Paul Johnson
Reply to CD in Wisconsin
September 2, 2022 2:15 pm
On a related topic, does this imply an Equilibrium Climate Sensitivity much lower than most models reflect?
leitmotif
Reply to Paul Johnson
September 2, 2022 3:54 pm
ECS is not indistinguishable from zero.
Mike
September 2, 2022 6:14 pm
Distinguishable. ECS is theoretical and current attempts at quantifying it are next to meaningless.
Last edited 1 month ago by Mike
Reply to Paul Johnson
September 2, 2022 9:49 pm
The models reflect whatever ECR is needed for a scary global warming prediction. It does not have to match observations because accurate predictions are not a goal.
leitmotif
Reply to CD in Wisconsin
September 2, 2022 3:52 pm
[Snipped: contained a personal attack only, not one scrap of science. w.]
Last edited 1 month ago by Willis Eschenbach
Reply to CD in Wisconsin
September 3, 2022 10:57 am
Do you have a reference to that study by Will?
CD in Wisconsin
September 3, 2022 12:27 pm
I should have posted it in my comment above. Apologies for not doing so.
https://www.heartland.org/news-opinion/news/study-suggests-no-more-co2-warming
“Happer and van Wijngaarden’s central conclusion is this:
“For the most abundant greenhouse gases, H2O and CO2, the saturation effects are extreme, with per-molecule forcing powers suppressed by four orders of magnitude at standard concentrations…””
My point here was that the conclusions drawn by the Happer/Wijngaarden paper and what Willis is saying here seem to complement each other. Namely, that CO2’s ability to affect the climate at 420 ppm and up is low/minimal.
Reply to CD in Wisconsin
September 3, 2022 6:52 pm
Thanks, Will’s paper states that a doubling of CO2 would result in ~3W/m^2 increase in forcing, and if other IR active gases are considered it could be about 5W/m^2. This is consistent with the conventional value so doesn’t support the Heartland Institute’s interpretation of the paper, fortunately the original preprint was linked.
Robert B
September 2, 2022 1:45 pm
“CERES TOA products are measured products, taking into account the solar radiation incident at the TOA. CERES surface products are a combination of CERES measured fluxes and atmospheric profiles (GEOS-5.4.1), NCEP SMOBA Ozone, MATCH aerosols and cloud cover properties derived from MODIS Collection 5 until March 2017”
I’m not convinced that it doesn’t say more about the methodology than physical reality.
mkelly
September 2, 2022 1:52 pm
WE says:”In other words, for every watt per square meter of solar input, we get ~ 1.7 watts per square meter of upwelling surface radiation.”
I am at a loss to understand what physical mechanism in nature multiplies energy. How does the dirt beneath my feet increase energy?
leitmotif
September 2, 2022 2:02 pm
It just gets more surreal every time Willis posts, mkelly.
Moritz Büsing
September 2, 2022 2:44 pm
It is not about multiplying energy. It is about the energy added to the system related to the energy in the sytem (or more precicely the flux)
There is an other example:
You may add 100W/m2 of heating in your insulated house in order to reach a temperature leading to 500W/m2 of heat radiaton from the inner walls.
leitmotif
Reply to Moritz Büsing
September 2, 2022 3:15 pm
mkelly refers to the energy of the dirt beneath his/her feet but he/she is really referring to the upwelling energy emitted by that dirt as in his/her quote “1.7 watts per square meter of upwelling surface radiation.”
Energy is measured in Joules and is a state property.
Power, as in upwelling energy, is measured in Watts which is Joules per second and is energy in motion.
mkelly wonders where the extra energy in the dirt beneath his/her feet comes from.
Don132
Reply to Moritz Büsing
September 3, 2022 4:24 am
If a 100 W/m2 heater warms up a room this is because the heater is continuously adding heat and the most of the heat isn’t allowed to leave.
Not so with the atmosphere. The heat is allowed to leave: balloon data tells us that very clearly.
leitmotif
Reply to Willis Eschenbach
September 2, 2022 4:38 pm
[SNIPPED—Personal attack only, science-free. w.]
Last edited 1 month ago by Willis Eschenbach
DonK
September 3, 2022 8:14 am
@mkelly. I’m glad that I’m not the only person a bit troubled/confused by the apparent violation of Conservation Of Energy in Willis’ numbers. I actually don’t think Willis methodology and conclusions are wrong. Neither do I think COE is violated. My guess is that we’ve got two similarly named quantities that you (and I) are confusing. FWIW Trenberth 2009 has downwelling radiation at 340,2 W/m2 and upwelling radiation at between 233.3 and 253.9 W/m2 — depending on which of a half dozen sources one chooses to believe. Those numbers seem likely to be not too far from what COE dictates. My plan is to go off and think about all this. Most likely the truth will eventually surface. … And I’ll learn something.
September 3, 2022 10:53 am
In order to maintain a stable atmospheric temperature only 1 W/m^2 of that upwelling surface radiation can leave the top of the atmosphere so 0.7 W/m^2 are recycled back to the planet/atmosphere.
September 2, 2022 1:56 pm
Very interesting figures Willis!
Looks like the negative feedback’s keep the earth’s temperature in narrow limits…
Richard M
Reply to Ferdinand Engelbeen
September 2, 2022 8:08 pm
I’d say the physics of the atmosphere limits the warming effect of GHGs. Well mixed GHGs have an almost constant warming effect.
john harmsworth
September 2, 2022 2:05 pm
I know there are a near infinite number of people who are smarter than I am on this site, and that’s before we even limit the discussion to the field of climate.
Would it not be possible to analyze data on nighttime cooling? If CO2 causes a lag ibn heat loos from the surface, it should show up as a higher morning temperature as CO2 accumulates and increases. That effect should show up at every level as one rises through the atmosphere.
Reply to john harmsworth
September 2, 2022 4:31 pm
Greenhouse gases mainly affect TMIN when people are sleeping, not TMAX in the afternoon. That has been true since the 1970s. Also, more effect in higher, colder N.H. latitudes in the coldest six months of the year. More CO2 has the most effect where there is little water vapor in competition. Think of warmer winter nights in Siberia as the global warming “poster child”, How is that a climate emergency? That sounds like good news to me.
Then think that the same warming pattern did not happen in the Southern Hemisphere since the 1970s= climate science is not settled. And there was no warming from 1940 to 197 as CO2 levels increased.
You don’t need science or scientists to know how it felt to live with global warming since 1975. We loved it here in SE Michigan and want more. A climate emergency would be if global warming stopped, and global cooling began. Those are the two trends — pick the one you like best and be happy if your favorite is global warming. The climate on our planet does not get much better than it is today. Not in the past 5,000 years. Celebrate the current climate ! Don’ fear the future climate.
Last edited 1 month ago by Richard Greene
Jim Davidson
September 2, 2022 2:07 pm
“ The Moon is much cooler than the Earth.” At mid day on the lunar equator the rocks reach a temperature of 130C. At night these rocks can radiate directly to apace and their temperature drops to -170C. The Moon is both colder and hotter than the Earth. The difference lies in the Eart’s two oceans: the ocean of water that covers 7/10ths of the Earth’s surface to an average depth of 4 kilometres: and the ocean of air which envelops the entire Earth to a depth of about 100 kilometres.
Macha
Reply to Willis Eschenbach
September 2, 2022 5:14 pm
Which shows that averaging temperatures says so little about living conditions..aka climate.
Rather have 20to30C than 0to50C.
September 2, 2022 2:20 pm
Ja. Ja. I told you. It is not the CO2. It is not the sun either as Tmax (global) is still going down.
Now I know there some who claim that the geothermal factor is only 90 milliW/m2.
I think something is wrong there. According to my old books T is going up 3K per km down. Come down in a goldmine here and soon you will have sweat pouring from your face.
That means I only need an internal shift of 1/3 = 334 meters of the inside of earth to get a raise of 1 degree Tmin in the NH.That is not much?
otoh
The movement of the magnetic northpole inherently means a lower Tmin in the SH.
THAT EXPLAINS THE RESULTS I AM SEEING
Last edited 1 month ago by HenryP
Ed Bo
September 3, 2022 8:12 am
Henry,
Let’s look at the basic equation for conduction heat transfer:
q = k * A * DeltaT / DeltaX
Re-arranging, we get
q/A = k * DeltaT / DeltaX
The thermal conductivity of most rocks is about 2 W/m/K
The geothermal gradient is about 30 (not 3) K per km.
So we get:
q/A = 2 (W/m/K) * 30 (K) / 1000 (m) = 0.06 W/m2 = 60 mW/m2
It does seem surprisingly small, doesn’t it?
September 3, 2022 10:19 am
You probably need to get some new books, in the S African goldmines the rock temperature get as high as 60ºC 4km down.
stinkerp
September 2, 2022 2:48 pm
Why is the Incoming Solar Radiation increasing over the last couple decades? I don’t understand. It seems like it might increase or decrease slightly due to orbital eccentricity throughout the year (a single orbit of the sun) and perhaps in relation to the 11-year solar cycle (roughly 0.07%). I can’t see why incoming radiation at the top of the atmosphere would increase over the last 22 years. Could you explain?
Last edited 1 month ago by stinkerp
Jason S.
September 2, 2022 5:32 pm
This is measuring incoming solar radiation after absorption/reflection. So these changes are not due available solar radiation (changes in sun output, orbits, etc.) but more likely changes to albedo. There has been literature showing a reduction in cloud cover over this same time period which correlates well to the increase in temperature.
This analysis from Willis seems to me to corroborate those findings. Changes is solar radiation driving temperature changes, not CO2.
Bob boder
September 2, 2022 5:50 pm
Changes in cloud coverage
Richard M
September 2, 2022 8:11 pm
Best to read Dubal/Vahrenholt 2021. The cloud changes appear to be related to ocean cycles.
davetherealist
September 2, 2022 2:51 pm
I 100% disagree with this statement. “We know the earth is warmer than expected.” Expected by who? and what is their basis of determining what the temperature should be.
Other than that, it is always back to the same answer: Its the SUN stoopid!.
Hubert
September 2, 2022 3:00 pm
In fact , this period of 20 years is too short to see a significant change in greenhouse multiplier !
The greenhouse effect has increased by 1 watt/m2 in 30 years , compared to the total Natural of 150 watts/ m2 .
Even the amount of 3.3 watts/m2 of anthropogenic greenhouse since the industrial revolution has a small impact on this multiplier !!
That’s not the right factor to analyse …
Eric Vieira
September 2, 2022 3:00 pm
Hi Willis
Just one comment: it looks like more energy is coming out of the system than is being put into it.
When you calculate the W/m2 is the wavelength and intensity distribution taken into account? The incoming (visible) light has more energy per photon than the outgoing (LW) radiation. Is this already inherent in the CERES data ? Does the W unit really reflect the energy coming in and out of the system ?
RickWill
September 2, 2022 4:02 pm
We have observational evidence that the temperature increase from 2000-2021 was not due to an increase in greenhouse gases, or any increase in the efficiency of the greenhouse effect from any cause. The efficiency has been very stable over the period, with a standard deviation of 0.2% and no significant trend.
I have always been clear on this. The “greenhouse effect” plays no role in Earth’s energy balance so why does it even come up in any discussion on climate?
leitmotif
September 2, 2022 4:18 pm
[SNIPPED—Contained personal attacks only, science-free. w.]
Last edited 1 month ago by Willis Eschenbach
leitmotif
September 2, 2022 4:41 pm
Snipped by Willis.
Don’t believe this man.
He snips or gets snipped those who disagree with him.
(No more discussion of moderation actions!) SUNMOD
Last edited 1 month ago by Sunsettommy
leitmotif
Reply to Willis Eschenbach
September 3, 2022 3:47 pm
WUWT cowards
(Hold for Administrator considerations and all the others below this moderated post) SUNMOD
RickWill
September 2, 2022 4:07 pm
Figure 4 shows a comparison of the upwelling longwave shown in Figure 3 and the Stefan-Boltzmann derived temperature. Basically identical in form.
Of course they are because the ULW is inferred from the temperature. It is not a measurement of power flux.
Like my wood moisture meter, it indicates moisture but actually reads a current and voltage to determine a resistance, which is correlated to moisture..
Last edited 1 month ago by RickWill
RickWill
Reply to Willis Eschenbach
September 2, 2022 5:29 pm
S-B is an approximation that can be applied in a constrained reference frame. It is not applicable when heat transport is dominated by other processes. For example an ocean warm pool has no SW radiation leaving the surface. All heat is transported by latent heat of evaporation.
You should not be surprised by the correlation – that is how the instrument is calibrated. It simply takes the temperature and applies the S-B equation. It is inferring a radiated power flux from temperature. Your chart simply verifies the calibration. But is is not measuring a radiated power flux.
There are other approximations used in appropriate reference frames. For example the gravity field. It is a field with time component as Einsien determined but we approximate the force between two bodies to a simple constant and the inverse of distance squared.
Michael Mishchenko is one of the best authors on the physics of EMR relative to Earth’s atmosphere. He understands field theory and has made efforts to apply it to climate science.
Ed Bo
September 3, 2022 8:40 am
Are you seriously claiming that if water is evaporating from the surface, it stops radiating away? (I presume you meant LW radiation). Really?
This afternoon, I will find a nearby rock that has been in the sun all day and is hot enough that I can feel the radiation from it. Then I will pour a little water on it, which will immediately start evaporating. I will check to see if it stops radiating. Not holding my breath…
September 3, 2022 7:06 pm
For example an ocean warm pool has no SW radiation leaving the surface. All heat is transported by latent heat of evaporation.”
I’m sure Mishchenko didn’t say anything this stupid!
RickWill
September 2, 2022 4:10 pm
The most important question to answer is why are there ever clear skies over oceans?
Once you can answer that, you begin to understand how the energy balance is controlled.
Bob boder
September 2, 2022 5:54 pm
So wouldn’t less cloud cover infer a cooling ocean?
Macha
September 2, 2022 4:54 pm
Because heat, manifested as temperature, is not entirely defined by quanitity W/m2. Intensity and emissivity play a significant part.
A few minutes of UV versus LWDR (15um) on your skin will prove.
Jeff Alberts
September 2, 2022 7:09 pm
The temperature has been generally rising over the period 2000-2021″
Which temperature?
richardw
Reply to Willis Eschenbach
September 2, 2022 11:39 pm
This is a question, not a criticism. Why not include and use the UAH record? From a long time (albeit as a non – scientist) reading WUWT I have reached the conclusion that UAH is the most reliable record as it avoids the various distortions inherent in land based temperature measurement.
Richard M
September 2, 2022 8:37 pm
It’s nice to see that Willis was able to reproduce essentially the same results seen in Dubal/Vahrenholt 2021. I like the greenhouse efficiency idea. It appears to support the Miskolczi 2010 constant opacity concept.
I’ve been bringing this up every now and then. Most skeptics seemed to accept the claims that Miskolczi was wrong. We can’t say he was precisely right, but it appears he had the general concept right.
Miskolczi analyzed 70 years of NOAA data and found the greenhouse effect was a constant. Willis has now found a similar result over the past 20 years. The chances both of these could be wrong is pretty small.
This means that climate science got it completely wrong. It also means luke-warmers got it wrong. There is no warming due to doubling of CO2.
There’s real physics involved. I’ve explained why boundary level feedback counters DWIR warming. I’ve also explained why the effective emission altitude is a constant. Time for skeptics to quit looking for feedbacks to warming and consider why the warming never occurs.
KcTaz
September 2, 2022 9:57 pm
“On the other hand, the change in incoming solar energy is both adequate to explain the increase in warming, and has the same shape as the change in surface radiation (blue LOWESS smooths in both panels in Figure 3). While there are undoubtedly other factors in play, the main cause of the warming is clearly the increase in the amount of solar energy after reflections from the clouds and the surface.
And once again, the clouds rule … go figure …”
There are other scientists whose work completely supports your statement, Willis.
Japanese researchers at the University of Kobe arrived at similar results as the Turku team, finding in a paper published in early July that cloud coverage may create an “umbrella effect” that could alter temperatures in ways not captured by current modeling.
A pdf (1.7MB) for download is available at
https://arxiv.org/pdf/1907.00165.pdf
Also to note is a recent paper called
‘No experimental evidence for the significant anthropogenic climate change’
Jyrki Kauppinen, Pekka Malmi
(Submitted on 29 Jun 2019)
Abstract
In this paper we will prove that GCM-models used in IPCC report AR5 fail to calculate the influences of the low cloud cover changes on the global temperature. That is why those models give a very small natural temperature change leaving a very large change for the contribution of the green house gases in the observed temperature. This is the reason why IPCC has to use a very large sensitivity to compensate a too small natural component. Further they have to leave out the strong negative feedback due to the clouds in order to magnify the sensitivity. In addition, this paper proves that the changes in the low cloud cover fraction practically control the global temperature.
CERN: Cosmic Rays Influence Cloud Formation
Aug 25, 2011
CERN
CLOUD discovers new way by which aerosols rapidly form and grow at high altitude
The resultant particles quickly spread around the globe, potentially influencing Earth’s climate on an intercontinental scale
18 MAY, 2022
Another Climate Scientist with Impeccable Credentials Breaks Ranks: “Our models are Mickey-Mouse Mockeries of the Real World” – Electroverse
September 26, 2019
http://bit.ly/33p7OSa
Dr. Mototaka Nakamura received a Doctorate of Science from the Massachusetts Institute of Technology (MIT), and for nearly 25 years specialized in abnormal weather and climate change at prestigious institutions that included MIT, Georgia Institute of Technology, NASA, Jet Propulsion Laboratory, California Institute of Technology, JAMSTEC and Duke University.
In his book The Global Warming Hypothesis is an Unproven Hypothesis, Dr. Nakamura explains why the data foundation underpinning global warming science is “untrustworthy” and cannot be relied on:
“Global mean temperatures before 1980 are based on untrustworthy data,” writes Nakamura. “Before full planet surface observation by satellite began in 1980, only a small part of the Earth had been observed for temperatures with only a certain amount of accuracy and frequency. Across the globe, only North America and Western Europe have trustworthy temperature data dating back to the 19th century.”
From 1990 to 2014, Nakamura worked on cloud dynamics and forces mixing atmospheric and ocean flows on medium to planetary scales. His bases were MIT (for a Doctor of Science in meteorology), Georgia Institute of Technology, Goddard Space Flight Center, Jet Propulsion Laboratory, Duke and Hawaii Universities and the Japan Agency for Marine-Earth Science and Technology.
He’s published 20+ climate papers on fluid dynamics.
There is no questioning his credibility or knowledge.
Doug Proctor
September 3, 2022 1:06 am
I just read the albedo article at the centre of this post
https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2021GL094888
The Earthshine project showed a 0.5W/m2 decrease over 20 years. The CERES satellite showed 1.5W/m2.
The authors seem to discount the satellite data and stick with the 0.5W/m2.
They say models say 0.6W/m2 increase from CO2 and pollution. So together 1.1W/m2.
Why discount 1.5? If really 1.5, then 0.6 models are wrong, just curve fitting to an alleged 0.5W/m2 albedo.
Also, the albedo change is said to be from melted ice loss in Arctic and less clouds over a warmer Pacific. But could this explanation be from confirmation bias? That they have cause and effect reversed?
If global warming models are curve fitting to an unknown reason for albedo changes, the whole global warming narrative fails. The models are based on an assumption we understand the variables heating the planet’s atmosphere.
A variable we don’t know we don’t know doesn’t get put it the model. The other variables are then tweaked to account for it.
Maybe I’m missing something.
Richard M
Reply to Doug Proctor
September 3, 2022 8:39 am
One of the key data items that supports CERES over Earthshine is the high correlation of the temperature computation Willis did from the LW data and the UAH data. Two different approaches and almost identical results.
Doug Proctor
September 3, 2022 1:13 am
The albedo change of 0.5 or 1.5 W/m2 is 15% or 42% of the alleged equivalent forced by 2X CO2 of 3.5W/m2. Which is supposed to be 3 or 4*C in the scary scenarios. Which strikes me as the entire reason for the planetary warming of the past 20 years.
Again, cause and effect inversion due to not knowing what we don’t know is going on?
Again, what am I missing?
Richard M
Reply to Doug Proctor
September 3, 2022 8:44 am
You’ve got it nailed. The CO2 increase over those 20 years would yield only about 20% of the 3.5 W/m2 claimed warming which is only half of the 1.5 W/m2 of solar warming. In fact, we’ve seen significant longwave cooling which reduced the amount of warming we have seen.
nobodysknowledge
September 3, 2022 3:55 am
Willis.
There is some problems with the “greenhouse effect”.
You should take all the energy in, and all the energy out from the earth surface. It is the energy budget that matters.
From Loeb et al 2021: Trend in EEI During the CERES Period
https://ceres.larc.nasa.gov/documents/STM/2021 05/35_Loeb_contrib_science_presentation.pdf
For the radiation at the earths surface we have the following numbers (Wild et al. 2019):<br />Solar radiation absorbed: 160 W/m2 with an increasing trend.<br />Longwave cooling from increased temperature: -56W/m2 with an increasing trend.<br />Evaporation, without presentation of trend: -82W/m2<br />Sensible heat, conduction/ convection from surface: -21W/m2 <br>Earth Energy Imbalance measurements tell us that there is a warming of 0,51 W/m2/dec from change in these variables, SWsurf down, LWsurf up, Evaporation, Sensible heat. The components behind these changes are Temperature change, Albedo change, Cloud radiation change, Water vapor Change, and Trace gases change. These are also the feedback components of climate change. <br>Loeb et al, 20 years of energy imbalance from 2000 to 2020:<br />Temperature surface radiation, Net LW cooling: -0,51 W/m2/dec<br />Albedo reduction. SW solar warming: 0,19 W/m2/dec<br />Cloud LW cooling (less clouds) -0,23 W/m2/dec<br />Cloud SW decreased absorption 0.44 W/m2/dec<br />Water vapor LW warming 0,33 W/m2/dec<br />Water vapor SW warming and latent heat. 0,05 W/m2/dec<br />Trace gas, aerosole LW warming 0,237 W/m2/dec<br />Trace gas, aerosole SW warming 0,002 W/m2/dec<br />If we assume that most trace gases and aerosols dont make much difference, and Methane stands for 22,9 % of trace gas warming, we get:
CO2 LW warming 0,185 W/m2/dec
Methane LW warming 0,055 W/m2/dec
Last edited 1 month ago by nobodysknowledge
nobodysknowledge
September 3, 2022 4:03 am
Sorry for the font shift
nobodysknowledge
September 3, 2022 4:12 am
The net LW cooling is change in the difference between longwave radiation up and the downwelling radiation, so there is a “backradiation” cooling. The downwelling doesn`t compensate for the surface warming radiation (Planck feedback).
Last edited 1 month ago by nobodysknowledge
Richard M
September 3, 2022 8:49 am
Much of the data used in Loeb et al 2021, which tries to compensate for the obvious problems the raw CERES data represents, is based on “estimates”, “guesses” and “models”. It looks very suspicious.
It reminds me of all the adjustments to surface data and the continued divergence of that data from satellite data.
Edim
September 3, 2022 5:44 am
Nice again Willis. As you probably know, there are already several papers showing that the recent warming is caused by increased absorbed solar radiation, but they speculate that it’s a feedback to CO2 warming (epicycles IMO).
https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2009GL037527
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4250165/
Btw, the Earth’s atmosphere has an insulating effect on the surface. Greenhouse effect just sounds dumb and unscientific.
September 3, 2022 6:49 am
So given the evidence above that the increase in upwelling surface radiation cannot be due to a change in greenhouse efficiency from increased CO2 or any other reason
≠==========
This is strong evidence that the so called greenhouse effect is NOT due to CO2.
Willis, maybe I missed something, but it sure looks to me like you have proven that the greenhouse multiplier is not the result of CO2. That something else must be the cause.
It only takes 1 confirmed false finding to prove a theory false.
September 3, 2022 7:11 am
Question. What about the energy that leaves and enters the surface via conduction and convection? This affects outgoing radiation and is unlikely to be net zero because it delays cooling and interacts with ghg at altitude.
nobodysknowledge
September 3, 2022 7:37 am
Conduction/convection from surface 21W per m2
Evaporation from land and ocean surface 82W per m2
Net longwave radiation, surface emission minus backradiation 56W per m2
All these are cooling the surface, giving energy to the atmosphere and radiated out to space.
From Martin Wild et al.
Richard M
September 3, 2022 8:55 am
Conduction, convection (and evaporation) are key to the negative feedback for increases in DWIR. They counter the warming effect from CO2 increases immediately. No warming at the surface. I call this boundary layer feedback.
Last edited 1 month ago by Richard M
September 3, 2022 7:33 am
Willis wrote: “..the increase in upwelling surface radiation cannot be due to a change in greenhouse efficiency from increased CO2 or any other reason….”
=========
Thus, given that GHG was increasing during that period, while the greenhouse efficiency did not, either the greenhouse effect is not caused by greenhouse gas or there is an error in your work.
There cannot be causation without correlation.
Richard M
September 3, 2022 8:58 am
The CO2/CH4 greenhouse effects are saturated which means their effect can no longer increase. The alarmist invention of an “enhanced greenhouse effect” is what Willis has shown to be pseudo-science.
September 3, 2022 7:55 am
At an altitude of 500mb, the air temperature is the predicted temperature of the earth without GHG.
50% of the atmospheric mass lies above 500mb and is cooler. 50% of the atmospheric mass lies below and is warmer.
Only the sun moves the 500mb line, so the only way to increase the greenhouse efficiency would be to reduce the water vapor in the atmosphere. See formula for lapse rate. Adding CO2 will not do it because it is non condensing.
Last edited 1 month ago by ferdberple
September 3, 2022 8:05 am
A decreasing greenhouse efficiency as Willis has identified at a time of increasing CO2 tells me that CO2 does not affect greenhouse efficiency because CO2 is non condensing. Rather, the decrease in greenhouse efficiency is due to an increase in atmospheric water vapor, flattening the lapse rate. The obvious cause is burning, agriculture, irrigation and land use changes.
Last edited 1 month ago by ferdberple
Richard M
September 3, 2022 9:00 am
No, it has little to do with water vapor. The greenhouse effect is saturated. Nothing else is required.
September 3, 2022 8:16 am
No new heat can be generated from a static pressure.
==≠======(
Misconception. The lapse rate is a result of convection. The pressure is not static from the point of view of an air molecule or parcel of air moving vertically. The pressure is truly only static if you remove the sun which would end all convection.
In any case new heat is not being generated. It is being pumped downwards using solar energy to move air and water vapor through a pressure gradiant, similat to a mechanical heat pump.
Last edited 1 month ago by ferdberple
Richard M
September 3, 2022 9:02 am
The lapse rate is actually due to well mixed GHGs controlling the energy levels allowed at varying densities.
Reply to Richard M
September 3, 2022 12:53 pm
The lapse rate is actually due to well mixed GHGs
========
Nowhere in rhe formula for lapse rate does well mixed GHG appear.
Richard M
September 3, 2022 2:18 pm
Not relationships are easy to see.
Well mixed GHGs means their concentration follows the changes in atmospheric density. What determines density? Good old gravity. And of course, gravity does appear in the lapse rate equation
September 3, 2022 8:34 am
It is a simple matter to generate 33C of warming by pumping a gas through a pressure gradiant. You are not creating energy. You are moving energy from one place to another.
Atmospheric convection driven by solar energy does this all day long.
Build a huge mechanical heat pump run via solar energy with the low pressue (cold) coils at altitude and the high pressure (warm) coils at the surface. That is convection.
Ps: this is a continuation of my earlier post based on Willis showing that greenhouse efficiency was not determined by CO2. It follows from that that the greenhouse effect must be based on some other gas other than CO2.
David Appleby
September 3, 2022 11:02 am
First of all, thank you for such a concise and informative post. The CERES data indicates an underlying fall in reflected sunlight of approximately 1.4% between 2000 and 2021. This could be due to a change in average cloud cover, but the reduction in average arctic sea ice & snow cover (from 10.5 to 9.5.10^6 km2 from AMSR data) could be having a significant effect. A very rough calculation, assuming a change in albedo from 0.8 to 0.15, gives an expected reduction in expected solar reflection of about 0.7%. It may be worth someone calculating a more accurate figure, accounting for ice cover & sun angle throughout the year.
Kevin kilty
September 3, 2022 12:21 pm
The efficiency used here is simply the inverse of the effective emissivity of the Earth: In this essay from 3 years ago, I calculated this effective emmissivity as 0.61 (ie 1.64 in Willis’s terms) but this presupposes an average albedo in the solar spectrum of 0.30. What goes on with regard to a secular trend in “efficciency of greenhouse effect” is actually a secular trend in the figure of merit of the Earth treated as a solar collector. What is good about the figure of merit is it takes into account both the effective emissivity and the effective solar absorptivity.
I’ll see here if LaTeX still works…figure of merit = $(\alpha_s/\epsilon)^{1/4}$
Kevin kilty
Reply to Kevin kilty
September 3, 2022 12:21 pm
It works!
September 3, 2022 1:10 pm
The lapse rate is actually due to well mixed GHGs
========
it this is a prediction of Greenhouse theory, then it is surprising it has not been dealt with.
On Earth there is no term for well mixed gas in the lapse rate.
The lapse rate is a function of gravity and the work required to compress air. This contradicts the notion that heating/cooling due to compression is a one time event.
In addition the lapse rate is a function of the condensation of water and the energy released via phase change.
This is all driven by low pressure in places where the sun is shinning and high pressure in places that are dark. These are continually changing because of orbital mechanics.
Last edited 1 month ago by ferdberple
Kevin kilty
September 3, 2022 7:13 pm
True, Mr. Ferdberple. It is very difficult to get people to understand that temperature (a proxy for internal energy at fixed composition) is a reflection of the first law of thermodynamics, towit, change in internal energy = heat in – work out; or, $dU=\delta Q-\delta W$
September 3, 2022 1:20 pm
It is generally recognized that the atmisphere below 500mb is opaque to IR and as a result there is little cooling of the surface due to outgoing radiation. The heavy lidting below 500mb is done by convection.
September 3, 2022 2:13 pm
No, the atmosphere is not opaque to IR, as shown in the spectrum below it’s fairly transparent at many wavelengths. The regions of the spectrum where it is opaque are where certain components (CO2, H2O, O3 etc) absorb.
Kevin kilty
September 3, 2022 2:19 pm
In most places, perhaps, but where I live cooling by radiation predominates after sundown and is substantial and quite apparent even without instruments.
Ulric Lyons
September 3, 2022 1:31 pm
“The earth is much warmer than the moon, which receives the same amount of solar energy. It’s generally accepted, including by me, that the warmth is from the very poorly named “greenhouse effect”
Earth’s sunlit side at any given time is cooler than on the Moon because of clouds and water vapour. Earth’s dark side at any given time is much warmer than on the Moon, primarily because of the sea surfaces barely cooling at night.
Last edited 1 month ago by Ulric Lyons
Reply to Ulric Lyons
September 3, 2022 7:36 pm
The Moon has an albedo of 0.12 as opposed to the Earth which has an albedo of 0.30 so the Moon receives about 26% more solar energy than the Earth. On the dark side of the Moon the surface has a lot longer to cool (~14 days) it gets a lot colder.
Ulric Lyons
September 4, 2022 4:27 am
The lunar equator cools about 270K in a quarter rotation from midday to dusk, and then cools about 40K from dusk to dawn for half a rotation. So in twice as long, the dark side cools much less.
Reply to Ulric Lyons
September 4, 2022 10:42 am
Yes the day-time temperatures are close to radiative equilibrium so the maximum temperature is at noon (~390K), due to the change in incidence angle the temperature drops rapidly before dawn (7 Earth days later) to about 200K (high std dev ~30K). By the Lunar midnight 7 Earth days later it reaches ~95K, since loss depends on T^4 further losses are slow. As I said, much longer cooling time than on Earth.
Ulric Lyons
September 4, 2022 12:58 pm
“due to the change in incidence angle the temperature drops rapidly before dawn (7 Earth days later) to about 200K”
Dusk comes after midday, by which time the equator has cooled down to around 120K. The night time cooling over ~14 days is only 40K.
Last edited 29 days ago by Ulric Lyons
Reply to Ulric Lyons
September 4, 2022 3:27 pm
Yes mistyped dawn instead of dusk.
Ulric Lyons
September 4, 2022 1:03 pm
Double or quadruple the lunar rotation rate, and the sunlit side will be virtually the same temperature, except for a slightly warmer dusk terminator and a slightly cooler dawn terminator. The dark side mean temperature would be virtually the same.
Reply to Ulric Lyons
September 4, 2022 4:04 pm
Well I was comparing it to the Earth which rotates 28 times faster!
Reaches its maximum after 7 days of heating (noon) and proceeds to cool for 21 days after that.
The moon surface at the equator is warmer than the Earth for about 6 days around noon, colder the rest of the time.
Ulric Lyons
September 4, 2022 5:12 pm
The lunar sunlit side is roughly in equilibrium with solar irradiance, it does not take days to heat up, the surface temperature is mostly dependent on the angle of incidence.
leitmotif
September 3, 2022 3:38 pm
This is totally hilarious!
griff, loydo and simon have been allowed to post here for years on WUWT despite being CAGWers and yet I have been placed on moderation for disputing the existence of the GHE.
All I ever asked for was evidence that the GHE exists and that it causes surface warming. I also asked for evidence that Equilibrium Climate Sensitivity was a true measurement.
What can I say?
WUWT is a cancel culture website.
If you want me to go just ban me. WUWT is not really worth the effort in its current lukewarmist stance.
It will just convince me and a few others on this website (not many it seems but also the more intelligent and informed ones, I’m sure) that WUWT is just a supporter of lukewarmists and not really an edge-cutting protester against government policy on climate change.
WUWT had a great platform for change but the platform moved so much WUWT just slid off into the mire.
Yours
Very Disappointed
(You have 1322 posts and allowed this one because it shows how far off the path YOU are when it comes to following the blog policy:
Trolls, flame-bait, personal attacks, thread-jacking, sockpuppetry, name-calling such as “denialist,” “denier,” and other detritus that add nothing to further the discussion may get deleted…
and,
For the same reasons as the absurd topics listed above, references to the “Slaying the Sky Dragon” Book and subsequent group “Principia Scientific” which have the misguided idea that the greenhouse effect doesn’t exist, and have elevated that idea into active zealotry, WUWT is a “Slayer Free Zone”. There are other blogs which will discuss this topic, take that commentary there.
No one here is forcing you to be here but this blog has a policy in place to help people stay on topic and be reasonably civil, you are in MODERATION now because our lack trust in you has fallen to the point that your future posts have to be in the moderation bin first to see if you are here to contribute to the debate without the baiting the trolling and the numerous personal attacks.
Deleted 10 posts to clean up the thread) SUNMOD
Last edited 29 days ago by Sunsettommy
Clyde Spencer
September 3, 2022 7:13 pm
… I have been placed on moderation for disputing the existence of the GHE.
It is unfortunate for you that you don’t realize that the problem is not “disputing the existence of the GHE,” but rather your uncivil behavior and unwillingness to abide by WE’s request to not ‘thread jack’ the topic. You have some issues and this is not the time to air them.
Clyde Spencer
September 3, 2022 6:14 pm
Nobody has ever come up with an explanation for that except the greenhouse effect.
Something to consider is that the moon has no liquid water. The surface of rocks are heated to high temperatures by direct sunlight, and because of the S-B 4th-power law, it is radiated away rapidly.
On the other hand, Earth with abundant water (with a specific heat capacity ~5X greater than rocks) only gets to about 1/5th the temperature and radiates at 1/625 (0.2%). Also, the evaporation of water and transpiration from plants keeps the surface at a low S-B base temperature, keeping the rate of radiation low.
Kevin kilty
September 3, 2022 9:24 pm
Willis,
I don’t necessarily have an argument about this work or your earlier essay, but you begin with the statement
I got to thinking about the oft-repeated claim that a doubling of CO2 increases top-of-atmosphere (TOA) radiative forcing by 3.7 watts per square meter (W/m2) … and that in turn, the additional 3.7 W/m2 of TOA forcing causes a ~3° warming of the temperature. In other words, they say that ~ 1.2 W/m2 of additional radiative forcing causes one degree of warming.”
When someone says TOA I take that to mean where the atmosphere ends. But at such a height the radiation leaving the Earth cannot be anything other than what is entering from the Sun, with perhaps some small adjustment for lack of equilibrium due to a slowly warming planet. There is nothing going on within the atmosphere that can increase a TOA radiative flux up or down by any amount.
Now it may be that when people say “greenhouse” effect they actually mean at top of the troposphere, or stratosphere, or perhaps truly out where the atmosphere ends. In your opening words, what does TOA mean?
Possibly by a forcing of 3.7 watts they mean that solar albedo has declined for some reason by this amount, so that this is an increase of solar reaching the ground surface, but LWIR increases by the same. Is this what is claimed?
For any greenhouse effect, its maximum magnitude will be obeserved at the surface — whether measured by temperature or downwelling radiation. Even saying “surface” has its pitfalls. They must mean that the surface sees an additional LWIR downwelling, which raises its temperature, and this in turn increases upwelling LWIR at the surface — but none of this has anything to do with TOA.
Howard Hayden has made a presentation about the IPCC and claims about three things (greenhouse effect, surface temperature, and stefan-boltzmann law) that cannot be made consistent. Are you aware of his work?
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2022-10-04 11:23:06
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https://mathoverflow.net/questions/220254/interesting-projective-varieties-being-quotients-of-mathbban-setminus-0?noredirect=1
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# “Interesting” projective varieties being quotients of $\mathbb{A}^n\setminus \{0\}$ by an action of an algebraic group?
The algebraic (multiplicative) group $G^m$ acts on $\mathbb{A}^n$ (diagonally) and the quotient of $\mathbb{A}^n\setminus \{0\}$ by $G_m$ is $\mathbb{P}^{n-1}$ (which is a proper variety). I would like to find a "more interesting" example of an action of an algebraic groups $G$ on $\mathbb{A}^n$ such that the quotient of $\mathbb{A}^n\setminus A$ by $G$ is proper, where $A$ is a certain "small" subvariety of $\mathbb{A}^n$ (say, over the field of complex numbers). Under these conditions does the quotient have to be toric (and does $G$ have to be a torus)?
I would be deeply grateful for any hints or references (and I know very little on algebraic groups and toric varieties)!
• Anything on "geometric invariant theory", e.g. the book with that title, will have lots more examples, where $A$ is the "unstable set". If $G$ contains dilation then the GIT quotient will be proper. For a non-torus example, let $GL(k)$ act on $k\times n$ matrices, and let $A$ be the set of matrices of rank $< k$. Then the quotient is $Gr(k,n)$. Of course, maybe this $A$ isn't "small" enough for you. – Allen Knutson Oct 7 '15 at 9:17
• Thank you!! Funnily enough, just before reading your comment I was wondering whether $A$ is small enough in this example.:) – Mikhail Bondarko Oct 7 '15 at 9:36
• Does "small" for you just mean that the quotient is proper? – Allen Knutson Oct 7 '15 at 10:28
• No, "small" means "of small dimension" (and I am thinking what does the latter "small" means). – Mikhail Bondarko Oct 7 '15 at 10:55
• – Lucas Kaufmann Oct 7 '15 at 13:35
By the way your question is phrased, it seems that you might be familiar with the following construction. But anyway, any toric variety can be realised as the quotient of $\mathbb{A}^n \setminus A$ by the action of an algebraic torus (for some $n$ and $A$). This is proved in the famous paper of Cox:
This theory has been generalised much by the theory of Cox rings and universal torsors. This says that any "suitably nice" variety $X$ (namely, a Mori dream space) is a quotient of $\mathrm{Spec}(\mathrm{Cox}(X)) \setminus A$ (for some $A$) by the action of the Néron-Severi torus of $X$. Moreover, $\mathrm{Spec}(\mathrm{Cox}(X))$ is isomorphic to an affine space if and only if $X$ is a toric variety.
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2021-06-17 07:32:29
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https://mindspace.arclind.com/learn/physics/?sort=new
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# Physics
Crash courses and sparks for students and explorers on the pillars of physical sciences with simplified straightforward lessons and insights.
It's empty here ...
Most people think that inhaling helium changes the pitch or frequency of the voice. No! That is no what happens. It’s the timbre of the voice that changes.
When you inhale helium, the medium inside the vocal cavities changes from a dense medium, air, to a lighter medium, helium. And we know that in air, any vibrations travel at a speed of 343 m/s. For helium, it’s 1007 m/s. The helium medium now increases the natural frequencies of the cavities. In other words, it changes the responsiveness of the cavities to higher frequencies. This results in the amplification of a higher range of frequencies compared to that of when the air was the medium—causing the squeaky voice!
The key observation here is that the frequency at which the sound is produced by the vocal folds doesn’t change. It’s the resonant frequencies that change, forcing a change in the timbre of the voice.
What’s a timbre of a human voice? Let’s start with the voice! The human voice is created when vocal folds vibrate. It’s this vibration of air molecules when travelling through different cavities like pharynx, sinuses, nose, and mouth, that’s converted to speech. And here is the interesting part that makes them unique to a person.
Like any physical objects, these cavities through which the sound travels have their own properties as well. They have their own distinct natural frequencies because of the geometries of the muscles that are unique to a person and the composition of the air.
And the sound from the vocal folds is made of not just a uniform sine wave with a fundamental frequency, but a composite of other distinct frequencies as well. So when certain frequencies of that sound wave hit the natural frequencies of the cavities, resonance happens, and those parts alone get amplified. The end result of all this is the distinct voice of a person. In other words, the lowest resonant frequency that’s modulated by the rest of the frequencies is what gives that unique tone to your voice. And that’s what we call the ‘timbre’ of the voice.
The voice you hear when you inhale helium, that’s because of the timbre change too. Not the frequency shift!
The longest vertical straw you can drink from is 10.3 metres. Even if you use a vacuum pump it won’t suck the liquid higher than that! Here is why!
Contrary to your intuition, when you drink from a straw you are not actually sucking up the fluid here. Just the air. So, when you do that, inside the straw, the pressure drops lower than that of the atmospheric pressure (101 kPa) outside. So, it’s the outside air pressure that pushes the water into the straw.
As the liquid moves up the straw, it is fighting against the gravity that is pulling it downwards. But it still keeps rising as long as the atmospheric pressure is greater than the pressure inside the straw due to gravity (weight of the liquid column).
The more liquid enters the column, the more it weighs. And at a certain height, there’d be enough water in the straw that’d exert the same pressure as that of the atmospheric pressure. That height, at sea level on earth, for water is 10.3 m.
$$p_{atm}= 101\;kPa$$
$$p_{straw}= \dfrac{F}{A} \Rightarrow \rho g h$$
$$\rho g h = 101 \times 10^3\;N/m^2$$
$$h = \dfrac{101 \times 10^3\;N/m^2}{10^3\;kg/m^3 \times 9.81\;m/s^2}$$
$$h = 10.3\; m$$
Temperature is not the measure of the heat. Most people think that they are the same. Take a pair of small and large vessels with water and expose it to the sun. You will find the smaller vessel becoming warm quicker than the larger one. Although an equal quantity of heat is supplied to the two vessels, due to the difference in the quantity of the water, the time it takes to raise the temperature varies. This must clarify the confusion between temperature and heat.
If you have poured tea from a mug you’ll intuitively know what a Coanda effect is. To put it simply, the Coanda effect is the phenomenon where fluids like water tend to follow and stick to a contour of an object.
So what happens here? When the water flows out of the mug, the water molecules encounter the air molecules and try to drag them along due to viscosity. As the air molecules under the mug get dragged off, the pressure at that spot, which is relatively constrained compared to the other side, decreases (Bernoulli’s principle). And as the pressure is higher at the top of the water than on any other side, the water reaches equilibrium by moving towards the low-pressure region, which is what makes it to stick to the surface of the mug.
Earth’s atmosphere is leaking a few grams of helium this very moment! Yep! As helium and hydrogen are the lightest elements of all, Earth’s gravity has little effect on them in the hydrostatic equilibrium. With higher kinetic energies, hydrogen and helium reach velocities greater than that of Earth’s escape velocity in the thermosphere and they shoot into space.
If you are in space and shine a torch in an arbitrary direction, the photons (although being massless) will impart a thrust on you that will propel you in the opposite direction. This is due to the conservation of momentum as well.
In other words, when photons are ejected out of the torch, they travel outward with a momentum*. And due to this, your momentum changes to conserve the total momentum of your initial state, pushing you in the opposite direction.
* Photons have a relativistic mass and they do have momentum, given by $p = \dfrac{h}{\lambda}$.
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2023-03-20 10:37:51
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http://mathoverflow.net/questions/24960/why-are-parabolic-subgroups-called-parabolic-subgroups
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# Why are parabolic subgroups called “parabolic subgroups”?
Over the years, I have heard two different proposed answers to this question.
1. It has something to do with parabolic elements of $SL(2,\mathbb{R})$. This sounds plausible, but I haven't heard a really convincing explanation along these lines.
2. "Parabolic" is short for "para-Borelic," meaning "containing a Borel subgroup."
Which answer, if either, is correct?
A related question is who first introduced the term and when. Chevalley perhaps?
-
I am certain that it's #1, but the terms "parahoric" (containing an Iwahori subgroup) and "mirabolic" (miracle parabolic) were so named to be consonant with "parabolic", which may have led to the folk etymology you've described. – Victor Protsak May 17 '10 at 3:38
Wow, the second one would be very creative. – user717 May 17 '10 at 9:00
The invention of "parahoric" (after Iwahori) is apparently due to Bruhat-Tits in their follow-up work on structure theory over local fields following fundamental work by Iwahori and Matsumoto. Tits has always been fond of this kind of wordplay. (The introduction of "Borel subgroup" in his 1965 paper with Borel was probably due to Tits, though they left that ambiguous in a famous footnote.) – Jim Humphreys May 17 '10 at 11:31
@Victor: What makes you so certain that it's #1? Any concrete evidence? I can imagine either definition being the original one and the other one being the folk etymology that was invented because it seemed plausible. – Timothy Chow May 17 '10 at 14:13
It appears that neither of the answers is fully correct. There is a great book, "Essays in the history of Lie groups and algebraic groups" by Armand Borel, when it comes to references of this type. To quote from chapter VI section 2:
...There was no nice terminology for the subgroups $P _I$ with lie algebra the $\mathfrak p _I$ until R. Godement suggested calling them parabolic subgroups. I shall therefore anachronistically call them that...
"The geometry of the finite simple groups" by F. Buekenhout is on the other hand the only paper that came up in a search for paraborelic, and the author mentions he is using this term instead of parabolic to distinguish from parabolic subgroups of Chevalley groups.
-
Borel's attribution of the terminology "parabolic subgroup" to Godement is reasonable, but Timothy Chow's first option probably comes closest to the rationale behind this choice. Study of the modular group by Fricke, Klein, and others distinguished several types of elements: "elliptic", "hyperbolic", "parabolic" (the latter typically coming from unipotent matrices). When Dan Mostow was asked about the origin of the naming convention back in 1977, I recall that he attributed it to the parallel with modular groups and parabolic elements. By 1962 Tits was using the term in his papers. – Jim Humphreys May 17 '10 at 11:25
P.S. A late instance of "parabolic" in connection with the modular group occurs in a 1974 thesis at NYU by the last student there of Wilhelm Magnus: Nonparabolic Subgroups of the Modular Group by Carol Tretkoff. But in line with Benoit's answer, the underlying rationale for the usage comes from study of homogeneous spaces such as $G/P$ in Lie theory. Borel himself didn't use the term "parabolic subgroup" in his 1956 Annals paper, but focused on complete/projective varieties starting with $G/B$. By 1962 he as well as Tits and others were using the term in print. – Jim Humphreys May 17 '10 at 13:03
My (completely non historical) point of view is the following. When you study non-compact symmetric spaces, e.g. the real hyperbolic space, isometries can be divided into three classes: elliptic (fixing a point in the space, so that it generates a relatively compact subgroup), hyperbolic (translates a geodesic, and acts like a dilation on the boudary of the space), and parabolic (none of the preceding type, but can be approximated both by elliptic and hyperbolic elements; always fixes a point on the boundary). In this context, a parabolic subgroup is the stabilizer of a point of the boundary, and contains many parabolic elements.
I guess that in a more algebraic (or should I say less geometrical?) context, this notion might generalize naturally to what is actually called a parabolic subgroup.
I hope this at least clarifies what is often meant by your answer #1.
-
This is roughly what I have heard before, but the reason it hasn't struck me as being a clincher is that the connection you give between parabolic subgroups and parabolic elements is not as crisp as I would have expected if this were the true motivation for the terminology. Is there a sharper theorem here than "a parabolic subgroup contains many parabolic elements"? – Timothy Chow May 17 '10 at 14:18
@Timothy: You may be expecting more rationality in the choice of terminology than exists. It's usually hard to come up with just the right word (standard or invented), so people may rely on (1) bland choices like "normal", (2) words transplanted from their original context like "parabolic", (3) names of people (appropriate or not) somehow associated with the concept --- the invented term "K3 surface" is one variant. After a while it's too late to go back and rethink the choices, as Freudentahl-deVries tried to do using highly nonstandard terminology in their 1969 book Linear Lie Groups. – Jim Humphreys May 17 '10 at 17:26
@Timothy Chow: in the stabilizer of a point $p$ of the boundary, there are: -- all hyperbolic elements whose translated geodesic has $p$ as an endpoint; there are many ways to deform the geodesic so that the other endpoint also tends to $p$, and the resulted isometries in the limit are parabolic, -- all elliptic elements whose fixed point set contains $p$ in its closure; for example in the real hyperbolic case, such a fixed point set is a totally geodesic subspace, that can be deformed to $\{p\}$, the deformed isometries in the limit being parabolic. – Benoît Kloeckner May 17 '10 at 18:32
Another point, in the real hyperbolic space: the stabilizer of a boundary point is isomorphic to the set of similarities of the euclidean space of one less dimension. Inside this set, the translations are the parabolic elements. This makes many of them. More significantly, you form a cusp by quotienting the space by a lattice of this euclidean space: parabolic elements play in this respect a prominent rôle. – Benoît Kloeckner May 17 '10 at 18:33
@Jim: Wow, the table of contents alone in "Linear Lie Groups" is a barrel of laughs. With crazy-sounding terms like trunks, tools ("Weyl tool"), dressings, wrappings, and virtual reality (really), I can't fathom what those guys were smoking when they wrote the book. – BCnrd May 18 '10 at 6:43
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2015-04-21 13:20:01
|
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|
https://dsp.stackexchange.com/questions/58651/finding-where-signal-is-showing-some-unknown-periodicity/58665
|
Finding where signal is showing some (unknown) periodicity
I have readings of the vertical displacement of someone over time, as in the picture below:
At some points in time, the person will be exercising or jumping up & down, creating more or less obvious periodic motions (yellow regions of the graph). I am now trying to find a way to automatically spot these yellow areas, by looking for periodicity in the recorded signal.
My initial attempt has been to use a spectrogram, using the specgram built-in function in Octave: segmentLength=150; window=hanning(150); overlap = 80% * segmentLength; specgram(x, segmentLength, 1, window, overlap)
Which results in the following output:
This seems to be hinting in an ok direction since I see for example that something gets picked up just before frame 2000, however it is not very conclusive. Given my limited understanding of the spectrogram, I am not sure how to proceed from here.
Is the spectrogram the right approach? If yes, how can I improve the output? If no, what approach would you recommend for this task? I am aware that some of the windows that I am trying to identify will be too difficult, but was expecting to at least pick up windows 1, 2 and 5 (starting from the left). Thanks for any pointers!
EDIT
I have tried to implement a simple ASDF in octave with the following code:
#Parameters for asdf analysis
N=400; %Window size
kmin= 12;
kmax= 40;
step = 50;
n0min = floor((N+kmax)/2);
n0max= size(x,2) - floor((N+kmax)/2) -1;
Q=zeros(kmax-kmin+1, floor((n0max-n0min)/step)+2);
i=1;
for k= kmin : kmax %i
j=1;
for n0= n0min : step : n0max %j
for n= 1 : N-1
Q(i, j) = Q(i,j) + ( (x(n+n0-n0min) - x(n+n0-n0min+k))^2 * hanning(N)(n+1) );
end
Q(i, j) = (2/N) * Q(i,j);
j++;
end
fprintf('Calculated %d out of %d periods \n', k-kmin+1, kmax-kmin+1);
fflush(stdout);
i++;
end
imagesc(Q)
Tweaking around with the parameters, I ended up with the ones shown in the code, that yield the following result:
This is already getting closer to what I expected. Other suggestions welcome :)
• "At some points in time, the person will be exercising or jumping up & down, creating more or less obvious periodic motions (yellow regions of the graph)." Well, there appears to me only two yellow regions (the first and the fifth) that display much periodicity. Normally to measure the degree of periodicity, something like the autocorrelation function is used. – robert bristow-johnson Jun 2 '19 at 9:09
• Here's another reference about measuring periodicity. The application area is more about musical pitch detection, but it's about measuring the period and measuring the degree of periodicity. – robert bristow-johnson Jun 2 '19 at 9:12
• Could you possibly make a dataset available? – Cedron Dawg Jun 2 '19 at 12:39
• I think you are defining the window wrong, in window = 25. The window variable is supposed to be samples from a window such as hanning or blackman. I think that, if you define a proper window, your spectrogram will give you much better information. – MBaz Jun 2 '19 at 17:33
• Thank you @robertbristow-johnson for your comments, I will look in more details into AMDF/ASDF. The theory in your linked posts starts to make sense, need to see how I can try this out in practice! – Duthopi Jun 3 '19 at 7:07
If you're gonna use ASDF to measure periodicity and you want to window the data with something other than rectangular window, do it after subtracting and squaring:
$$Q_x[k, n_0] \triangleq \frac{2}{N} \sum\limits_{n=0}^{N-1} \left(x[n+n_0-\left\lfloor \tfrac{N+k}{2}\right\rfloor] \ - \ x[n+n_0-\left\lfloor \tfrac{N+k}{2}\right\rfloor + k] \right)^2 w\left(\tfrac{n}{N}\right)$$
where
$$\left\lfloor \cdot \right\rfloor$$ is the floor() function and, if $$k$$ is even then $$\left\lfloor \frac{k}{2}\right\rfloor = \left\lfloor \frac{k+1}{2}\right\rfloor = \frac{k}{2}$$
and $$w\left(\tfrac{n}{N}\right)$$ is a window function of non-zero width $$N$$ samples centered at $$n=\tfrac{N}{2}$$. A Hann window would be
$$w(u) \triangleq \begin{cases} \tfrac12 - \tfrac12 \cos(2\pi u) \qquad & 0 \le u < 1 \\ \\ 0 & \text{otherwise} \\ \end{cases}$$
To make this ASDF into "autocorrelation" (in the neighborhood of sample $$x[n_0]$$) defined from the ASDF:
$$R_x[k,n_0] = R_x[0,n_0] - \tfrac12 Q_x[k, n_0]$$
where
$$R_x[0, n_0] \triangleq \frac{2}{N} \sum\limits_{n=0}^{N-1} \Big(x[n+n_0-\left\lfloor \tfrac{N}{2}\right\rfloor]\Big)^2 w\left(\tfrac{n}{N}\right)$$
Since $$Q_x[0, n_0] = 0$$ and $$Q_x[k, n_0] \ge 0$$ for all lags $$k$$, that means that $$R_x[k, n_0] \le R_x[0, n_0]$$ for all lags $$k$$.
Suppose for a minute that $$x[n]$$ is periodic with period $$P$$ (and $$P$$ happens to be an integer), then
$$x[n+P] = x[n] \quad \forall n$$
and $$Q_x[mP, n_0] = 0$$ and $$R_x[mP, n_0] = R_x[0, n_0] \ge R_x[k, n_0]$$ for any integer number of periods ($$m$$ is an integer). So you get a peak at $$k=0$$ and at $$k$$ equal to any other multiple of $$P$$ if $$x[n]$$ is periodic. If $$x[n]$$ is not perfectly periodic, what we might expect is the biggest peak at $$k=0$$, another peak (but slightly smaller) at $$k=P$$ (the period we are looking for) and progressively smaller peaks for larger multiples of $$P$$.
The measure of periodicity in the neighborhood of $$n_0$$ would be
$$\frac{R_x[P, n_0]}{R_x[0, n_0]}$$
Hmmm, I just thought of something. Because of my audio/music-centric POV, I have been assuming no DC bias (i.e. $$x[n]$$ should maybe come out of a DC blocking high-pass filter). But I don't think that is the case with your application. So maybe just stick with ASDF, but somewhere you will need a threshold for how low $$Q_x[k, n_0]$$ can go (for $$k \ne 0$$) for you to consider that $$x[n]$$ is periodic in the neighborhood of $$n_0$$.
• Many thanks for the detailed answer. I edited my post to show my attempt to implement it, and am getting closer to what I need! I was just not sure about what you mean by 'you need a threshold for how low Q can go'? If you have any advice on how to find the optimal parameters for window, step and k-values, I am also interested, as this took me a lot of trying out... – Duthopi Jun 4 '19 at 16:38
• $Q_x[k, n_0]$ is always non-negative and $Q_x[P, n_0]$ gets close to zero when there is a match of periodicity. I am not sure how low it must go for you to be able to call it "periodic" in the neighborhood of $n_0$. – robert bristow-johnson Jun 4 '19 at 20:09
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2021-04-20 03:10:29
|
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|
https://aatila.com/tag/oxide-glasses/
|
Oxide Glasses
Atomistic insights into the mixed-alkali effect in phosphosilicate glasses
Oxide glasses have proven useful as bioactive materials, owing to their fast degradation kinetics and tunable properties. Hence, in recent years tailoring the properties of bioactive glasses through compositional design have become the subject of …
Atomistic insights into the structure and elasticity of densified 45S5 bioactive glass
Glasses have applications in regenerative medicine due to their bioactivity, enabling interactions with hard and soft tissues. Soda-lime phosphosilicate glasses, such as 45S5, represent a model system of bioactive glasses. Regardless of their …
Oxide Glasses
Study the properties of oxide glasses
Atomic Structure and Modifiers Clustering in Silicate glasses: Effect of Modifier Cations
Oxide glasses are generally formed by a network of glass former polyhedra, and network modifiers having the role, either to neutralize/ stabilize the charge of the glass former polyhedra or depolymerize the glass network. The effect of the modifier …
Ionic Self-Diffusion and the Glass Transition Anomaly in Aluminosilicates
The glass transition temperature (T$_g$) is the temperature, after which the supercooled liquid undergoes a dynamical arrest. Usually, the glass network modifiers (e.g., Na$_2$O) affect the behavior of T$_g$. However, in aluminosilicate glasses, the …
Ionic Self-Diffusion and the Glass Transition Anomaly in Aluminosilicates
The glass transition temperature (Tg) is the temperature, after which the supercooled liquid undergoes a dynamical arrest. Usually, the glass network modifiers (e.g., Na2O) affect the behavior of Tg. However, in aluminosilicate glasses, the effect of …
Alumina effect on the structure and properties of calcium aluminosilicate in the percalcic region: A molecular dynamics investigation
We rely on molecular dynamics simulations to investigate and discuss the effect of alumina content on the thermodynamic, elastic and structural properties of calcium aluminosilicate glasses in the light of available experimental data. The alumina …
Atomistic insights into the impact of charge balancing cations on the structure and properties of aluminosilicate glasses
Ternary aluminosilicate glasses are of great interest in glass and earth sciences. The structural role of the non-network cations is not fully understood until now. Understanding the structural effect of the non-network cations is necessary for …
Computational insights into the structure of barium titanosilicate glasses
Understanding the role of TiO$_2$ in BaO‐TiO$_2$‐SiO$_2$ (BTS) glasses is one of the keys to develop new glasses and glass‐ceramics for different technological applications. For the first time, molecular dynamics simulations were conducted to get new …
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2022-05-24 12:40:42
|
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http://geoexamples.blogspot.com/2014/07/
|
## Monday, July 7, 2014
### Using the D3 trail layout to draw the Hayian tracks
I wrote many examples (1, 2, 3 and 4) and some entries in the blog (1 and 2) showing how to draw animated paths on a map using the D3 library.
But since then, Benjamin Schmidt wrote a D3 layout, called trail layout, that simplifies a lot doing this kind of stuff.
Since the layout is new, and hasn't got many examples (actually, two made by the author), I'll try to show how to work with it.
### The trail layout
How does the trail layout work? The author defines it as:
This is a layout function for creating paths in D3 where (unlike the native d3.svg.line() element) you need to apply specific aesthetics to each element of the line.
Basically, the input is a set of points, and the layout takes them and creates separate segments to join them. This segments can be either line or d SVG elements.
#### Let's see the simplest example:
var width = 600,
height = 500;
var points = [{"x":0,"y":0}, {"x":200,"y":200}, {"x":0,"y":400}, {"x":200,"y":100}];
var svg = d3.select("body").append("svg")
.attr("width", width)
.attr("height", height);
var trail = d3.layout.trail().coordType('xy');
var trail_layout = trail.data(points).layout();
paths = svg.selectAll("line").data(trail_layout);
paths.enter()
.append('line')
.style("stroke-width",3)
.style("stroke","black")
.attr("x1",function(d) {return d.x1})
.attr("y1",function(d) {return d.y1})
.attr("y2",function(d) {return d.y2})
.attr("x2",function(d) {return d.x2})
• In this case, the points are defined as an array of objects with the x and y properties. If the x and y are named this way, the layout takes them directly. If they are called for instance lon and lat, the layout must be told how to get them.
• Line 10 creates the SVG
• Line 14 initializes the layout. In this case, the layout is using the coordType xy, which means that as a result will give the initial and end point for each segment, convenient for drawing SVG line elements. The other option is using the coordinates value, which is convenient for drawing d elements, as we will see later.
• Line 15 is where the data is set and the layout is retrieved
• The last step is where the lines are actually drawn.
• For each data element, the line svg is added
• The styles are applied
• The extremes of the line are set using the attributes x1, y1, x2, y2
#### How to use coordinates as the coordType:
The following example created the trail as a set of SVG line elements, but the trail layout has an option for creating it as a set of SVG d elements (paths).
You can see the example here. The data, in this case, is the Hayian track. As you can see, it's quite similar as the former example, with the following differences:
• Since in this case we are using geographical coordinates, a projection must be set, and also a d3.geo.path to convert the data into x and y positions, as usual when drawing d3 maps
• When initializing the trail layout, coordinates must be set as the coordType.
• Since the data elements do not store the positions with the name x and y, the layout has to be told how the retrieve them using the positioner:
.positioner(function(d) {return [d.lon, d.lat];})
• When drawing the trail, a path element is appended instead the line element, and the d attribute is set with the path function defined above.
### Creating the map with the trail
Once the basic usage of the trail layout is known, let's reproduce the Hayian path example (simplified for better understanding):
.map {
fill: none;
stroke: #777;
stroke-opacity: .5;
stroke-width: .5px;
}
.land {
fill: #999;
}
.boundary {
fill: none;
stroke: #fff;
stroke-width: .5px;
}
var width = 600,
height = 500;
var projection = d3.geo.mercator()
.scale(5*(width + 1) / 2 / Math.PI)
.translate([width / 2, height / 2])
.rotate([-125, -15, 0])
.precision(.1);
var path = d3.geo.path()
.projection(projection);
d3.json("/mbostock/raw/4090846/world-50m.json", function(error, world) {
d3.json("track.json", function(error, track) {
var color_scale = d3.scale.quantile().domain([1, 5]).range(colorbrewer.YlOrRd[5]);
var svg = d3.select("body").append("svg")
.attr("width", width)
.attr("height", height);
var trail = d3.layout.trail()
.positioner(function(d) {return projection([d.lon,d.lat]);})
.coordType('xy');
var trail_layout = trail.data(track).layout();
svg.insert("path", ".map")
.datum(topojson.feature(world, world.objects.land))
.attr("class", "land")
.attr("d", path);
svg.insert("path", ".map")
.datum(topojson.mesh(world, world.objects.countries, function(a, b) { return a !== b; }))
.attr("class", "boundary")
.attr("d", path);
var hayan_trail = svg.selectAll("d").data(trail_layout);
hayan_trail.enter()
.append('line')
.attr("x1",function(d) {return d.x1})
.attr("x2",function(d) {return d.x1})
.attr("y1",function(d) {return d.y1})
.attr("y2",function(d) {return d.y1})
.attr("class","line")
.style("stroke-width",4)
.attr("stroke", function(d){return color_scale(d.class);})
.transition()
.ease("linear")
.delay(function(d,i) {return i*500})
.duration(500)
.attr("x2",function(d) {return d.x2})
.attr("y2",function(d) {return d.y2})
;
});
});
• The map creation is as usual (explained here)
• Lines 49 to 51 create the trail layout as in the former example
• Line 67 creates the trail, but with some differences:
• the beginning and the end of the line are the same point at the beginning, so the line is not drawn at this moment (lines 69 to 72)
• The stroke colour is defined as a function of the typhoon class using the colour scale (line 75)
• A transition is defined to create the effect of the line drawing slowly
• The ease is defined as linear, important in this case where we join a transition for each segment.
• The delay is set to draw one segment after the other. The time (500 ms) must be the same as the one set at duration
• Finally, the changed values are x2 and y2, that is, the final point of the line, which are changed to their actual values
• The complete example, with the typhoon icon and the date is also available
It's possible to use paths instead of lines to draw the map, as in the first version. The whole code is here, but the main changes are in the last section:
hayan_trail.enter()
.append('path')
.attr("d", path)
.style("stroke-width",7)
.attr("stroke", function(d){return color_scale(d.class);})
.style('stroke-dasharray', function(d) {
var node = d3.select(this).node();
if (node.hasAttribute("d")){
var l = d3.select(this).node().getTotalLength();
return l + 'px, ' + l + 'px';
}
})
.style('stroke-dashoffset', function(d) {
var node = d3.select(this).node();
if (node.hasAttribute("d"))
return d3.select(this).node().getTotalLength() + 'px';
})
.transition()
.delay(function(d,i) {return i*1000})
.duration(1000)
.ease("linear")
.style('stroke-dashoffset', function(d) {
return '0px';
});
• The strategy here is to change the stroke-dasharray and stroke-dashoffset style values as in this example, and changing it later so the effect takes place.
• At the beginning, both values are the same length as the path. This way, the path doesn't appear. The length is calculated using the JavaScript function getTotalLength
• After the transition, the stroke-offset value will be 0, and the path is fully drawn
### Conclusion
I recommend using the trail layout instead of the method from my old posts. It's much cleaner, fast, easy, and let's changing each segment separately.
The only problem I find is that when the stroke width gets thicker, the angles of every segment make strange effects, because the union between them doesn't exist.
This didn't happen with the old method. I can't imagine how to avoid this using lines, but using the coordinates option could be solved transforming the straight lines for curved lines.
|
2019-01-17 02:51:47
|
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|
https://wiseodd.github.io/techblog/2017/01/29/infogan/
|
# InfoGAN: unsupervised conditional GAN in TensorFlow and Pytorch
Generative Adversarial Networks (GAN) is one of the most exciting generative models in recent years. The idea behind it is to learn generative distribution of data through two-player minimax game, i.e. the objective is to find the Nash Equilibrium. For more about the intuition and implementation of GAN, please see my previous post about GAN and CGAN.
Note, the TensorFlow and Pytorch code could be found here: https://github.com/wiseodd/generative-models.
One natural extension of GAN is to learn a conditional generative distribution. The conditional could be anything, e.g. class label or even another image.
However, we need to provide those conditionals manually, somewhat similar to supervised learning. InfoGAN, therefore, attempted to make this conditional learned automatically, instead of telling GAN what that is.
## InfoGAN intuition
Recall, in CGAN, the generator network has an additional parameter: $$c$$, i.e. $$G(z, c)$$, where $$c$$ is a conditional variable. During training, $$G$$ will learn the conditional distribution of data $$P(X \vert z, c)$$. Although principally what CGAN and InfoGAN learn is the same distribution: $$P(X \vert z, c)$$, what different is how they see $$c$$.
In CGAN, $$c$$ is assumed to be semantically known, e.g. labels, so during training we have to supply it. In InfoGAN we assume $$c$$ to be unknown, so what we do instead is to put a prior for $$c$$ and infer it based on the data, i.e. we want to find posterior $$P(c \vert X)$$.
As $$c$$ in InfoGAN is inferred automatically, InfoGAN could assign it to anything related to the distribution of data, depending to the choice of the prior. For example, although we could not specify what $$c$$ should encodes, we could hope that InfoGAN captures label information into it by assigning a Categorical prior. Another example, if we assign a Gaussian prior for $$c$$, InfoGAN might assign a continuous propery for $$c$$, e.g. rotation angle.
So how does InfoGAN do that? This is when information theory takes part.
In information theory, if we want to express the knowledge about something if we know something else, we could use mutual information. So, if we maximize mutual information, we could find something that could contribute to the knowledge of another something the most. In our case, we want to maximize the knowledge about our conditional variable $$c$$, if we know $$X$$.
The InfoGAN mutual information loss is formulated as follows:
where $$H(c)$$ is the entropy of the prior $$P(c)$$, $$G(z, c)$$ is the generator net, and $$Q(c \vert X)$$ is a neural net that takes image input and producing the conditional $$c$$. $$Q(c \vert X)$$ is a variational distribution to model the posterior $$P(c \vert X)$$, which we do not know and as in any Bayesian inference, it is often hard to compute.
This mutual information term fits in the overall GAN loss as a regularization:
where $$V(D, G)$$ is GAN loss.
## InfoGAN training
During training, we provide a prior $$P(c)$$, which could be any distribution. In fact, we could add as many priors as we want, and InfoGAN might assign different properties to them. The author of InfoGAN called this as “disentangled representations”, as it kind of breaking down the properties of data into several conditional parameters.
The training process is similar for discriminator net $$D(X)$$ and generator net $$G(z, c)$$ is quite similar to CGAN, which could be read further here. The differences, however, are:
• instead of $$D(X, c)$$, we use discriminator as in vanilla GAN: $$D(X)$$, i.e. unconditional discriminator,
• instead of feeding observed data for the $$c$$, e.g. labels, into $$G(z, c)$$, we sample $$c$$ from prior $$P(c)$$.
In addition to $$D(X)$$ and $$G(z, c)$$, we also train $$Q(c \vert X)$$ so that we could compute the mutual information. What we do is to sample $$c \sim P(c)$$ and use it to sample $$X \sim G(z, c)$$ and finally pass it to $$Q(c \vert X)$$. The result, along with prior $$P(c)$$ are used to compute the mutual information. The mutual information is then backpropagated to both $$G$$ and $$Q$$ to update both networks so that we could maximize the mutual information.
## InfoGAN implementation in TensorFlow
The implementation for vanilla and conditional GAN could be found here: GAN, CGAN. We will focus on the additional implementation for InfoGAN in this section.
We will implement InfoGAN for MNIST data, with $$c$$ categorically distributed, i.e. one-hot vector with ten elements.
As seen in the loss function of InfoGAN, we need one additional network, $$Q(c \vert X)$$:
Q_W1 = tf.Variable(xavier_init([784, 128]))
Q_b1 = tf.Variable(tf.zeros(shape=[128]))
Q_W2 = tf.Variable(xavier_init([128, 10]))
Q_b2 = tf.Variable(tf.zeros(shape=[10]))
theta_Q = [Q_W1, Q_W2, Q_b1, Q_b2]
def Q(x):
Q_h1 = tf.nn.relu(tf.matmul(x, Q_W1) + Q_b1)
Q_prob = tf.nn.softmax(tf.matmul(Q_h1, Q_W2) + Q_b2)
return Q_prob
that is, we model $$Q(c \vert X)$$ as a two-layer net with softmax on top. The choice of softmax is because $$c$$ is categorically distributed, and softmax could pose as its parameter. If we choose $$c$$ to be Gaussian, then we could design the network so that the outputs are mean and variance.
Next, we specify our prior:
def sample_c(m):
return np.random.multinomial(1, 10*[0.1], size=m)
which is a categorical distribution, with equal probability for each of the ten elements.
As training $$D$$ and $$G$$ is not different than vanila GAN and CGAN, we will omit it from this section. To train $$Q$$, as seen in the regularization term above, we first sample $$c$$ from $$P(c)$$, and use it to sample $$X$$ from $$Q(X \vert z, c)$$:
G_sample = generator(Z, c)
Q_c_given_x = Q(G_sample)
during runtime, we will populate $$c$$ with values from sample_c().
Having all ingredients in hands, we could compute the mutual information term, which is the conditional entropy of the prior and our variational distribution, plus the entropy of our prior:
cond_ent = tf.reduce_mean(-tf.reduce_sum(tf.log(Q_c_given_x + 1e-8) * c, 1))
ent = tf.reduce_mean(-tf.reduce_sum(tf.log(c + 1e-8) * c, 1))
Q_loss = cond_ent + ent
Then, we optimize both $$G$$ and $$Q$$, based on that:
Q_solver = tf.train.AdamOptimizer().minimize(Q_loss, var_list=theta_G + theta_Q)
We initialized the training as follows:
for it in range(1000000):
""" Sample X_real, z, and c from priors """
X_mb, _ = mnist.train.next_batch(mb_size)
Z_noise = sample_Z(mb_size, Z_dim)
c_noise = sample_c(mb_size)
""" Optimize D """
_, D_loss_curr = sess.run([D_solver, D_loss],
feed_dict={X: X_mb, Z: Z_noise, c: c_noise})
""" Optimize G """
_, G_loss_curr = sess.run([G_solver, G_loss],
feed_dict={Z: Z_noise, c: c_noise})
""" Optimize Q """
sess.run([Q_solver], feed_dict={Z: Z_noise, c: c_noise})
After training, we could see what property our prior $$c$$ encodes. In this experiment, our $$c$$ will encode label property nicely, i.e. if we pass c = [0, 0, 1, 0, 0, 0, 0, 0, 0, 0], we might get this:
Note, naturally, there is no guarantee on the ordering of $$c$$.
We could try different values for $$c$$:
We could see that our implementation of InfoGAN could capture the conditional variable, which in this case is the labels, in unsupervised manner.
## Conclusion
In this post we learned the intuition of InfoGAN: a conditional GAN trained in unsupervised manner.
We saw that InfoGAN learns to map the prior $$P(c)$$, together with noise prior $$P(z)$$ into data distribution $$P(X \vert z, c)$$ by adding maximization of the mutual information between $$c$$ and $$X$$ into GAN training. The rationale is that at the maximum mutual information between those two, they can explain each other well, e.g. $$c$$ could explain why $$X \sim P(X \vert z, c=c)$$ are a images of the same digit.
We also implemented InfoGAN in TensorFlow, which as we saw, it is a simple modification from the original GAN and CGAN.
The full code, both TensorFlow and Pytorch implementations are available in: https://github.com/wiseodd/generative-models.
## References
1. Chen, Xi, et al. “Infogan: Interpretable representation learning by information maximizing generative adversarial nets.” Advances in Neural Information Processing Systems. 2016.
|
2018-06-24 22:28:31
|
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|
https://diffgeom.subwiki.org/w/index.php?title=Frenet-Serret_frame&oldid=569
|
# Frenet-Serret frame
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Definition
Let $\gamma$ be a regular curve (for convenience, unit speed-parametrized) in $\R^3$. The Frenet-Serret frame or Serret-Frenet frame of $\gamma$ associates, to each point on $\gamma$, an orthonormal basis at that point. The orthonormal basis comprises the followign unit vectors: the unit tangent, the unit normal and the unit binormal.
The Frenet-Serret frame keeps changing in direction as we move along the curve, and this change in direction is characterized by the Frenet-Serret equations, which show that the relative rate of change depends only on the curvature and torsion. Thus, the geometry of a unit-speed curve depends only on the values of curvature and torsion, as scalar functions.
|
2020-06-02 14:48:06
|
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|
https://quantumcomputing.stackexchange.com/questions/15554/how-to-describe-the-state-of-a-qubit-passing-through-two-hadamard-gates/15556
|
# How to describe the state of a qubit passing through two Hadamard gates? [duplicate]
Describe the state of the qubit at points A, B and C. How does this demonstrate that we need the “ket” (or the vector) representation of qubits, rather than just describing them in terms of probabilities (e.g. “this is a qubit with a 50% probability of being a 1”).
By definition, the Hadamard gate can be written in the computational basis as:
$$H = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix}$$
Now, note that applying a Hadamard gate to the state, says, $$|\psi \rangle = |0\rangle$$, then it is equivalent to be doing the following rotation:
So what if you apply another Hadamard gate? What would happen? Well, firs note that he inverse of Hadamard gate is itself. That is:
$$H^{-1} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix} = H$$
Therefore, you apply another Hadamard gate then you will get back to the same spot as indicated by the picture below:
The takeaway is:
$$HH|\psi\rangle = H H^{-1} |\psi \rangle = I |\psi \rangle = |\psi\rangle \ \ \ \textrm{where I is the Identity operator.}$$
Hence, applying the Hadmard gate twice to the state $$|\psi\rangle$$ will keep the state as $$|\psi\rangle$$.
This doesn't change even if $$|\psi \rangle$$ is in some superposition state. That is, suppose, $$|\psi \rangle = \sqrt{\dfrac{2}{3}}|0\rangle + \sqrt{\dfrac{2}{3}}|1\rangle$$,
and if you apply a Hadamard gate, then geometrically, it is equivalent as doing:
and if you apply another Hadamard gate you will get back to the original vector/state $$|\psi \rangle = \sqrt{\dfrac{2}{3}}|0\rangle + \sqrt{\dfrac{2}{3}}|1\rangle$$
So if starting at the stage $$A$$, you have arbitrary state $$|\psi_A \rangle = \begin{pmatrix} \cos \dfrac{\theta}{2} \\ e^{i \phi} \sin \dfrac{\theta}{2} \end{pmatrix}$$ then by applying Hadamard gate you get:
$$H |\psi \rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix} \begin{pmatrix} \cos \dfrac{\theta}{2} \\ e^{i \phi} \sin \dfrac{\theta}{2} \end{pmatrix} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} \cos \dfrac{\theta}{2} + e^{i \phi} \sin \dfrac{\theta}{2} \\ \cos \dfrac{\theta}{2} - e^{i \phi} \sin \dfrac{\theta}{2} \end{pmatrix}$$
So at stage $$B$$, your qubit is in the state $$|\psi_B \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} \cos \dfrac{\theta}{2} + e^{i \phi} \sin \dfrac{\theta}{2} \\ \cos \dfrac{\theta}{2} - e^{i \phi} \sin \dfrac{\theta}{2} \end{pmatrix}$$
and like what we discussed above, applying another Hadamard gate will get it back to the starting vector $$|\psi_A\rangle$$. That is, the qubit at stage $$C$$ have the state $$|\psi_C \rangle = |\psi_A\rangle$$
• So what is the quit at B and C? – Oliver Custance Jan 17 at 17:06
• It depends on what $|\psi\rangle$ is, but the state of the qubit at $C$ is the same as the state of the qubit at $A$. – KAJ226 Jan 17 at 17:21
• Perfect, thanks very much for answering this. – Oliver Custance Jan 17 at 17:24
• @Oliver You are welcome. – KAJ226 Jan 17 at 17:35
Qubit states at points A, B and C
Let the input pure state be $$|\psi\rangle = a|0\rangle + b|1\rangle$$. At point A the state is simply
$$|\psi_A\rangle = |\psi\rangle = a|0\rangle + b|1\rangle$$
since there are no gates that change the input state before the point A. At point B the state is
$$|\psi_B\rangle = H|\psi\rangle = \frac{a}{\sqrt{2}}(|0\rangle + |1\rangle) + \frac{b}{\sqrt{2}}(|0\rangle - |1\rangle) = \frac{a+b}{\sqrt{2}}|0\rangle + \frac{a-b}{\sqrt{2}}|1\rangle$$
due to the action of the first Hadamard
$$H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$
Finally, at point C the state is
$$|\psi_C\rangle = HH|\psi\rangle = a|0\rangle + b|1\rangle$$
because Hadamard is self-inverse, i.e. $$HH = I$$, as can be checked by matrix multiplication.
Why superpositions cannot be described as probabilistic mixtures?
The behavior of the quantum circuit in the question rules out description of quantum superpositions as probabilistic mixtures (e.g. the qubit is in the $$0$$ state with probability 50% and in the $$1$$ state with probability 50%), because the process of forming a probabilistic mixture is not invertible. In particular, applying it twice does not return back to the initial state as it does in the case of the Hadamard gate. There are two compatible explanations of this fact: one from purely mathematical perspective and the other from the physics point of view.
Mathematically, a process of forming an equal probabilistic mixture is described by the stochastic matrix
$$\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}.$$
This does not describe a quantum gate because - unlike the Hadamard above - it is not invertible. In particular, applying it twice to a vector does not return it to its original state. Note that by the postulates of quantum mechanics evolution of any closed quantum system is unitary and hence invertible.
Physically, the description of superpositions as probabilistic mixtures turns out to be insufficient because probabilities cannot interfere destructively whereas amplitudes of a quantum states can. Indeed, we can see destructive interference in action as it cancels out the amplitude of the $$|1\rangle$$ state when we compute $$HH|0\rangle$$ step by step
\begin{align} HH|0\rangle &= \frac{1}{\sqrt{2}}H|0\rangle + \frac{1}{\sqrt{2}}H|1\rangle\\ &= \frac{1}{2}(|0\rangle +|1\rangle)+ \frac{1}{2}(|0\rangle -|1\rangle)\\ &= \left(\frac{1}{2} + \frac{1}{2}\right)|0\rangle + \left(\color{red}{\frac{1}{2} - \frac{1}{2}}\right)|1\rangle\\ &= 1|0\rangle + \color{red}{0}|1\rangle =\\ & = |0\rangle. \end{align}
Probabilities of terms in a mixture are never subtracted this way, because probabilities cannot be negative.
• Awesome, thanks very much for answering my question. – Oliver Custance Jan 17 at 18:01
|
2021-04-16 16:19:03
|
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|
http://www.bmj.com/content/335/7633/1290
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Mixed Messages
# Screening programme evaluation applied to airport security
BMJ 2007; 335 (Published 20 December 2007) Cite this as: BMJ 2007;335:1290
1. Eleni Linos, doctoral student1,
2. Elizabeth Linos, research assistant23,
3. Graham Colditz, associate director4
1. 1Department of Epidemiology, Harvard School of Public Health, Boston, MA 02115, USA
2. 2Department of Economics, Harvard University, Littauer Center, Cambridge, MA, USA
3. 3J-Poverty Action Laboratory, Massachusetts Institute of Technology, Cambridge, MA 02139-4307, USA
4. 4Prevention and Control, Siteman Cancer Center, Washington University School of Medicine, Campus Box 8109, St Louis, MO 63110, USA
1. Correspondence to: E Linos elinos{at}hsph.harvard.edu
Eleni Linos, Elizabeth Linos, and Graham Colditz investigate whether airport security screening would pass the National Screening Committee’s criteria for an effective screening test
### The tests and evidence of benefit
We systematically reviewed the literature on airport security screening tools. A systematic search of PubMed, Embase, ISI Web of Science, Lexis, Nexis, JSTOR, and Academic Search Premier (EBSCOhost) found no comprehensive studies that evaluated the effectiveness of x ray screening of passengers or hand luggage, screening with metal detectors, or screening to detect explosives. When research teams requested such information from the US Transportation Security Administration they were told that evaluating new screening programmes might be useful, but it was overshadowed by “time pressures to implement needed security measures quickly.”16 In addition, we noticed that new airport screening protocols were implemented immediately after news reports of terror threats (fig 1).
Fig 1 Timeline of changes to airport screening protocols, costs, and news events related to terrorist threats
The little we do know about airport security screening comes from investigations of the factors that influence the sensitivity of visual screening of x ray images. These studies conclude that sensitivity depends on the screener’s experience, rather than the precision of the machine. Practice improves the screener’s performance, but unfamiliar or rare objects are hard to identify regardless of experience.171819 Mammography radiologists realise this and undergo years of specialised training after medical school.20
Even without clear evidence of the accuracy of testing, the Transportation Security Administration defended its measures by reporting that more than 13 million prohibited items were intercepted in one year.21 Most of these illegal items were lighters. The screening literature shows that length time and lead time bias produce misleading interpretations of screening studies because of earlier detection of more benign cases that would not necessarily become clinically apparent (overdiagnosis). A similar problem arises with the above reasoning—although more than a million knives were seized in 2006, we do not know how many would have led to serious harm.
## The questions
The absence of scientific evaluations of the screening tools currently in place and the vast amount of money spent by governments worldwide on airport security have led us to muse over current airport security protocols and wonder about their optimal implementation. What is the sensitivity of the screening question, “Did you pack all your bags yourself?” and has anyone ever said no? Can you hide anything in your shoes that you cannot hide in your underwear? What are the ethical implications of preselecting high risk groups? Are new technologies that “see” through clothes acceptable? What hazards should we screen for? Guns and explosives certainly, but what about radioactive materials or infectious pathogens? Concerns about cost effectiveness—including the indirect costs of passengers’ time spent in long queues—will be central to future decisions, but first we need solid evidence of benefit.
## An experiment
If we were to evaluate the effectiveness of airport screening, we would start by assessing the accuracy of current tests for illegal objects in passengers’ luggage. This would yield only preliminary information on screening test performance; we would need to reapply for funding to evaluate the overall benefit of security screening on mortality and calculate the number needed to screen to prevent the death of one traveller.22 After informing the airport managers, gaining approval from research ethics committees and police, and registering our trial with one of the acceptable International Committee of Medical Journal Editors trial registries, we would select passengers at random at the check-in desks and give each traveller a small wrapped package to put in their carry-on bags. (We would do this after they have answered the question about anyone interfering with their luggage.) A total of 600 passengers would be randomised to receive a package, containing a 200 ml bottle of a non-explosive liquid, a knife, or a bag of sand of similar weight (control package) in a 1:1:1 ratio. Investigators and passengers would be blinded to the contents of the package. Our undercover investigators would measure how long it takes to get through security queues and record how many of the tagged customers are stopped and how many get through. A passenger who is stopped and asked to open the wrapped box would be classed as a positive test result, and any unopened boxes would be considered a negative test result. We would use the number of true and false positives and true and false negatives to estimate the sensitivity and specificity of the current screening process and pool the waiting times to estimate an average waiting time for each passenger (fig 2).
Fig 2 Study design flow chart for evaluation of current screening test for hand luggage
We have heard rumours that this sort of thing actually goes on—that agents occasionally carry illicit items through airport screening units to “test” them and identify gaps in security. Perhaps the evidence we are searching for is strong, but secret. And of course rigorous airport screening may have other benefits. It certainly deters the transport of any illicit object, such as less dangerous but equally unwanted plants, animals, or drugs. In addition, in the midst of mounting reports of thwarted terrorist attacks on airports, the process is comforting to frequent flyers and their families. Nevertheless, the absence of publicly available evidence to satisfy even the most basic criteria of a good screening programme concerns us.
## Conclusion
Of course, we are not proposing that money spent on unconfirmed but politically comforting efforts to identify and seize water bottles and skin moisturisers should be diverted to research on cancer or malaria vaccines. But what would the National Screening Committee recommend on airport screening? Like mammography in the 1980s, or prostate specific antigen testing and computer tomography for detecting lung cancer more recently, we would like to open airport security screening to public and academic debate. Rigorously evaluating the current system is just the first step to building a future airport security programme that is more user friendly and cost effective, and that ultimately protects passengers from realistic threats.
## Footnotes
• Thanks to Lorelei Mucci, Monica McGrath, Mike Stoto, and Pat Cox for useful discussions.
• Contributors and sources: Eleni L and GC conceived and designed the study. All authors helped collect data and write and edit the manuscript. Eleni L is guarantor. GC has worked extensively on breast and colorectal cancer screening and advises the American Cancer Society on implementation of screening programmes.
• Funding: NIH grant R25 CA098566 provided salary support for Eleni L. The funder had no role in study design; in the collection, analysis, and interpretation of data; in the writing of the report; and in the decision to submit the article for publication.
• Competing interests: None declared.
• Provenance and peer review: Not commissioned; externally peer reviewed.
## References
View Abstract
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2018-03-23 11:47:36
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http://www.ams.org/mathscinet-getitem?mr=2423999
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MathSciNet bibliographic data MR2423999 (2009d:35264) 35Q35 (35P05 35P15 47F05 76N10) Pribylʹ, M. A. Spectral analysis of linearized stationary equations of a viscous compressible fluid, defined in \$\Bbb R\sp 3\$$\Bbb R\sp 3$ with periodic boundary conditions. (Russian) Algebra i Analiz 20 (2008), no. 2, 149--177; translation in St. Petersburg Math. J. 20 (2009), no. 2, 267–288 Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
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2014-09-30 22:42:24
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https://www.physicsforums.com/threads/expression-for-magnitude-of-electric-field-by-dipole-integral.877414/
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# Expression for Magnitude of Electric Field by dipole integral
## Homework Statement
Consider the electric dipole seen in the notes. (a) Using integration, derive an expression for the magnitude of the electric field produced by the dipole at any point along the x-axis.
Electric Dipole: http://labman.phys.utk.edu/phys136/modules/m5/images/electr5.gif
## Homework Equations
Electric Field Equation, Differential Form $${d \vec E} =\frac 1 {4\pi\epsilon_0} \frac {dq} {r^2} \hat {\mathbf r}$$
Linear Charge Density $$dq = \lambda dx$$
electric dipole: $$\vec p = q\vec d$$
## The Attempt at a Solution
I am completely confused as to how to get started or which equation to integrate to obtain the equation. I am aware that the y-components of the point charges cancel each other out, hence only the charge along the x-axis matters. As well as the general integration needed to be done to obtain the formula.$$\int_{-\infty}^{\infty} F$$ where F is the function to integrate.
Last edited:
blue_leaf77
Homework Helper
The picture you posted doesn't seem to correspond to the problem you are trying to solve. That picture is more inclined for a problem about electric dipole motion under external electric field and has nothing to do with a charged rod. Moreover, where is the mentioned x axis in that picture?
I believe the problem would be attuning to the effects of two point charges creating a dipole along the x-axis, in the picture it would be the E line. Im trying to figure out how to properly set the problem up for integration as I can't seem to figure out how to relate dE and p, so it can be proven with integration all I've been able to find is the derivation using binomial expansion theorems
blue_leaf77
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2021-02-28 10:10:20
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https://nadporamiroku.pl/2018-12-11/24695.html
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# heat resistance weight of 30 m high pressure hose
#### Hosecraft USA Metal Hoses of Every Style
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#### HIGH-TEMPERATURE CHARACTERISTICS OF STAINLESS STEELS
9/3/1972· high-temperature service, strength at temperature is related to time at temperature. Allowable Deformation Another factor to consider in designing for high-temperature service is the amount of deformation that can be permitted during the total service life.
#### Air Conditioning Trouble Shooting FAQ''s from Vintage Air …
1) High Side Pressure (160-250 PSI) * Note - general rule of thu is two times the aient daytime temperature, plus 15 - 20%. 2) Low Side Pressure (6-12 PSI in a steady state). 3) Center Duct Temperature (36-46 Degrees F).
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#### Pressure Washer Hoses | Northern Tool
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#### 5. Thermal insulation materials, technical characteristics …
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#### Characteristic properties of Silicone Rubber Compounds
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#### Masterduct: Lightweight hose, flexible hose, abrasive …
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#### Static Pressure vs. Head in Fluids - Engineering ToolBox
Δp = change in pressure (Pa, psi) Δh = change in height (m, in) γ = specific weight of fluid (N/m 3, lb/ft 3) The pressure gradient in vertical direction is negative - the pressure decrease upwards. Specific Weight Specific Weight of a fluid can be expressed as:
#### Hose, Piping, Tubing and T-Slot Framing | Parker NA
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#### High-density polyethylene - Wikipedia
HDPE is known for its high strength-to-density ratio. The density of HDPE can range from 930 to 970 kg/m 3. Although the density of HDPE is only marginally higher than that of low-density polyethylene, HDPE has little branching, giving it stronger intermolecular forces and …
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#### Outlet Pipe Size for Pump- is a Bigger Pipe Better?
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#### Properties and Appliions of Ni-Resist and Ductile Ni-Resist …
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#### Wire Braid Hydraulic Hose with Single or Double Wire Braid
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#### DESIGN GUIDELINES FOR THE SELECTION AND USE OF STAINLESS …
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#### The Difference Between Pressure and Flow - GPM …
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#### Fundamentals of leak detection - Leybold
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#### 14.7 Viscosity and Turbulence | University Physics Volume 1
A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1.62\,×\,{10}^{6}\,{\text{N/m}}^{2}$. The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Calculate the Reynolds nuers for flow in the fire hose and nozzle to show that the flow in each must be turbulent.
#### : Coleman High-Pressure Propane Hose …
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2021-06-20 16:33:45
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http://mathonline.wikidot.com/the-centralizer-of-a-subset-a-of-a-group-g-cg-a
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The Centralizer of a Subset A of a Group G, CG(A)
# The Centralizer of a Subset A of a Group G, CG(A)
Recall from The Center of a Group G, Z(G) page that if $G$ is a group then the center of $G$ denoted $Z(G)$ is defined to be the set of all elements of $g$ that commute with every element of $G$, that is:
(1)
\begin{align} \quad Z(G) = \{ a \in G : ag = ga\quad \: \forall g \in G \} \end{align}
We proved that $G$ is an abelian group if and only if $G = Z(G)$. Furthermore, we proved that $Z(G)$ is always an abelian subgroup of $G$.
We now generalize this concept.
Definition: Let $G$ be a group and let $A$ be a nonempty subset of $G$. The Centralizer (or Commutant) of $A$ in $G$ denoted $C_G(A)$ is defined to be the set of all elements of $G$ that commute with every element of $G$, that is, $C_G(A) = \{ g \in G : gag^{-1} = a \quad \: \forall g \in G \}$.
Note that $Z(G) = C_G(G)$. By convention, if $A$ is a singleton set, say $A = \{ a \}$ then we denote the centralizer of $A$ in $G$ by $C_G(a)$.
Proposition 1: Let $G$ be a group. Then $Z(G) = \bigcap_{a \in G} C_G(a)$.
• Proof: Let $g \in Z(G)$. Then $ga = ag$ for all $a \in G$. Since $C_G(a) = \{ h \in G : ha = ah \}$ we see that $g \in C_G(a)$ for each $a \in G$}]. Thus [[$g \in \bigcap_{a \in G} C_G(a)$. So $Z(G) \subseteq \bigcap_{a \in G} C_G(a)$.
• Let $g \in \bigcap_{a \in G} C_G(a)$. Then $g \in C_G(a)$ for each $a \in G$. So $ga = ag$ for each $a \in G$. Thus $g \in Z(G)$. So $\bigcap_{a \in G} C_G(a) \subseteq Z(G)$.
• Therefore $Z(G) = \bigcap_{a \in G} C_G(a)$. $\blacksquare$
The following proposition tells us that the centralizer of $A$ in $G$ is always an abelian subgroup of $G$.
Proposition 2: Let $G$ be a group and let $A$ be a nonempty subset of $G$. Then $C_G(A)$ is a subgroup of $G$.
Note that $C_G(A)$ MIGHT NOT BE ABELIAN! For example, if $G$ is a nonabelian group with identity $1$ then $C_G(1) = \{ g \in G : g1g^{-1} = 1 \} = G$, which is nonabelian.
• Proof: Clearly $(C_G(A), \cdot)$ is closed under the operation $\cdot$, for if $g_1, g_2 \in C_G(A)$ then $g_1ag_1^{-1} = a$ and $g_2ag_2^{-1} = a$ for all $a \in A$, so:
(2)
\begin{align} \quad g_1g_2a(g_1g_2)^{-1} = g_1(g_2ag_2^{-1})g_1^{-1} = g_1ag_1^{-1} = a \end{align}
• So $g_1g_2 \in C_G(A)$.
• Now clearly $1 \in C_G(A)$ since by definition $1a1^{-1} = a$ for all $a \in A \subseteq G$.
• Lastly, let $g \in C_G(A)$. Then $gag^{-1} = a$. This equation can be rewritten as $g^{-1}ag = a$. So $g^{-1} \in C_G(A)$.
• So $C_G(A)$ is a subgroup of $G$. $\blacksquare$
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2019-08-20 13:02:03
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https://www.physicsforums.com/threads/convolution-fourier-transform.927745/
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# Convolution - Fourier Transform
## Homework Statement
An LTI system has an impulse response h(t) = e-|t|
and input of x(t) = ejΩt
## Homework Equations
Find y(t) the system output using convolution
Find the dominant frequency and maximum value of y(t)
## The Attempt at a Solution
I have tried using the Fourier transform to get y(t) but when you try to find X(Ω), I get infinity
as X(Ω) = ∫ x(t) * e-jΩtdt = ∫ e jΩt*e-jΩtdt = ∫ 1 dt = t between inf and -inf
h(t) I could find as 2/(1+Ω2)
Any ideas on how to solve this?
Last edited:
Related Engineering and Comp Sci Homework Help News on Phys.org
Homework Helper
Gold Member
2020 Award
You need to write the transform as ## X(\omega)=\int x(t) e^{-i \omega t} dt ##. The result is ## X(\omega)=2 \pi \, \delta(\omega-\Omega) ##. You need to read about delta functions. If you google it, you should find some useful formulas like the one I just gave you.
You need to write the transform as ## X(\omega)=\int x(t) e^{-i \omega t} dt ##. The result is ## X(\omega)=2 \pi \, \delta(\omega-\Omega) ##. You need to read about delta functions. If you google it, you should find some useful formulas like the one I just gave you.
So then using that you would get that Y(w) = 4π/(1+w2) * δ(w-2), but then how would you get that back into the time domain?
Homework Helper
Gold Member
2020 Award
## H(\omega)=\int\limits_{0}^{+\infty} h(t) e^{-i \omega t} dt ##, since ## h(t)=0 ## for ## t<0 ##. (Perhaps they didn't tell you this (## h(t)=0 ## for ## t<0 ##) in the problem statement, but it is clear that that's what they want). Try recomputing ## H(\omega) ##.(You have it incorrect). ## \\ ## Now ## Y(\omega)=H(\omega)X(\omega) ##. (That part you have correct.) ## \\ ## Use an inverse transform to get ## y(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} Y(\omega) e^{i \omega t} \, d \omega ##. ## \\ ## Wait until you process everything to put in the value for ## \Omega ##.
## H(\omega)=\int\limits_{0}^{+\infty} h(t) e^{-i \omega t} dt ##, since ## h(t)=0 ## for ## t<0 ##. Try recomputing ## H(\omega) ##.(You have it incorrect). ## \\ ## Now ## Y(\omega)=H(\omega)X(\omega) ##. (That part you have correct.) ## \\ ## Use an inverse transform to get ## y(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} Y(\omega) e^{i \omega t} \, d \omega ##. ## \\ ## Wait until you process everything to put in the value for ## \Omega ##.
Sorry the reason why I got what i did for H(w), was because I forget to put in an absolute around the t. I've changed in now
Homework Helper
Gold Member
2020 Award
Sorry the reason why I got what i did for H(w), was because I forget to put in an absolute around the t. I've changed in now
An ## h(t) ## with an absolute value would be unphysical. That is saying it responds before the impulse. These functions always begin at ## t=0 ##. If they gave you the problem in such a fashion, it is unphysical.
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2021-01-21 15:28:12
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http://books.duhnnae.com/2017/jul/149887591624-Virtual-Compton-Scattering-measurements-in-the-Nto-transition-Nuclear-Experiment.php
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# Virtual Compton Scattering measurements in the $γ^* N o Δ$ transition - Nuclear Experiment
Virtual Compton Scattering measurements in the $γ^* N o Δ$ transition - Nuclear Experiment - Download this document for free, or read online. Document in PDF available to download.
Abstract: We report on new H$e,e^\prime p\gamma$ measurements in the $\Delta1232$resonance at $Q^2=0.06$ GeV-c carried out simultaneously with H$e,e^\primep\pi^0$. It is the lowest $Q^2$ for which the virtual Compton scattering VCSreaction has been studied in the first resonance region. The VCS measured crosssections are well described by dispersion-relation calculations in which themultipole amplitudes derived from H$e,e^\prime p\pi^0$ data are used asinput, thus confirming the compatibility of the results. The derived resonantmagnetic dipole amplitude $M^{3-2} {1+} = 40.60 \pm0.70 {stat+sys}10^{-3}-m {\pi^+}$ at $W=$ 1232 MeV is in excellent agreementwith the value extracted from H$e,e^\prime p\pi^0$ measurements.
Author: N.F. Sparveris, P. Achenbach, C. Ayerbe Gayoso, D. Baumann, J. Bernauer, A.M. Bernstein, R. Böhm, D. Bosnar, T. Botto, A. Christ
Source: https://arxiv.org/
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2017-10-23 00:58:23
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https://proofwiki.org/wiki/Definition:Multiplicative_Identity
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Definition:Multiplicative Identity
Definition
Let $\left({F, +, \times}\right)$ be a field.
Then the identity element of the multiplicative group $\left({F^*, \times}\right)$ of $F$ is called the multiplicative identity of $F$.
It is often denoted $e_F$ or $1_F$, or, if there is no danger of ambiguity, $e$ or $1$.
Note that the multiplicative identity of $F$ is the unity of the ring that $\left({F, +, \times}\right)$ is by definition of a field.
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2020-01-18 22:34:30
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https://www.konradvoelkel.com/2011/
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# Mindmap on complex analysis in one variable
Monday, November 14th, 2011 | Author:
Here is my mind-map for first-course complex analysis. It contains some well-known theorems and "arrows" between them.
Here it is, and of course you can download it as a PDF or as a SVG (vector graphics) as well (click on the image to enlarge it):
The license is CC-BY-NC-SA (if you redistribute, put my name on it, don't make profit, share alike).
There are some aspects which require an explanation:
Category: English, Mathematics | Comments off
# Properties of Scheme Morphisms
Sunday, November 06th, 2011 | Author:
To prepare for my oral exams in algebraic geometry (covering Hartshorne's book "Algebraic Geometry" Chapter II and III) I sketched an overview diagram of morphism properties in the category of noetherian schemes. Maybe this is a good cheat sheet to keep with you while reading the book for the first or second time (ok, and I dropped a "Nisnevich" for no good reason, you can ignore it).
You can get a PDF version of the image or click on it to get a readable version.
I'm still in the process of writing down examples and counter-examples to these properties, maybe that list will be online some day (another kind of "counterexamples in algebraic geometry").
As always, I'm happy to hear any comments (did I miss an important arrow, did I get anything wrong) -- but I should stress that the diagram works in Hartshorne-world, not in EGA-terms (this kind of confusion cost me almost one entire day trying to prove wrong statements..)
UPDATE (2011-11-18): improved diagram (more information, less colour) and higher quality PNG file.
Category: English, Mathematics | Comments off
# Export purchased books list from Amazon
Sunday, September 18th, 2011 | Author:
If you happened to buy books from Amazon.com (or, in my case, Amazon.de) and maybe used the recommendation engine and the wishlist (and and and ...) then there will be lots of data about your books on the Amazon website. Have you ever thought about organizing your library with a different tool? May it be Google Books or LibraryThing or Shelfari, you will have to export this precious big amount of data from Amazon to the other service. Luckily, some intelligent people invented ISBN, so you basically need to extract a list of ISBNs to identify the books (neglecting your reviews and tags for now). Not that luckily, Amazon doesn't offer such export functionality to the layman. Searching the internet yields a Greasemonkey script that enables you to export wishlist content - but no ISBNs, so import into other services is not so easy.
The solution is to save each website of "your purchases" (or other such lists of books) as HTML file and let a smart script do the extraction work. This way, you're not violating Amazon's terms of service (which most likely don't allow any robots scraping the website) and on the positive side, it works.
# Essential manifolds
Saturday, August 13th, 2011 | Author:
Now I'll explain a little bit what essential manifolds are and what they're good for.
Definition
A (connected closed orientable topological) n-manifold $M$ is called essential, if there exists a continuous map $f : M \to K(\pi_1(M,\ast),1)$ such that the induced morphism on the top homology $f_\ast : H_n(M,\mathbb{Z}) \to H_n(K(\pi_1(M,\ast),1),\mathbb{Z})$ maps the fundamental class $[M] \in H_n(M,\mathbb{Z})$ to some non-zero element $f_\ast([M]) \neq 0 \in H_n(K(\pi_1(M,\ast),1),\mathbb{Z})$.
Category: English, Mathematics | Comments off
# Aspherical manifolds
Wednesday, August 10th, 2011 | Author:
In this post I want to sketch the idea of aspherical manifolds - manifolds which don't admit higher homotopically non-trivial spheres - and the related concepts of Eilenberg-MacLane-spaces and classifying spaces for groups.
Definition
A topological space $M$ is called aspherical if all higher homotopy groups vanish, i.e. $\pi_n(M,m_0) = 0 \quad \forall n > 1$ where $m_0 \in M$ is an arbitrary basepoint and $M$ is assumed to be connected.
Since manifolds admit universal covers, you could equivalently define a manifold to be aspherical if and only if its universal cover is contractible.
Category: English, Mathematics | Comments off
# Diploma thesis (in german)
Tuesday, August 09th, 2011 | Author:
Now this is a slightly corrected (although still somewhat messy) version of my diploma thesis - in german:
Matsumotos Satz und A¹-Homotopietheorie.
You can read something about the content in this blog post, containing an extended abstract in english.
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2019-01-16 20:17:00
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https://www.physicsforums.com/threads/circumference-of-a-4-sphere.675800/
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# Circumference of a 4-sphere
What is the circumference of a four dimensional sphere?
$2\pi R$, I guess.
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2020-10-24 04:09:09
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https://jackrobinson.com/.tmb/2emo1/page.php?30603a=statsmodels-logistic-regression
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statsmodels logistic regression
The formula specifying the model. “Econometric Analysis,” 5th ed., Pearson, 2003. This example file shows how to use a few of the statsmodels regression diagnostic tests in a real-life context. result = model.fit(), 0 1 The summary is as follows. We perform logistic regression when we believe there is a relationship between continuous covariates X and binary outcomes Y. The package contains an optimised and efficient algorithm to find the correct regression parameters. Change ). We will begin by importing the libraries that we will be using. A logistic regression model provides the ‘odds’ of an event. We do logistic regression to estimate B. MacKinnon. This is great. The initial part is exactly the same: read the training data, prepare the target variable. Multiple Regression Using Statsmodels. $$\Sigma=\Sigma\left(\rho\right)$$. Note that most of the tests described here only return a tuple of numbers, without any annotation. we will use two libraries statsmodels and sklearn. Logistic regression with Python statsmodels. ( Log Out / To test our model we will use “Breast Cancer Wisconsin Dataset” from the sklearn package and predict if the lump is benign or malignant with over 95% accuracy. A p x p array equal to $$(X^{T}\Sigma^{-1}X)^{-1}$$. You can learn about more tests and find out more information about the tests here on the Regression Diagnostics page.. $$\Psi$$ is defined such that $$\Psi\Psi^{T}=\Sigma^{-1}$$. It is approximately equal to Is y base 1 and X base 0. ( Log Out / This is equal to p - 1, where p is the Sorry, your blog cannot share posts by email. Why this name? R-squared: 0.353, Method: Least Squares F-statistic: 6.646, Date: Thu, 29 Oct 2020 Prob (F-statistic): 0.00157, Time: 16:00:02 Log-Likelihood: -12.978, No. specific results class with some additional methods compared to the Results class for Gaussian process regression models. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. number of regressors. Then, we’re going to import and use the statsmodels Logit function: You get a great overview of the coefficients of the model, how well those coefficients fit, the overall fit quality, and several other statistical measures. Parameters formula str or generic Formula object. This was done using Python, the sigmoid function and the gradient descent. RollingRegressionResults(model, store, …). The model degrees of freedom. The confidence interval gives you an idea for how robust the coefficients of the model are. An implementation of ProcessCovariance using the Gaussian kernel. The logistic regression function () is the sigmoid function of (): () = 1 / (1 + exp (− ()). ==============================================================================, Dep. You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. I ran an OLS regression using statsmodels. number of observations and p is the number of parameters. Here, we are using the R style formula. Results class for a dimension reduction regression. Post was not sent - check your email addresses! Using the statsmodels package, we perform a series of regressions between life expectancy and Census data. Adapted by R. Jordan Crouser at Smith College for SDS293: Machine Learning (Spring 2016). and can be used in a similar fashion. Peck. All regression models define the same methods and follow the same structure, Fitting a Multiple Linear Regression Model. The whitened design matrix $$\Psi^{T}X$$. Depending on the properties of $$\Sigma$$, we have currently four classes available: GLS : generalized least squares for arbitrary covariance $$\Sigma$$, OLS : ordinary least squares for i.i.d. Tot_percpaid_bin 0.300069 0.490454 X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, random_state=0) Please help, import statsmodels.formula.api as sm Estimate AR(p) parameters from a sequence using the Yule-Walker equations. Depending on the properties of Σ, we have currently four classes available: GLS : generalized least squares for arbitrary covariance Σ. OLS : ordinary least squares for i.i.d. We can now see how to solve the same example using the statsmodels library, specifically the logit package, that is for logistic regression. The following are 14 code examples for showing how to use statsmodels.api.Logit().These examples are extracted from open source projects. I am confused looking at the t-stat and the corresponding p-values. The binary value 1 is typically used to indicate that the event (or outcome desired) occured, whereas 0 is typically used to indicate the event did not occur. Fitting a linear regression model returns a results class. errors with heteroscedasticity or autocorrelation. Earlier we covered Ordinary Least Squares regression with a single variable. The p x n Moore-Penrose pseudoinverse of the whitened design matrix. Is it Maximum Likelihood Estimation. My question is how to interpret the meaning of the coefficient? In stats-models, displaying the statistical summary of the model is easier. if the independent variables x are numeric data, then you can write in the formula directly. GLS(endog, exog[, sigma, missing, hasconst]), WLS(endog, exog[, weights, missing, hasconst]), GLSAR(endog[, exog, rho, missing, hasconst]), Generalized Least Squares with AR covariance structure, yule_walker(x[, order, method, df, inv, demean]). results class of the other linear models. W.Green. endog is an 1-d vector of the endogenous response. specific methods and attributes. This module allows I am not getting intercept in the model? Note that the intercept is not counted as using a The n x n covariance matrix of the error terms: Apply the logistic regression as follows: logistic_regression= LogisticRegression() logistic_regression.fit(X_train,y_train) y_pred=logistic_regression.predict(X_test) Then, use the code below to get the Confusion Matrix: In this case is the final cost minimised after n iterations (cost being – in short – the difference between the predictions and the actual labels). Edu -0.278094 0.220439 The value of the likelihood function of the fitted model. intercept is counted as using a degree of freedom here. But I have issue with my result, the coefficients failed to converged after 35 iterations. Assuming that the model is correct, we can interpret the estimated coefficients as statistica… 10 min. Logitic regression is a nonlinear regression model used when the dependent variable (outcome) is binary (0 or 1). See Module Reference for commands and arguments. Remember that, ‘odds’ are the probability on a different scale. statsmodels.formula.api.logit¶ statsmodels.formula.api.logit (formula, data, subset = None, drop_cols = None, * args, ** kwargs) ¶ Create a Model from a formula and dataframe. We have seen an introduction of logistic regression with a simple example how to predict a student admission to university based on past exam results. Y = X β + μ, where μ ∼ N ( 0, Σ). “Introduction to Linear Regression Analysis.” 2nd. Parameters endog array_like. We have seen an introduction of logistic regression with a simple example how to predict a student admission to university based on past exam results. February 15, 2014. by. Credits: Fabio Rose Introduction. In this post, we’re going to build our own logistic regression model from scratch using Gradient Descent. This class summarizes the fit of a linear regression model. Delay_bin 0.992853 1.068759 Note that the What is the definition of “current function value” ? Logistic Regression using Statsmodels. Change ), You are commenting using your Google account. ( Log Out / Basically y is a logical variable with only two values. This notebook uses the dateframes technique when performing the regression. The whitened response variable $$\Psi^{T}Y$$. Some of them contain additional model My thoughts are that the treatment X 0 is .47% less likely to show positive savings? Hi you have a wonderful Posting site It was very easy to post good job, Pingback: Multi-class logistic regression – Look back in respect, Hi you have a user friendly site It was very easy to post I enjoyed your site, Pingback: Logistic regression using SKlearn – Look back in respect. The following is more verbose description of the attributes which is mostly The independent variables should be independent of each other. Also, I’m working with a complex design survey data, how do I include the sampling unit and sapling weight in the model? The example for logistic regression was used by Pregibon (1981) “Logistic Regression diagnostics” and is based on data by Finney (1947). How can I increase the number of iterations? OLS has a Logistic regression is the type of regression analysis used to find the probability of a certain event occurring. RollingWLS(endog, exog[, window, weights, …]), RollingOLS(endog, exog[, window, min_nobs, …]). Peter Prettenhofer. Age_bin 0.169336 0.732283, Pingback: Classification metrics and Naive Bayes – Look back in respect, What does MLE stands for? X’B represents the log-odds that Y=1, and applying g^{-1} maps it to a probability. Fit a Gaussian mean/variance regression model. Econometrics references for regression models: R.Davidson and J.G. We will be using the Statsmodels library for statistical modeling. For 'var_1' since the t-stat lies beyond the 95% confidence interval (1.375>0.982), shouldn't the p-value be less than 5%? Ed., Wiley, 1992. Change ), You are commenting using your Twitter account. autocorrelated AR(p) errors. This is my personal blog, where I write about what I learned, mostly about software, project management and machine learning. It is the best suited type of regression for cases where we have a categorical dependent variable which can take only discrete values. This is equal n - p where n is the endog can contain strings, ints, or floats or may be a pandas Categorical Series. “Econometric Theory and Methods,” Oxford, 2004. We assume that outcomes come from a distribution parameterized by B, and E(Y | X) = g^{-1}(X’B) for a link function g. For logistic regression, the link function is g(p)= log(p/1-p). The residual degrees of freedom. Though StatsModels doesn’t have this variety of options, it offers statistics and econometric tools that are top of the line and validated against other statistics software like Stata and R. When you need a variety of linear regression models, mixed linear models, regression with discrete dependent variables, and more – StatsModels has options. This was done using Python, the sigmoid function and the gradient descent. D.C. Montgomery and E.A. Variable: y R-squared: 0.416, Model: OLS Adj. statsmodels.discrete.discrete_model.MNLogit¶ class statsmodels.discrete.discrete_model.MNLogit (endog, exog, check_rank = True, ** kwargs) [source] ¶ Multinomial Logit Model. In stats-models, displaying the statistical summary of the model is easier. A simple data science+journalism tutorial. Note: this post is part of a series about Machine Learning with Python. model = sm.Logit(endog=y_train,exog= X_train) Linear regression is used as a predictive model that assumes a linear relationship between the dependent variable (which is the variable we are trying to predict/estimate) and the independent variable/s (input variable/s used in the prediction).For example, you may use linear regression to predict the price of the stock market (your dependent variable) based on the following Macroeconomics input variables: 1. degree of freedom here. However, if the independent variable x is categorical variable, then you need to include it in the C(x)type formula. You can follow along from the Python notebook on GitHub. To build the logistic regression model in python. The blog should help me to navigate into the future using (and not forgetting) the past experiences. Odds are the transformation of the probability. In this posting we will build upon that by extending Linear Regression to multiple input variables giving rise to Multiple Regression, the workhorse of statistical learning. I think that statsmodels internally uses the scipy.optimize.minimize() function to minimise the cost function and that method is generic, therefore the verbose logs just say “function value”. It explains the concepts behind the code, but you'll still need familiarity with basic statistics before diving in. common to all regression classes. $$\mu\sim N\left(0,\Sigma\right)$$. We can now see how to solve the same example using the, Logistic regression with Python statsmodels, a series about Machine Learning with Python, Classification metrics and Naive Bayes – Look back in respect, Multi-class logistic regression – Look back in respect, Logistic regression using SKlearn – Look back in respect, An introduction to logistic regression – Look back in respect, Follow Look back in respect on WordPress.com. $$\Psi\Psi^{T}=\Sigma^{-1}$$. Lab 4 - Logistic Regression in Python February 9, 2016 This lab on Logistic Regression is a Python adaptation from p. 154-161 of \Introduction to Statistical Learning with Applications in R" by Gareth James, Daniela Witten, Trevor Hastie and Robert Tibshirani. Change ), You are commenting using your Facebook account. The statistical model is assumed to be. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. We'll build our model using the glm() function, which is part of the formula submodule of (statsmodels). The n x n upper triangular matrix $$\Psi^{T}$$ that satisfies Chapter 11: Regression of Think Stats (Allen B. Downey) - This chapter covers aspects of multiple and logistic regression in statsmodels. GLS is the superclass of the other regression classes except for RecursiveLS, errors Σ = I. PredictionResults(predicted_mean, …[, df, …]), Results for models estimated using regularization, RecursiveLSResults(model, params, filter_results). Each student has a final admission result (1=yes, 0= no). ( Log Out / LIMIT_BAL_bin 0.282436 0.447070 errors $$\Sigma=\textbf{I}$$, WLS : weighted least squares for heteroskedastic errors $$\text{diag}\left (\Sigma\right)$$, GLSAR : feasible generalized least squares with autocorrelated AR(p) errors Linear models with independently and identically distributed errors, and for We can now see how to solve the same example using the statsmodels library, specifically the logit package, that is for … Let’s proceed with the MLR and Logistic regression with CGPA and Research predictors. In this guide, the reader will learn how to fit and analyze statistical models on quantitative (linear regression) and qualitative (logistic regression) target variables. In this lab, we will fit a logistic regression model in order to predict Direction using Lag1 through Lag5 and Volume. estimation by ordinary least squares (OLS), weighted least squares (WLS), Compute Burg’s AP(p) parameter estimator. y=data_final.loc[:,target] © Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers. $$Y = X\beta + \mu$$, where $$\mu\sim N\left(0,\Sigma\right).$$. ProcessMLE(endog, exog, exog_scale, …[, cov]). Interest Rate 2. Based on this formula, if the probability is 1/2, the ‘odds’ is 1 X=data_final.loc[:,data_final.columns!=target]
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2021-01-23 08:02:37
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http://gmatclub.com/forum/if-ab-2-3ab-18-a-1-a-2-0-where-a-and-b-are-integers-144196.html
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# If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers
Author Message
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Status: struggling with GMAT
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GMAT Date: 04-06-2013
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Kudos [?]: 180 [0], given: 46
If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers [#permalink] 15 Dec 2012, 11:48
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Difficulty:
95% (hard)
Question Stats:
37% (03:01) correct 63% (01:39) wrong based on 73 sessions
If $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0$$ where a and b are integers,which of the following could be the value of b?
I. 1
II. 2
III. 3
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only
[Reveal] Spoiler: OA
Last edited by Bunuel on 16 Dec 2012, 06:20, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If \frac{(ab)^2+3ab-18}{(a-1)(a+2)}=0 [#permalink] 16 Dec 2012, 03:22
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Ans:
from the fraction we get that “a” cannot be 1 or -2, so putting the value of a in numerator we get the the values of “b” to 3,-6, -3/2 , therefore b cannot be 3 so the answer is (C).
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers [#permalink] 16 Dec 2012, 06:29
From the Q,
Numerator = 0 whereas Denominator <> (cannot be) 0
Numerator = 0 provides: ab = -6 or ab = 3
Denominator <> 0 provides: a <> 1 and a <> -2
Put values of b in to ab to check for a:
1) b = 1; a = -6 or a = 3
2) b = 2; a = -3 or a = 3/2
3) b = 3; a = -2 or a = 1
But a <> 1 and a <> -2,
Hence, b can not be 3 and could be either 1 or 2.
Ans: C
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers [#permalink] 20 Dec 2012, 12:50
mun23 wrote:
If $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0$$ where a and b are integers,which of the following could be the value of b?
I. 1
II. 2
III. 3
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only
For the $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0$$ = 0
numerator has to be zero
and a # 1 and a # -2 (Since these two values would make denominator zero and fraction undefined)
Now $$(ab)^2+3ab-18$$
Put ab = x
$$x^2 + 3x - 18$$
factorized to
(x-3)(x+6)
Case I
x = ab = 3, => a =1 , b = 3 -- NOT POSSIBLE as we cannot have a = 1
x = ab = 3, => a =3 , b = 1 -- POSSIBLE
Case II
x = ab = -6, => a = -1, 2,+3,+6 & b = +2, -3, +6
Hence we can note that b can assume values 1 & 2 but not 3.
Hence C
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers [#permalink] 20 Dec 2012, 20:25
Expert's post
mun23 wrote:
If $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0$$ where a and b are integers,which of the following could be the value of b?
I. 1
II. 2
III. 3
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only
For $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0$$ to hold, $$(ab)^2+3ab-18 = 0$$
You need 'a' to be an integer so put in the values of b to check whether you get integral values for 'a'
b = 1 => a^2 + 3a - 18 = 0 => (a + 6)(a - 3) = 0 => Integral values so acceptable
b = 2 => 4a^2 + 6a - 18 = 0 => (2a + 6)(2a - 3) = 0 => We get a = -3 (an integer) hence acceptable
b = 3 => 9a^2 + 9a - 18 = 0 => (a + 2)(a - 1) = 0 => We get a = -2 or 1. a can take neither of these values since they make the denominator 0. Not acceptable
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers [#permalink] 20 Dec 2012, 20:25
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Display posts from previous: Sort by
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2015-08-05 02:31:59
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https://mathhelpboards.com/threads/maximize-an-integral.8291/
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# Maximize an Integral
#### anemone
##### MHB POTW Director
Staff member
Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
#### HallsofIvy
##### Well-known member
MHB Math Helper
Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
I presume you know that to determine maximum or minimum values for a differentiable function, you set the derivative equal to 0. You should also know that if $$f(y)= \int_a^y g(x,y) dx$$ then $$df/dy= g(y,y)$$. So you should find y such that $$\sqrt{y^4+ (y- y^2)^2}= 0$$. (If that y is not between 0 and 1 then look at the value at 0 and 1.)
#### Klaas van Aarsen
##### MHB Seeker
Staff member
I presume you know that to determine maximum or minimum values for a differentiable function, you set the derivative equal to 0. You should also know that if $$f(y)= \int_a^y g(x,y) dx$$ then $$df/dy= g(y,y)$$. So you should find y such that $$\sqrt{y^4+ (y- y^2)^2}= 0$$. (If that y is not between 0 and 1 then look at the value at 0 and 1.)
That doesn't look quite right.
Suppose $g(x,y) = x+y$.
Then:
\begin{aligned}\frac{d}{dy}\int_0^y g(x,y)dx
&= \frac{d}{dy}\int_0^y (x+y)dx \\
&= \frac{d}{dy}\left(\frac 1 2 x^2 + xy \Big|_0^y\right) \\
&= \frac{d}{dy}\left(\frac 3 2 y^2\right) \\
&= 3y
\end{aligned}
But:
$$g(y,y) = y+y = 2y \ne 3y$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
We cannot differentiate with respect to $y$ while it is inside the integral so we have first to separate them then
$$\displaystyle \int^y_0 (x+y)dx = \int^y_0 xdx +y\int^y_0 dx$$
Now if we differentiate and according to the product rule we have
$$\displaystyle \frac{df}{dy}\int^y_0 (x+y)dx=\frac{df}{dy} \left( \int^y_0 xdx +y\int^y_0 dx \right) =y +\int^y_0 dx+y = 3y$$
So the FTC doesn't apply directly here.
#### Random Variable
##### Well-known member
MHB Math Helper
If $\displaystyle f(y)= \int_a^y g(x,y) \ dx$, then $\displaystyle \frac{df}{dy}= \int_{a}^{y} g_{y}(x,y) \ dx + g(y,y)$.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
If $\displaystyle f(y)= \int_a^y g(x,y) \ dx$, then $\displaystyle \frac{df}{dy}= \int_{a}^{y} g_{y}(x,y) \ dx + g(y,y)$.
That integral doesn't seem to be solvable in terms of elementary functions. The W|A returns an elliptic integral.
#### anemone
##### MHB POTW Director
Staff member
Thank you all for the feedback!
If $\displaystyle f(y)= \int_a^y g(x,y) \ dx$, then $\displaystyle \frac{df}{dy}= \int_{a}^{y} g_{y}(x,y) \ dx + g(y,y)$.
I believe you're right, and if one wants to attack the problem using this definition, then that would be welcome!
That integral doesn't seem to be solvable in terms of elementary functions.
Hmm...that's not quite right, Zaid! But speaking of more advanced integration field, you and the rest of the members are the experts, not me. I want to let you know that one of the solutions that I have solved it through the inequality route and I actually can't wait to share it with MHB!
Having said so, I will only post the solutions days later with the hope that others may want to take a stab at it.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Hmm...that's not quite right, Zaid!
Well, I don't know whether I am messing something but I still cannot think how to find an elementary anti-derivative for that integral.
#### anemone
##### MHB POTW Director
Staff member
Well, I don't know whether I am messing something but I still cannot think how to find an elementary anti-derivative for that integral.
Most probably that I am wrong, Zaid. Please don't take what I said seriously because really I am no comparison to you when it comes to the territory of advance integration.
#### Opalg
##### MHB Oldtimer
Staff member
Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
Let $$\displaystyle f(y) = \int_0^y \sqrt{x^4+(y-y^2)^2} dx.$$ My instinct is that the maximum value of $f$ over the interval $[0,1]$ must be $$\displaystyle f(1) = \int_0^1 x^2\, dx = 1/3.$$ Following anemone's hint about using an inequality, it occurred to me that if $a,b \geqslant 0$ then $\sqrt{a^2+b^2} \leqslant a+b.$ Applying that with $a=x^2$ and $b= y-y^2$, you see that $$\displaystyle f(y) \leqslant \int_0^y(x^2 + y - y^2)\,dx = \tfrac13y^3 + y^2 - y^3 = y^2 - \tfrac23y^3.$$ But $y^2 - \tfrac23y^3$ is an increasing function on the interval $[0,1]$, with a maximum value $1/3$ when $y=1$. Therefore the maximum value of $f(y)$ is $f(1) = 1/3.$
#### anemone
##### MHB POTW Director
Staff member
Let $$\displaystyle f(y) = \int_0^y \sqrt{x^4+(y-y^2)^2} dx.$$ My instinct is that the maximum value of $f$ over the interval $[0,1]$ must be $$\displaystyle f(1) = \int_0^1 x^2\, dx = 1/3.$$ Following anemone's hint about using an inequality, it occurred to me that if $a,b \geqslant 0$ then $\sqrt{a^2+b^2} \leqslant a+b.$ Applying that with $a=x^2$ and $b= y-y^2$, you see that $$\displaystyle f(y) \leqslant \int_0^y(x^2 + y - y^2)\,dx = \tfrac13y^3 + y^2 - y^3 = y^2 - \tfrac23y^3.$$ But $y^2 - \tfrac23y^3$ is an increasing function on the interval $[0,1]$, with a maximum value $1/3$ when $y=1$. Therefore the maximum value of $f(y)$ is $f(1) = 1/3.$
Well done, Opalg, and thanks for participating!
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2021-01-24 04:47:51
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http://denkovacs.com/
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Catmull-Rom splines are interpolating, piece-wise cubic splines: a spline of this class passes through each of its control points, and restricted to each knot interval it is a cubic function.
How you choose the knot intervals has a significant effect on the spline’s quality. Long story short: you get the best results by setting them to the Euclidean distance between the two corresponding control points. Splines with such knot assignments are called centripetal splines.
For more information, this Wikipedia page has a nice summary and comparison plots for different choices of knot intervals. Cem Yuksel’s project page has more interesting details as well as many applications.
Computing a spline’s value $C(t)$ at time $t$ is easy using a pyramidal formulation described here.
You can use the same pyramid structure to evaluate the tangent $C'(t)$ as well:
Here is a little MATLAB script that computes and plots a centripetal Catmull-Rom spline and its tangents.
Most iPad styluses are bulky. I always hated those. Of course the touchscreen sensor has been optimized for the thickness of a finger and not that of a pencil tip. But eventually clever people started figuring out ways to have the same surface contact area, without blocking the view with a finger-thick stylus. From DIY stylus projects to commercial variants like the Adonit Jot, the GoSmart Stylus, or the pressure-sensitive Hex3 Jaja, the stylus form factor of a thin tip with an attached wider disk of conductive material is now well established. I still wasn’t happy though. My own DIY stylus was too unreliable and flimsy, and I really wanted a stylus that felt as natural as a pencil, and not some strange artsy piece of aluminum.
The final project (chapter 1) of my PhD thesis became a behemoth of a paper. This was unavoidable but unfortunate, because the original idea and the final code are relatively simple and straight-forward. So in a few easy-to-digest blog posts, I will highlight the motivation, approach, and final code that might otherwise be difficult to extract from the paper.
A while ago I replaced my old IKEA Kilby bookshelf with two Billy shelves. Kilby is the low-cost intro model, but it has served me well, and so I thought this would be a great time to do my first own IKEA Hack. I wanted to turn it into a desk shelf for my Mikael desk so I can place my MIDI keyboard under the monitors and speakers.
Of course the spacing between the screwed shelves did not match the exact width of the Mikael desk, and so originally I thought I would cut the side boards to the right width and then drill some holes near the cut end. But as it turned out the spacing between the two screwed shelves was wide enough for my keyboard, and so I simply left one side as an overhang:
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2017-07-28 00:32:01
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http://openstudy.com/updates/508aa8a3e4b077c2ef2e3f50
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## Biancao9o3o12 Group Title In set-builder notation, how do you write the solutions of the inequality? 3x + 10 ≥ 4 (1 point) A.{x | x ≥ –2} B. {x | x ≤ 2} C.{x | x ≤ –2} D.{x | x ≥ 2} one year ago one year ago
1. UnkleRhaukus Group Title
$3x + 10 ≥ 4$ minus ten from both sides, then divide both sides by three, what do you get
2. Biancao9o3o12 Group Title
okay I got -7+10>4
3. UnkleRhaukus Group Title
try again
4. Biancao9o3o12 Group Title
C.{x | x ≤ –2} ?
5. UnkleRhaukus Group Title
what do you get after you minus ten from both sides, and divide both sides by three,
6. Biancao9o3o12 Group Title
I get 3
7. Biancao9o3o12 Group Title
Loook I don't know how to do this and I have this test that was due 4 days ago and I haven't finished because I thought I could do it but I can't
8. UnkleRhaukus Group Title
$3x+10≥4$ minusing ten from both sides, $3x+10-10≥4-10$
9. Biancao9o3o12 Group Title
Okay 3x+10=13-10>4-10=5
10. UnkleRhaukus Group Title
$3x+10−10≥4−10$simplifying $3x≥-6$
11. UnkleRhaukus Group Title
now, divide both sides by three,
12. Biancao9o3o12 Group Title
3>1
13. UnkleRhaukus Group Title
$3x≥−6$dividing both sides by three $\frac{3x}3≥\frac{−6}3$ simplify
14. Biancao9o3o12 Group Title
2>1
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2014-09-02 17:16:23
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https://stats.stackexchange.com/questions/187015/interpreting-dispersion-of-glm-nb
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# Interpreting dispersion of glm.nb
First, I have read this post, this post and this post. All have very useful information. I have three other more specific questions.
I have estimated a negative binomial model using the glm.nb function of MASS and discovered the following parameters Theta: 9.0487, S.E: 0.444
1. Is it correct to assume that dispersion parameter has a standard deviation of 20.38?
2. Does this value correspond to the Poisson overdispersion that is corrected by the negative binomial model or is my model still overdispersed?
3. Joseph Hilbe states in his book that R's glm.nb function employs an inverted relationship of the dispersion parameter, theta. Thus a Poisson model results when theta approaches infinity. Suppose now that my second glm.nb model had estimates of Theta: 19.0487, S.E: 0.444. Would this model be less overdispersed than the first model?
• Can you explain where the value of 20.38 in part 1 is coming from? – Aniko Dec 16 '15 at 15:21
• I am dividing the value of Theta with the standard error – user3218416 Dec 16 '15 at 18:45
1. Standard error is the standard deviation of an estimate of a parameter (see eg Wikipedia). So the standard deviation of the estimate $\hat\theta$ of the dispersion parameter $\theta$ is $0.444$. The ratio of an estimate to its standard error $\hat\theta/SE(\hat\theta)=20.38$ is often used as a test statistic for the null-hypothesis $\theta=0$. A large value here suggests that this null-hypothesis would probably be rejected. However,
• as you noted, $\theta\rightarrow\infty$ corresponds to no overdispersion, so testing $\theta=0$ is not very meaningful
2. The function fits a negative binomial distribution, and $\theta$ is one of its parameters. The negative binomial distribution is always overdispersed compared to the Poisson distribution (unless $\theta=\infty$). Since by modifying $\theta$ for a fixed $\mu$ the negative binomial distribution can achieve any variance ($Var(NB) = \mu +\frac{\mu^2}{\theta}$), there is no such thing as "overdispersion relative to the negative binomial distribution". Of course, it is possible that the negative binomial does not provide a good fit to the data, but the concept of overdispersion does not apply.
3. If the means are the same, then yes, if you have a larger $\theta$, then that model is less overdispersed compared to the Poisson distribution. If you change $\mu$ as well, then you have to think about quantifying overdispersion. For the Poisson distribution $E(X)=Var(X)$, but do you want to quantify deviations as $Var(X)/E(X)$? or $Var(X)-E(X)$? or some other way? Your question will have different answers for different ways to quantify overdispersion.
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2019-11-17 03:30:30
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https://www.emathhelp.net/notes/differential-equations/applications-of-first-order-ode/temperature-problems/
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# Temperature Problems
Newton's law of cooling, which is equally applicable to heating, states that the time rate of change of the temperature of a body is proportional to the temperature difference between the body and its surrounding medium. Let T denote the temperature of the body and T_m denote the temperature of the surrounding medium. Then, the time rate of change of the temperature of the body is (dT)/(dt), and Newton's law of cooling can be formulated as (dT)/(dt)=-k(T-T_m), or as (dT)/(dt)+kT=kT_m, where k is a positive constant of proportionality. Once k is chosen positive, the minus sign is required in Newton's law to make (dT)/(dt) negative in a cooling process, when T is greater than T_m, and to make it positive in a heating process, when T is smaller than T_m.
Example 1. A body at an unknown temperature is placed in a room which is held at a constant temperature of 30° F. If after 10 minutes the temperature of the body is 0° F and after 20 minutes the temperature of the body is 15° F, find the unknown initial temperature.
We have that T_m=30; so, the differential equation is (dT)/(dt)+kT=30k. This is a first-order linear differential equation. The integrating factor is I=e^(int kdt)=e^(kt). After multiplying the equation by the integrating factor, we obtain that e^(kt)(dT)/(dt)+kTe^(kt)=30ke^(kt), or (d(Te^(kt)))/(dt)=30ke^(kt).
Integrating both sides gives Te^(kt)=30e^(kt)+C, or T=Ce^(-kt)+30.
We are given that T(10)=0, or 0=Ce^(-10k)+30. Also, T(20)=15, or 15=Ce^(-20t)+30.
Thus, we have a system of two equations:
{(Ce^(-10k)=-30),(Ce^(-20k)=-15):}
Dividing the first equation by the second gives e^(10k)=2. Now, from the first equation, we have that C=-30e^(10k)=-30*2=-60.
Finally, T_0=T(0)=Ce^(-k*0)+30=C+30=-60+30=-30.
Let's take a look at another interesting example.
Example 2. A body at a temperature of 50° F is placed in an oven whose temperature is kept at 150° F. If after 10 minutes the temperature of the body is 75° F, find the time required for the body to reach a temperature of 100° F.
We have that T(0)=50, T(10)=75, T_m=150.
So, the differential equation is (dT)/(dt)+kT=150k. Again, as in example 1, this is a linear first-order differential equation. Its solution is T=Ce^(-kt)+150.
Since T(0)=50, we have that 50=Ce^(-k*0)+150, or C=-100.
Now, the equation has the form T=-100e^(-kt)+150.
Since T(10)=75, we have that 75=-100e^(-10k)+150, or k=-(ln(0.75))/10.
Finally, the equation has the following form: T=-100e^(ln(0.75)/10t)+150.
Now, we need to find such t that T(t)=100:
100=-100e^((ln(0.75))/10t)+150, or t=10 ln(0.5)/(ln(0.75))~~24.0942 minutes.
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2020-09-30 10:11:30
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https://www.physicsforums.com/threads/help-finding-constants-for-taylor-series.637803/
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# Help finding Constants for Taylor Series
jtleafs33
## Homework Statement
The Taylor expansion of ln(1+x) has terms which decay as 1/n.
Show, that by choosing an appropriate constant 'c', the Taylor series of
(1+cx)ln(1+x)
can be made to decay as 1/n2
## Homework Equations
f(x)=$\sum$$^{n=\infty}_{n=0}$ f(n)(0) $\frac{x^{n}}{n!}$
## The Attempt at a Solution
I used Maple to differentiate this function and find values at x=0 for several derivative:
f(0)(0) = 0
f(1)(0) = 1
f(2)(0) = 2c-1
f(3)(0) = -3c+2
f(4)(0) = 8c-6
f(5)(0) = -30c+24
f(x)=$\frac{(0)x^{0}}{0!}$+$\frac{(1)x^{1}}{1!}$+$\frac{(2c-1)x^{2}}{2!}$+$\frac{(-3c+2)x^{3}}{3!}$+$\frac{(8c-6)x^{4}}{4!}$+$\frac{(-30c+24)x^{5}}{5!}$ ....
This is where I'm stuck... In order to get the terms decaying as 1/n2, I get different values of c for each term...
c0=1
c1=1
c2=$\frac{3}{4}$
c3=$\frac{8}{9}$
c4=$\frac{15}{16}$
c5=$\frac{24}{25}$
And I need one constant c that will do it all. Any help would be greatly appreciated.
Homework Helper
Did you construct the Taylor series for ln(1+x)?
In what way do the coefficients decay as 1/n?
Homework Helper
Gold Member
Try simplifying the coefficients in your Taylor series to see if you can find a trend. Also, you can avoid the laborious task of calculating derivatives if you approach this another way. Hint: try substituting the Taylor series for ln(1+x) into the expression (1+cx)ln(1+x).
P.S. There's no point trying to set each coefficient equal to 1/n^2. You won't get the coefficients to EQUAL 1/n^2, but you should be able to get them to decay at the same rate as 1/n^2.
jtleafs33
I did the differentiation method to more easily see the trend myself.
I know ln(1+x)=$\sum$$^{n=\infty}_{n=1}$(-1)$^{n+1}$$\frac{x^{n}}{n}$
Also, the taylor expansion of a polynomial is just that polynomial
So, (1+cx)ln(1+x)=$\sum$$^{n=\infty}_{n=1}$(-1)$^{n+1}$$\frac{(1+cx)x^{n}}{n}$
Rearranging this, I can get a general expression for each coefficient:
an=(-1)n+1($\frac{c}{n-1}$-$\frac{1}{n}$)
But I'm stuck and don't know how to go about choosing this c. I'd imagine I'm going to need an equation which somehow relates an to an+1 and then solve for c, but I don't know what to do.
I need to figure this out and really understand it, because I also have to do the same thing to make the function
(1+ax+bx2)ln(1+x) decay as 1/n3
Homework Helper
Gold Member
Rearranging this, I can get a general expression for each coefficient:
an=(-1)n+1($\frac{c}{n-1}$-$\frac{1}{n}$)
OK, this looks promising. Let's rearrange it a bit:
$$a_n = (-1)^{n+1}\left( \frac{nc - (n-1)}{n(n-1)} \right) = (-1)^{n+1}\left(\frac{n(c-1) + 1}{n^2 - n}\right)$$
This should give you a pretty good idea what $c$ should be.
Last edited:
jtleafs33
So, now I'm trying to solve this expression, substituting your equation into
$\frac{1/n^2}{1/(n+1)^2}$=an/an+1
But this will still give different C's for different n's. I don't understand how I can find one exact c that will work for all n's. I still don't understand what to do with the equations I have.
Homework Helper
Gold Member
So, now I'm trying to solve this expression, substituting your equation into
$\frac{1/n^2}{1/(n+1)^2}$=an/an+1
You won't be able to achieve this for every $n$, but you don't need to. "Decays as $1/n^2$" means that this is true asymptotically (in the limit).
Suppose I take $c = 1$. Then
$$|a_{n}| = \frac{1}{n^2 - n}$$
Clearly if $n$ is large, then the $n^2$ in the denominator is the dominant term, so asymptotically, $|a_n|$ decays like $\frac{1}{n^2}$.
Equivalently,
$$|a_{n}| = \frac{1}{n(n-1)}$$
For large $n$, the distinction between $n$ and $n-1$ is negligible, so $\frac{1}{n(n-1)}$ is almost the same as $\frac{1}{n^2}$.
Now see if you can make this precise. Hint: try to quantify the relative error between $\frac{1}{n(n-1)}$ and $\frac{1}{n^2}$.
Last edited:
jtleafs33
Okay, that's exactly what I needed.
When I started the post and found a few values of c for various n, I immediately saw that c approached 1 as n approached infinity. Basically, I've been trying to make things work exactly, but I didn't realize I really just need to make things approach that behavior in the limit.
Thanks!
Homework Helper
Gold Member
Okay, that's exactly what I needed.
When I started the post and found a few values of c for various n, I immediately saw that c approached 1 as n approached infinity. Basically, I've been trying to make things work exactly, but I didn't realize I really just need to make things approach that behavior in the limit.
Thanks!
Right, terminology like "such and such decays as so and so" pretty much universally means "in the limit". I edited my above comment with a hint regarding how to quantify this.
Homework Helper
Wasn't there a clue in the expansion for ln(1+x) to that effect? Does the expansion decay exactly as 1/n or just sort-of (in the limit) like that?
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2022-09-28 08:57:28
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https://thegoodbean.com/shop/4ebg6r/5ccz0rk.php?id=faf339-galaxy-holm-15a
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7. Handy & Telefon (19) Art. Dort sitzt inmitten der zentralen Galaxie Holm 15A ein schwarzes Loch, das 40-Milliarden-fach schwerer ist als unsere Sonne. Freitag um 23:46 #4 @holms … Jul 03, 2020. 100% Upvoted. Kreis Pinneberg. maik005 Urgestein. Close. I wonder if it's possible to get better image of the black hole with newly discovered supermassive Holm 15A*. Huge Black Hole That Is Eating One Sun Everyday Discovered. Astronomers used data gathered by the Very Large Telescope in Chile’s Atacama Desert to run simulations that map out this distant galaxy. Rekord im Galaxienhaufen: Abell … FEATURED Latest SCIENCE . Am Ende konnte sich im Samsung galaxy a20 Vergleich unser Testsieger auf den ersten Platz hiefen. A new study conducted by an international team of astronomers found that the fastest-growing black hole, J2157 known to humans is 34 billion times the mass of our Sun and is extremely hungry. Samsung Galaxy S20 5G Forum ... Freitag um 23:15 #3 @maik005 Doch, das Ausrufezeichen bedeutet, dass man zwar mit dem WLAN verbunden ist, aber keine Verbindung zum Internet hat. Holm 15A is a huge elliptical galaxy at the center of a cluster of galaxies called Abell 85. Holmberg 15A Galaxy . Schleswig-Holstein. Auf der Website findest du jene wichtigen Fakten und die Redaktion hat die Samsung galaxy a20 getestet. A supermassive black hole has just been spotted in a galaxy 700 million light years from Earth. Its cusp radius, r γ = 4.57 ± 0.06 kpc (4 26 ± 0 06), is more than 18 times larger than the mean for BCGs and 1 kpc larger than A2261-BCG, hitherto the largest-cored BCG. Holm 15A is a central elliptical galaxy within the Abell 85 cluster, which contains more than 500 galaxies. Um den möglichen Unterschieden der Produkte gerecht zu werden, messen wir bei der Auswertung diverse Eigenarten. save hide report. Is it bigger than M87? Holm 15A holds the record for the heaviest black hole in the nearby universe. Holm 15A. Videos. A 2019-es év már egészen biztosan úgy vonul majd be a tudomány történetébe, hogy ez volt az év, amikor a fekete … For this, Holm 15A represents an ideal opportunity for testing the SMBH “scouring” scenario for the creation of BCG cores. Despiteitshighoverall … We model the one-dimensional (1D) light profile, and also the two-dimensional (2D) image (using GALFIT-CORSAIR, a tool for fitting the core-Sérsic model in 2D).Wefind good agreement between the 1D and 2D analyses, with minor discrepancies attributable to intrinsic ellipticity gradients. What the angular diameter of Holm 15A*? Of consequences unknown. Éppen ezért megnézték, mi lehet ennek az oka. Gründe können sehr verschieden sein. 0. This new research marks the first direct measurement; the paper has been submitted to The Astrophysical Journal, and awaits peer review. Galaxy Holm 15A. In all the upper echelons bright, And the far reaching designs, out of my sight. Ultra-massive Black Hole 40 Billion Times the Mass of The Sun Discovered . UltraMassive Black Hole Measured in Elliptical Galaxy Holm 15A Second Largest Ever Seen Space Fan News is Sponsored by OPT Telescopes and Patreon Patrons: https://bit.ly/2SwhmVB Earlier this month astronomers revised some mass estimates of a black hole at the center of the elliptical galaxy Holmberg 15A, an enormous galaxy 700 million TAG: Holm 15A. Holm 15A is a huge elliptical galaxy at the center of a cluster of galaxies called Abell 85. An ultramassive black hole clocking in at around 40 billion solar masses is at the heart of the galaxy Holm 15A, around 700 million light-years away. Kategorien . What Are The Main Challenges In India’s Real Estate Sector? Adrian Gabor-aug. 16, 2019, 3:06 PM. This is the Large Magellanic Cloud, a nearby satellite galaxy to our Milky Way. It's 10.4 billion light-years away. Aceasta este Cea mai MARE Gaura Neagra Descoperita Vreodata. @Kloopy Ist das in allen WLANs so? See more ideas about galaxies, astronomy, nebula. Elektronik. It was discovered c. 1937 by Erik Holmberg. share . Scientists Just Found the Smallest Black Hole Yet Scientists Baffled by Sudden Brightness of Our Galaxy's Supermassive Black Hole This target, lying at 13.74” from the center of Holm 15A, is a quasar candidate with z_phot ~ 0.9. A Holm 15A galaxis 700 millió fényévre van tőlünk, de a kutatók szerint nagyon furcsán viselkedik. Holm 15A, the brightest cluster galaxy of Abell 85. August 6, 2019 August 6, 2019 The Notitia 0 Comments Abell 85 galaxy cluster, Astronomy, axisymmetric Schwarzschild models, Holm 15A, Holmberg 15A Galaxy, quasar TON 618, research, space, space exploration, Ultra-Massive Black Hole. Itisaveryluminous(MV = 24:8mag,Kluge etal.2019)early-typegalaxy(ETG)withahighstellar massofM?& 2 1012 M . Here we investigate the unusually large ({R} γ \prime =0.5 = 4.57 kpc) depleted core recently reported for Holm 15A, the brightest cluster galaxy of Abell 85. Posted by. Holmberg 15A is a supergiant elliptical galaxy and the central dominant galaxy of the Abell 85 galaxy cluster in the constellation Cetus, about 700 million light-years from Earth. Holm 15A, the brightest cluster galaxy (BCG) of the cool-core cluster Abell 85, has an ultra-diffuse central region, ∼$2mag$ fainter than the faintest depleted core of any early-type galaxy (ETG) that has been dynamically modelled in detail. 40km=s andsmallcomparedtothe velocitydispersion˙˘350km=s. We model the one-dimensional (1D) light profile, and also the two-dimensional (2D) image (using Galfit-Corsair, a tool for fitting the core-Sérsic model in 2D). It is the brightest galaxy in Abell 85, and one of the brightest in our corner of the universe. A supermassive black hole (SMBH) is the largest type of black hole, on the order of hundreds of thousands to billions of solar masses (M ☉), and is theorized to exist in the center of almost all massive galaxies.In some galaxies, there are even binary systems of supermassive black holes, see the OJ 287 system. The galaxy, Holm 15A, is one of several that make up the Abell 85 galaxy cluster. In space, black holes appear in different sizes and masses. NGC 4567/8 oder KPG 347 oder VV 219 sind zwei Spiralgalaxien im Virgo-Cluster im Sternbild Virgo.Die Galaxien werden auch "The Siamese Twins" oder "The Butterfly Galaxies" genannt. Wenn nur bei deiner Fritzbox, dann würde ich dort die Ursache suchen. Most Read . Handy & Telefon in Holm 1 - 19 von 19 gebrauchte Handys, Smartphones & Telefone in Holm - Kreis Pinneberg. | NAR India 2019. We have found that the brightest cluster galaxy (BCG) in A85, Holm 15A, displays the largest core known so far. The Holm 15A bright cluster galaxy has a central region that’s far fainter than any other early-type galaxy… Samsung galaxy a20 - Betrachten Sie unserem Gewinner. Therotationalvelocityof Holm15Aisvrot. Holm 15A* is one of those things. Holm 15A, the brightest cluster galaxy (BCG) of the galaxy cluster Abell 85, has an ultra-diffuse central region, ~ 2 mag fainter than the faintest depleted core of any early-type galaxy (ETG) that has been dynamically modelled in detail. Holm 15A hosts the luminous amorphous radio source 0039-095B and has the optical signature of a LINER. 4 months ago. It briefly shot to fame when it was reported to have the largest core ever observed in a galaxy, spanning some 15,000 light years, however this was subsequently refuted. What the angular diameter of Holm 15A*? Holm 15A, the brightest cluster galaxy of the galaxy cluster Abell 85, has an ultradiffuse central region, ∼ 2 {mag} fainter than the faintest depleted core of any early-type galaxy (ETG) that has been dynamically modeled in detail. Deutschland. Partially depleted cores, as measured by core-Sérsic model “break radii,” are typically tens to a few hundred parsecs in size. And then there's the ultramassive black hole powering the quasar TON 618 - an absolute beast at 66 billion solar masses. Is it bigger than M87? A team of astronomers captured a snapshot of Holm 15A’s stars in orbit around the galaxy’s central black hole and created a model to help them calculate the black hole’s mass. Of the Meanwhile, 700 million light-years away from Earth, a galaxy called Holm 15A contains the largest known black hole in the observable universe. Thisisverycommon among massive ETGs (e.gEmsellem et al.2011;Cap- pellari2016;Vealeetal.2017). May 11, 2020 - Explore Patricia Holm's board "Galaxies" on Pinterest. Faint glow: This diagram shows the distribution of the surface brightness of the central cluster galaxy Holm 15A. Previous calculations based on the dynamics of the galaxy and the cluster had resulted in Holm 15A* mass estimates of up to 310 billion times the mass of the Sun. If confirmed, it would be the largest in the local universe, which spans a billion light years. The black holes of Holm 15A and TON 618 are pretty difficult to understand. … Es wird angenommen, dass sich das Paar in einem frühen Stadium der Interaktion befindet. Here we investigate the unusually large (${R}_{\gamma \prime =0.5}$ = 4.57 kpc) depleted core recently reported for Holm 15A, the brightest cluster galaxy of Abell 85. u/Another__one. 1 comment. Holm 15A is the brightest cluster galaxy of Abell 85 with ~2 mag fainter central region SuperMassive Black Hole(SMBH) of 4.0±0.8x1010 solar masses in the center of the galaxy Seeking confirmation that cores are generated from SMBH binaries Believed to be formed A follow-up study of the cD galaxy Holm 15A in order to Um dies herauszufinden, haben Forschende der Arbeitsgruppe von Ralf Bender am Max-Planck-Institut für extraterrestrische Physik und an der Universitäts-Sternwarte München fotometrische Daten sowie spektrale Beobachtungen ausgewertet. Apple (3) Samsung (6) Siemens (4) Sony (2) Weitere (2) Preis - Ort. Alle Kategorien. The record is now held by a specimen in the Abell 85 cluster of galaxies, where an ultra-massive black hole with 40 billion times the mass of our Sun sits in the middle of the central galaxy Holm 15A. Holm 15A is the brightest cluster galaxy (BCG) of Abell85. However, these were all indirect measurements of the black hole.
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2021-01-27 17:38:07
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https://www.physicsforums.com/threads/spring-pendulum-kinetic-energy.806665/
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# Spring pendulum, Kinetic Energy
1. Apr 4, 2015
### KEVmathematics
In the included picture, I don't get how they get to the kinetic energy part. I would say, that the travelled distance is equal to (l + x(t))*θ. Then I would take the time derivative, resulting in dx(t)/dt * θ + (l + x(t))* dθ/dt. Then I would square this result and multiply that with 1/2 m. But then I would get a totally different kinetic energy.
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2. Apr 4, 2015
### Staff: Mentor
What does that mean? The travelled distance depends on the trajectory. You'll need a position (as vector) to get a meaningful derivative.
3. Apr 5, 2015
### Vishwaas
You must also consider the radial component. The velocity component in the radial direction is the time derivative of x(t) (since it is a spring, it can compress or expand). Hence the total kinetic energy is the sum of energies in both radial and angular directions.
4. Apr 5, 2015
### vanhees71
In polar coordinates, with the polar angle relative to the direction poiinting downwards, you have
$$\vec{x}=r \begin{pmatrix} \sin \varphi \\ \cos \varphi \end{pmatrix}.$$
Then, after straight-forward algebra, you have
$$T=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} (\dot{r}^2+r^2 \dot{\varphi}^2).$$
Now in the textbook, they set
$$r=l+x,$$
where $l$ is the length of the relaxed spring.
For the potential energy you have the part from the gravitational field of the earth and the spring:
$$V=-m g x_2+\frac{k}{2} x^2 = -m g (x+l) \cos \varphi+\frac{k}{2} x^2.$$
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2018-07-23 04:41:22
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http://dataspace.princeton.edu/jspui/handle/88435/dsp01bz60cw28q
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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01bz60cw28q
Title: Resonant Wave-Particle Manipulation Techniques Authors: Zhmoginov, Andrey Igorevich Advisors: Fisch, Nathaniel J Contributors: Plasma Physics Department Keywords: alpha channelingmirror machinenegative mass effectparticle diffusionplasma waveswave-particle interaction Subjects: Plasma physics Issue Date: 2012 Publisher: Princeton, NJ : Princeton University Abstract: Charged particle dynamics can be altered considerably even by weak electromagnetic waves if some of the particles are in resonance. Depending on the wave parameters, the resonances in the phase space can either be well separated, in which case the particle dynamics is regular almost everywhere, or they can overlap leading to stochastic particle motion in a large volume of the phase space. Although different, both of these regimes allow one to manipulate particle ensembles by arranging resonant interactions with appropriate waves. This thesis is devoted to studying two wave-particle manipulation techniques having potential applications in fusion and laser-plasma interaction research. Specifically, we study the alpha-channeling effect (which relies on stochastic diffusion of resonant particles) and the so-called negative-mass effect (NME) (which involves the conservation of the adiabatic invariant). The alpha-channeling effect entails the use of radio-frequency waves to expel and cool high-energetic alpha particles born in a fusion reactor; the device reactivity can then be increased even further by redirecting the extracted energy to fuel ions. Recently, the alpha-channeling technique, originally proposed for tokamaks, was shown to be suitable for application in mirror machines as well. In the first part of this thesis, we deepen the understanding of issues and possibilities of the alpha-channeling implementation in open-ended reactors. We verify the feasibility of this technique and identify specific waves and supplementary techniques, which can potentially be used for implementing the alpha-channeling in realistic mirror devices. We also propose a new technique for using the alpha-channeling wave energy to catalyze fusion reaction by employing minority ions as a mediator species. In the second part of this thesis, the NME manifesting itself as an unusual response of a resonant particle to external adiabatic perturbations mimicking the behavior of a particle with a negative mass, is discussed. Using the Hamiltonian perturbation theory, the calculation of the effective parallel mass is extended to the non-vacuum waves and the NME is shown to be robust. Also, the consequences of radiation friction and collisions with the background particles on the NME are studied and new collective phenomena emerging in plasmas with negative-mass particles are considered. URI: http://arks.princeton.edu/ark:/88435/dsp01bz60cw28q Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog: http://catalog.princeton.edu/ Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: Plasma Physics
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2014-12-22 21:57:18
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https://math.stackexchange.com/questions/2014980/integer-solutions-star-and-bar
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# Integer solutions (star and bar)
Consider the equation $x_1+x_2+x_3+x_4+x_5=10$.
How many non-negative integer solutions if $x_1 > x_2$?
Apparently counting $x_1$ and $x_2$ one by one is too slow, and impractical if the sum is not 10 but 100 instead. Is there a general way to solve this kind of problems?
• There are infinitely many non-integer solutions. You probably mean integer solutions. In that case, you may rewrite equation as $x_1'+2x_2+x_3+x_4+x_5 = 9$. – Abstraction Nov 15 '16 at 10:59
• non-integer solutions are $\infty^5$ . Are you looking for integral or non integral solutions ? and maybe for integral non-negative? – G Cab Nov 15 '16 at 11:00
• @Abstraction I'm sorry for the typo. I meant non-negative integers :( – Lon Edwards Nov 15 '16 at 11:10
Hint. By Stars-and-Bars, the number of non-negative integer solutions of $$x_1+x_2+x_3+x_4+x_5=10$$ is $\binom{10+(5-1)}{5-1}$. The numbers of non-negative integer solutions of $$k+k+x_3+x_4+x_5=10$$ that is $$x_3+x_4+x_5=10-2k$$ for $k=0,1,\dots ,5$ is $\binom{10-2k+(3-1)}{3-1}$. Can you take it from here?
Finally you should find that the number you are looking for is $420$.
• That means by finding out $\binom{14}{4}$ first, and subtract by the number of solutions for $x_3+x_4+x_5=10-2k$, since if two integers have different values it must mean that one of them is larger than the other. Is my interpretation correct? – Lon Edwards Nov 15 '16 at 11:20
• @Lon Edwards Yes. Then after the subtraction you have to divide the result by two. In one half $x_1>x_2$, in the other one $x_2>x_1$. – Robert Z Nov 15 '16 at 11:28
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2020-08-09 20:27:28
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https://math.stackexchange.com/questions/3154942/area-inside-a-loop-formed-by-parametric-equations
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# Area Inside A Loop Formed By Parametric Equations
We are given:
$$x=49-t^2$$
$$y=t^3-16t$$
The curve apparently makes a loop which lies along the x-axis. I need help finding total area inside the loop. I don't know where to even start.
If it helps, in previous parts of the question, I found that
(a) the tangent line is horizontal at $$t=\sqrt{16/3}$$ and $$x = 43.6666666666667$$
(b) the tangent line is vertical at $$t=0$$
Thank you!
I think this is a Green's theorem problem. The loop is traced as $$t$$ goes from $$-4$$ to $$4$$ (but clockwise.) So the area is
$$-\frac{1}{2} \int_{-4}^4 x \; dy - y \; dx$$ $$= -\frac{1}{2} \int_{-4}^{4} (49-t^2)(3t^2-16)-(t^3-16t)(-2t) \; dt$$ $$= \frac{8192}{15} = 546.13\ldots.$$
The extra minus sign is because of the clockwise orientation.
Have you plotted it? I don't find a loop. I find an arc below the $$x$$ axis from $$t=0$$ to $$t=4$$. This is from $$x=33$$ to $$x=49$$. Maybe you are supposed to find the area between the $$x$$ axis and the curve in this region. Here is my plot
$$t=0$$ is the right hand end, at $$(49,0).\ \ t=4$$ is the point where it crosses the axis at $$(33,0)$$. If you are to find the area between the curve and the $$x$$ axis you can just solve the $$x$$ equation for $$t$$ and plug into the $$y$$ equation $$x-49-t^2\\t=\sqrt{49-x}\\y=(49-x)^{3/2}-16(49-x)^{1/2}$$
and you can integrate the last from $$x=49$$ to $$x=33$$
• I understand what you've done and I've tried doing it this way. However, I've failed to come up with the correct answer (I got -273.0666). Thank you for trying though! – CodingMee Mar 20 at 3:13
• I believe the answer should be positive, but that is the answer Alpha gets. Why do you think it is wrong? – Ross Millikan Mar 20 at 3:35
• I tried both negative and positive just to be sure, and they both come up incorrect. I think it is wrong because my assignment is online and automatically checks our answers. – CodingMee Mar 20 at 3:40
• The answer key could be wrong, it could be expecting the exact answer of $\frac {4096}{15}$, or I could be interpreting the problem wrong. – Ross Millikan Mar 20 at 3:44
Making a parametric plot, there is effectively a loop which is symmetric with respect to the $$x$$ axis; the points where the curve intersect the axis correspond to $$x=33$$ and $$x=49$$ as @Ross Millikan already answered.
The major issue is to compute $$I=\int \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt=\int\sqrt{4 t^2+\left(3 t^2-16\right)^2}\,dt$$ which would lead to nasty elliptic integrals.
So, the simplest is to do what @Ross Millikan already answered, that is to say $$t=\pm \sqrt{49-x} \implies y=\pm (x-33)\sqrt{49-x}$$ So, the total area enclosed by the loop is $$A=2\int_{33}^{49}(x-33)\sqrt{49-x}\,dx$$ Using $$\int(x-33)\sqrt{49-x}\,dx=-\frac{2}{15} (49-x)^{3/2} (3 x-67)$$ we end with $$A=2 \times \frac{4096}{15}$$.
Notice that $$\frac{4096}{15}\approx 273.067$$
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2019-06-24 22:20:15
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https://zbmath.org/?q=an:1014.65115
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# zbMATH — the first resource for mathematics
A priori convergence theory for reduced-basis approximations of single-parameter elliptic partial differential equations. (English) Zbl 1014.65115
The authors consider an elliptic system parameterized by a scalar $$\mu\in[0,\mu_0]$$ of the form $a_0(u(\mu),v)+\mu a_1(u(\mu),v)=f(v)\qquad \forall v\in Y\tag{*}$ where $$Y$$ is an appropriate function space, $$a_0$$ and $$a_1$$ are continuous and symmetric, $$a_0$$ is coercive and $$a_1$$ is postitive semi-definite. For each choice of $$\mu$$ it is possible to approximate $$u(\mu)$$ to arbitrary accuracy by a member $$u^{\mathcal N}(\mu)$$ of an approximating subspace of $$Y^{\mathcal N}\subset Y$$ of sufficiently large but finite dimension $$\mathcal N$$.
The authors prove that it is possible to choose $$N\ll {\mathcal N}$$ sample values $$\mu_n$$, $$n=1,2,\ldots,N$$, logarithmetically distributed in the interval $$[0,\mu_0]$$, with the following property. For each $$\mu_n$$, denote an approximate solution to $$(*)$$ in $$Y^{\mathcal N}$$ by $$u^{\mathcal N}_n$$, and denote the span of these approximates as $$W^{\mathcal N}_N$$. If $$N$$ is larger than a critical value $$N_0$$, then an approximation to $$u(\mu)$$ can be found in $$W^{\mathcal N}_N$$ so that $|||u(\mu)-u^{\mathcal N}_N(\mu)|||\leq |||u(\mu)-u^{\mathcal N}(\mu)|||+C|||u(0)|||e^{-N/N_0}$ where $$C$$ denotes a constant depending only on $$a_0$$, $$a_1$$, and $$\mu_0$$, and $$|||\cdot|||$$ denotes the norm induced by $$a_0(u,v)$$.
Numerical testing indicates that the logarithmic distribution is optimal and that a similar result might hold in more than one dimension.
##### MSC:
65N12 Stability and convergence of numerical methods for boundary value problems involving PDEs 65N30 Finite element, Rayleigh-Ritz and Galerkin methods for boundary value problems involving PDEs 35J25 Boundary value problems for second-order elliptic equations
Full Text:
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2021-02-25 03:14:35
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https://www.gamedev.net/topic/649534-windowed-mode-beyond-desktop-resolution-issue/
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• Create Account
## Windowed mode beyond desktop resolution issue
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9 replies to this topic
### #1Tispe Members
Posted 29 October 2013 - 05:05 PM
Hello
If I set my game to a windowed resolution (1920x1080) which makes the window(client area + border area) larger than the monitor resolution(1920x1080) then Windows will automatically "crop" down the window size to fit inside the desktop. This causes the client area to be compressed causing image distortions and positioning errors.
Is there a way to allow "larger then desktop" resolution windows without this client area compression?
### #2Tom KQT Members
Posted 30 October 2013 - 12:26 AM
That's still the "better" behaviour, for me IIRC it was crashing on some computers (not on all).
I don't know how do force windows to allow you to do this (if it even is possible), but are you sure you need it? If you just want to make a client area of the same size as your resolution, you can create a window without border and the title bar.
### #3Tispe Members
Posted 30 October 2013 - 12:40 AM
I don't really need it because most players would not use a window that is larger then the desktop. But, if the user accidentally choses a larger then desktop resolution I don't want a distorted client area. My Gui code also kinda breaks because I assume the width/height to be the value they were set to be, not the cropped values Windows forces them to be.
I can probably "check" the client area size after changing the resolution and see if they match up with what I set it be and do a new resolution change to those values. But that seems way big of a hassle for something Windows messes up. Plus, if the user wants a window with 1920x1080 client resolution I should not force him to have a "1912x1056" resolution. Yes the window will be offscreen but atleast the buttons will work properly and the client area won't be distorted.
### #4Tom KQT Members
Posted 30 October 2013 - 01:31 AM
It seems to be doable, look here for example: http://stackoverflow.com/questions/445893/create-window-larger-than-desktop-display-resolution. (Search for the answer that contains SWP_NOSENDCHANGING.)
Anyway, Windows don't allow windows larger than screen (unless you bypass it using the above-mentioned solution - which I personaly haven't tried!) and that's probably for a good reason, so I personally would follow that restriction and prevent the user to choose a larger window than possible. Regardless how complicated the code would be ;)
A window larger than screen may actually behave badly, because the OS doesn't expect to have such windows. You won't be able to resize it by dragging the borders at the very least.
Edited by Tom KQT, 30 October 2013 - 01:32 AM.
### #5N.I.B. Members
Posted 30 October 2013 - 03:12 AM
I can probably "check" the client area size after changing the resolution
Doing that is about 5 lines of code. It's a good practice anyway, why are you so reluctant to do it?
But that seems way big of a hassle for something Windows messes up.
Windows doesn't mess up anything. You assume something that is not correct, it's not Windows fault.
Edited by satanir, 30 October 2013 - 03:13 AM.
### #6Tispe Members
Posted 30 October 2013 - 04:03 AM
If I set some parameters to a window using SetWindowPos() I expect that window to do what I say. If I have to double check that every simple function I call actually does its job then the application would be overly bloated. I can see that most HRESULTs need checking before proceeding but not in this case.
Going to try the SWP_NOSENDCHANGING flag later when I get home. I think I should have to send a flag if I WANTED windows to resize, not the other way around.
### #7Tom KQT Members
Posted 30 October 2013 - 04:54 AM
Actually everything is working correctly. You called SetWindowPos and it did its job without errors. When the function tries to change position or size of a window, it's said that it will send the WM_WINDOWPOSCHANGING message. And the default handler of that message checks the size and fixes it if it's out of limits (it uses the WM_GETMINMAXINFO message). That actually is what that message was made for - to give you (or the system) a way how to easily check the validity of newly set window size - and to do whatever you need as a reaction to it.
But nobody prevents you from handling WM_WINDOWPOSCHANGING yourself. And you even can use a flag when calling SetWindowPos to disable sending WM_WINDOWPOSCHANGING.
You want something what's not allowed by default. But you still have ways how to do it. So I don't understand why are you complaining ;)
Edited by Tom KQT, 30 October 2013 - 04:57 AM.
### #8Tispe Members
Posted 30 October 2013 - 05:02 AM
My only complaint is that it tries to restrict the window size unless I tell it not to. It complicates things in my opinion.
### #9Tispe Members
Posted 30 October 2013 - 08:42 AM
Ok, SWP_NOSENDCHANGING works, but it is not a cure! It only prevents WM_WINDOWPOSCHANGING from being sent from SetWindowPos(), it does not prevent WM_WINDOWPOSCHANGING from being sent at another time by the OS.
You see, I can SetWindowPos() with this flag and get a 1920x1080 client area window. But as soon as I want to drag that window around on the desktop the OS sends another WM_WINDOWPOSCHANGING message messing it up again.
I have to capture the message in the winproc function and prevent it from doing anything.
LRESULT CALLBACK WindowProc(HWND hWnd, UINT message, WPARAM wParam, LPARAM lParam)
{
switch(message){
case WM_WINDOWPOSCHANGING:
return 0; //We stop the message here!
break;
.
.
.
.
.
.
}
Edited by Tispe, 30 October 2013 - 08:43 AM.
### #10Tom KQT Members
Posted 30 October 2013 - 11:01 AM
My only complaint is that it tries to restrict the window size unless I tell it not to. It complicates things in my opinion.
Complicates for you because you want to allow something what's not allowed by default and what programmers usually don't want. The point is that it simplifies things for most people and complicates for a minority ;)
Ok, SWP_NOSENDCHANGING works, but it is not a cure! It only prevents WM_WINDOWPOSCHANGING from being sent from SetWindowPos(), it does not prevent WM_WINDOWPOSCHANGING from being sent at another time by the OS.
You see, I can SetWindowPos() with this flag and get a 1920x1080 client area window. But as soon as I want to drag that window around on the desktop the OS sends another WM_WINDOWPOSCHANGING message messing it up again.
I have to capture the message in the winproc function and prevent it from doing anything.
That's exactly what I said. That trick allows you to create a large window. And I warned you that the window will bring you more problems (I named resizing, that was the most obvious one). I also said that you can process the message on your own and disable the resizing ;)
Old topic!
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2017-01-23 11:09:08
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https://wesmckinney.com/blog/apache-arrow-pandas-internals/
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This post is the first of many to come on Apache Arrow, pandas, pandas2, and the general trajectory of my work in recent times and into the foreseeable future. This is a bit of a read and overall fairly technical, but if interested I encourage you to take the time to work through it.
In this post I hope to explain as concisely as I can some of the key problems with pandas's internals and how I've been steadily planning and building pragmatic, working solutions for them. To the outside eye, the projects I've invested in may seem only tangentially-related: e.g. pandas, Badger, Ibis, Arrow, Feather, Parquet. Quite the contrary, they are all closely-interrelated components of a continuous arc of work I started almost 10 years ago.
Side note: consider making a tax-deductible donation to support pandas development
Some background
I started building pandas in April, 2008. It started out as a skunkworks that I developed mostly on my nights and weekends. I didn't know much about software engineering or even how to use Python's scientific computing stack well back then. My code was ugly and slow. I figured things out as I went and learned as much from others as I could. I didn't start doing serious C development until 2013 and C++ development until 2015. I appreciate C++ a lot more now than I would have 9 years ago.
Python was a comparatively more inhospitable place for what we might now call data science development. The problems that pandas solves for people in 2017 were not problems that people generally solved with Python at all. They generally used R, SAS, SPSS, Stata, or MATLAB, in no particular order of preference.
So maybe it's not a surprise that pandas's internal architecture has some warts. In Summer 2011, I devised a contraption known as the BlockManager, a memory management object that uses NumPy arrays internally, for managing the internal columns of data inside a pandas.DataFrame. You can see me writing about it all the way back in July 2011.
While the BlockManager and pandas's overall tight internal coupling to NumPy has served the project well historically, these things are some of the root causes of problems that plague pandas users working with larger datasets in modern times.
To put it simply, we weren't thinking about analyzing 100 GB or 1 TB datasets in 2011. Nowadays, my rule of thumb for pandas is that you should have 5 to 10 times as much RAM as the size of your dataset. So if you have a 10 GB dataset, you should really have about 64, preferably 128 GB of RAM if you want to avoid memory management problems. This comes as a shock to users who expect to be able to analyze datasets that are within a factor of 2 or 3 the size of their computer's RAM.
pandas rule of thumb: have 5 to 10 times as much RAM as the size of your dataset
There are additional, hidden memory killers in the project, like the way that we use Python objects (like strings) for many internal details, so it's not unusual to see a dataset that is 5GB on disk take up 20GB or more in memory. It's an overall bad situation for large datasets.
I started DataPad in 2013 with Chang She, my longtime friend and pandas collaborator. We wanted to use the nascent PyData stack to power the visual analytics application we were building, but we ran into some serious performance issues, especially in the cloud. The responsiveness of analytics queries from the DataPad application weren't great with pandas out of the box.
So I pared down the pandas feature set to the bare essentials and created a small new implementation which we called Badger. I found that through using contiguous, immutable columnar data structures optimized for data locality, that I could get 2-20x better performance in a wide variety of operations. The biggest wins were in string processing, but there were huge gains across the board. You can see a demo of DataPad here.
Badger was definitely "startup code". When we were acquired by Cloudera in 2014, I contemplated open sourcing Badger, but felt that it would be a lot of work to clean up the code (mostly written in C, with far too many macros) for human consumption and I wanted to build a more future-proof implementation that would still be useful 10 years down the road. Releasing it as-is would have been distracting for pandas users, and I didn't want to keep developing that codebase. It's not a good idea to release codebases only to abandon them. In light of the fact that basically a rewrite was needed, I left Badger on the shelf.
I gave a talk in November 2013 with the subtitle 10 Things I Hate About Pandas, which has had almost 100,000 slide views 4 years later. It's a summary of the things that I'd learned throughout 2013 and battle scars from the first 5 years of pandas development.
The 10 (really 11) things are (paraphrasing my own words):
1. Internals too far from "the metal"
2. No support for memory-mapped datasets
3. Poor performance in database and file ingest / export
4. Warty missing data support
5. Lack of transparency into memory use, RAM management
6. Weak support for categorical data
7. Complex groupby operations awkward and slow
8. Appending data to a DataFrame tedious and very costly
10. Eager evaluation model, no query planning
11. "Slow", limited multicore algorithms for large datasets
I had begun to solve some of these problems in Badger, but the solutions were narrow in scope to the problems we were solving at DataPad. Luckily, I moved to Cloudera where there were a lot of database and big data system developers for me to learn from.
At Cloudera, I started looking at Impala, Kudu, Spark, Parquet, and other such big data storage and analysis systems. Since Python and pandas had never been involved with any of these projects, building integrations with them was difficult. The single biggest problem was data interchange, particularly moving large tabular datasets from one process's memory space to another's. It was extremely expensive, and there was no standard solution for doing it. RPC-oriented serialization protocols like Thrift and Protocol Buffers were too slow and too general purpose.
As I dug through the different points of contact between different systems, I saw a lot of commonality with the problems I'd been working on above in Badger. Zero-copy data access was the biggest thing; you need to be able to memory map complex tables to make accessing 1 terabyte of data on disk as fast and easy as 1 megabyte.
By early 2015, I was yearning for what I was then calling a "columnar data middleware" which provided zero-copy access, with rich enough support for strings, nested types, and all the other hairy JSON-like data found in the wild. Like the prototype Badger runtime, this format needed to be optimized for data locality so that we could evaluate queries at maximum speeds.
I was lucky to bump into a collection of like-minded people across many big data projects, especially folks from Apache Drill, Impala, Kudu, Spark, and others. In late 2015, to create a neutral "safe space" free from software vendor affiliation (which can make industry collaborations more complex), we worked with the Apache Software Foundation to establish Apache Arrow.
On paper, Apache Arrow was everything I had been wanting for years. But, in late 2015, all I had (as far as Python is concerned) were some Markdown specification documents. These specifications weren't even final; we set up the Apache project to create a venue for the broader community to have a dialogue about the specs and the problems that Arrow solves. We had to buckle down and build real software to make the vision real and useful. Now that I've been working on the project for almost 2 years, we've made huge progress in realizing the things that we set out to accomplish.
I strongly feel that Arrow is a key technology for the next generation of data science tools. I laid out my vision for this recently in my JupyterCon keynote.
Also in late 2015, I wrote a long set of design documents to start discussions about building a faster, cleaner core pandas implementation, which we may call pandas2. pandas is a community project that governs itself based on consensus (with me as the BDFL to break impasses). I wanted to see if the rest of the core developers agreed with my assessment of what is wrong with pandas's internals. It's been 2 years since then, and by and large there has been general agreement on the problems, but how to solve them all without disrupting the existing pandas user community is an open question. Over this time I have focused on building computational infrastructure that will largely go unseen by pandas users.
Does Arrow solve the "10 Things"?
Arrow doesn't solve all of the 10 things quite yet, but it's made huge strides toward doing so.
Arrow's C++ implementation provides essential in-memory analytics infrastructure for projects like pandas:
• A runtime column-oriented memory format optimized for analytical processing performance
• A zero-copy, streaming / chunk-oriented data layer designed for moving and accessing large datasets at maximum speeds
• Extensible type metadata for describing a wide variety of flat and nested data types occurring in real-world systems, with support for user-defined types
What's missing from the Arrow C++ project at the moment (but not for too much longer) is:
• A comprehensive analytical function "kernel" library
• Logical operator graphs for graph dataflow-style execution (think TensorFlow or PyTorch, but for data frames)
• A multicore schedular for parallel evaluation of operator graphs
I'll write more about the roadmap for building an analytics engine for Arrow memory (that we can use in projects like pandas) in a follow up post.
In the rest of this post, I'm going to go deeper into the "10 Things" and how they're addressed by the Arrow project.
1. Getting closer to the metal
All memory in Arrow on a per column basis, whether strings, numbers, or nested types, is arranged in contiguous memory buffers optimized for random access (single values) and scan (multiple values next to each other) performance. The idea is that you want to minimize CPU or GPU cache misses when looping over the data in a table column, even with strings or other non-numeric types.
In pandas, an array of strings is an array of PyObject pointers, and the actual string data lives inside PyBytes or PyUnicode structs that live all over the process heap. As developers, we are hamstrung by the bloated, memory-bound nature of processing these objects. In Python, the simple string 'wes' occupies 52 bytes of memory. '' occupies 49 bytes. For a great discussion of issues around this, see Jake Vanderplas's epic exposé on Why Python is Slow.
In Arrow, each string is right next to the previous one in memory, so you can scan all of the data in a column of strings without any cache misses. Processing contiguous bytes right against the metal, guaranteed.
Arrow's C/C++ API means that applications which know nothing about Python can consume or produce pristine Arrow tables and share them either in-process or via shared memory / memory maps. pandas's lack of a C or Cython API for data frames has been another big problem over time.
2. Memory mapping huge datasets
Perhaps the single biggest memory management problem with pandas is the requirement that data must be loaded completely into RAM to be processed. pandas's internal BlockManager is far too complicated to be usable in any practical memory-mapping setting, so you are performing an unavoidable conversion-and-copy anytime you create a pandas.DataFrame.
Arrow serialization design provides a "data header" which describes the exact locations and sizes of all the memory buffers for all the columns in a table. This means you can memory map huge, bigger-than-RAM datasets and evaluate pandas-style algorithms on them in-place without loading them into memory like you have to with pandas now. You could read 1 megabyte from the middle of a 1 terabyte table, and you only pay the cost of performing those random reads totalling 1 megabyte. With modern solid state drives, this is generally a good strategy.
Arrow's memory-mapping capability also allows multiple processes to work with the same large dataset without moving it or copying it in any way. We've seen this applied to great effect in the Plasma Object Store (now part of Arrow) used in the Ray project at UC Berkeley.
3. High speed data ingest and export (databases and file formats)
Arrow's efficient memory layout and rich type metadata make it an ideal container for inbound data from databases and columnar storage formats like Apache Parquet.
One of Arrow's primitive constructs is the concept of a "record batch stream", a sequence of atomic tables together comprising a large dataset. This stream processing data model is an idea for databases which serve streams of records from a database cursor.
We have been developing a high-speed connector with Parquet format. We've also seen the optimized turbodbc project for ODBC-based database connections.
I aspire to build Arrow-native connectors for many other file formats and databases, such as:
• SQLite
• PostgreSQL
• Apache Avro
• Apache ORC
• CSV (a better version of pandas.read_csv)
• JSON
4. Doing missing data right
All missing data in Arrow is represented as a packed bit array, separate from the rest of the data. This makes missing data handling simple and consistent across all data types. You can also do analytics on the null bits (AND-ing bitmaps, or counting set bits) using fast bit-wise built-in hardware operators and SIMD.
The null count in an array is also explicitly stored in its metadata, so if data does not have nulls, we can choose faster code paths that skip null checking. With pandas, we cannot assume that arrays do not have null sentinel values and so most analytics has extra null checking which hurts performance. If you have no nulls, you don't even need to allocate the bit array.
Because missing data is not natively supported in NumPy, over time we have had to implement our own null-friendly versions of most key performance-critical algorithms. It would be better to have null-handling built into all algorithms and memory management from the ground up.
5. Keeping memory allocations in check
In pandas, all memory is owned either by NumPy or the Python interpreter, and it can be difficult to measure exactly how much memory is used by a given pandas.DataFrame. It's not unusual for a line of code to double or triple the memory footprint of a process due to temporary allocations, sometimes causing a MemoryError.
In Arrow's C++ implementation, all memory allocations are carefully tracked in a central "memory pool", so you know exactly how much Arrow memory is in RAM at any given time. By using "subpools" with parent-child relationships, you can precisely measure the "high water mark" in algorithms to understand the peak memory usage of analytical operations. This technique is common in databases to monitor or limit memory usage in operator evaluation. If you know that you are going to exceed available RAM, you can apply mitigation strategies like spilling to disk (where the ability to memory-map on-disk datasets is of course key).
In Arrow memory is either immutable or copy-on-write. At any given time, you know if another array references a buffer that you can see. This enables us to avoid defensive copying.
6. Supporting categorical data well
When I gave my talk in 2013, pandas did not have the pandas.Categorical type; that was implemented afterwards. But pandas's workarounds for data types not in NumPy has always been a bit warty. If you step outside pandas, you can't work with pandas Categoricals. The way that extension dtypes are implemented works, but is a bit bolted-on due to pandas's tight coupling to NumPy.
In Arrow, categorical data is a first-class citizen, and we have prioritized having an efficient and consistent representation both in-memory and on the wire or in shared memory. We support sharing categories (called dictionaries in Arrow) between multiple arrays.
pandas has other user-defined types: datetime with time zone and periods. We intend to be able to support logical data types (having a particular physical memory representation) in Arrow gracefully so that a particular system can faithfully transport its data using Arrow without having to make changes to the Arrow format documents.
7. Better groupby(...).apply operations
The way that Arrow helps is by enabling easier parallelization of groupby operations; due to other problems listed here, it is difficult or impossible to fully parallelize a df.groupby(...).apply(f) operation.
At some point, we will also want to improve the API for complex apply operations in pandas.
8. Appending to data frames
In pandas, all of the data in a column in a DataFrame must reside in the same NumPy array. This is a restrictive requirement, and frequently results in memory-doubling and additional computation to concatenate Series and DataFrame objects.
Table columns in Arrow C++ can be chunked, so that appending to a table is a zero copy operation, requiring no non-trivial computation or memory allocation. By designing up front for streaming, chunked tables, appending to existing in-memory tabler is computationally inexpensive relative to pandas now. Designing for chunked or streaming data is also essential for implementing out-of-core algorithms, so we are also laying the foundation for processing larger-than-memory datasets.
There are multiple layers of complexity to adding new data types:
• Creating dynamic dispatch rules to operator implementations in analytics
For example, a "currency" type could have a currently type a string, with the data physically represented as a float64 or decimal. So you could treat the currency computationally like its numeric representation, but then carry through the currency metadata in numeric operations.
The rules about preserving metadata may be operator-dependent, so it can get complicated.
In Arrow we have decoupled the metadata representation from the details of computation and metadata nannying. In the C++ implementation, we have been planning ahead for user-defined types, so when we are focusing more on building an analytics engine it is a goal to enable the creation of user-defined operator dispatch and metadata promotion rules.
10/11. Query planning, multicore execution
When you write df[df.c < 0].d.sum(), pandas creates a temporary DataFrame df[df.c < 0] then sums the d column of that temporary object. If df contains a lot of columns, this is ridiculously wasteful. Of course you can write df.d[df.c < 0].sum(), but even that produces a temporary Series, which is then summed!
Clearly, if you know the whole expression you are evaluating you can do better and avoid these temporary allocations altogether. Additionally, many algorithms (including this example) can be parallelized amongst all the processors cores on your computer.
As part of building an analytics engine for Arrow, we also plan to build a lightweight physical "query planner" with a multicore in-process scheduler to enable many kinds of algorithms to be parallelized and evaluated efficiently. There is substantial prior art in the domain of graph data flow execution (particularly in the ML world lately, like TensorFlow and PyTorch), so this amounts to creating a graph data flow engine whose primitive unit of data is an Arrow table.
To plan ahead for this use case, in 2015, I started the Ibis project (still under active development) to create a pandas-friendly deferred expression system for static analysis and compilation these types of operations. Since an efficient multithreaded in-memory engine for pandas was not available when I started Ibis, I instead focused on building compilers for SQL engines (Impala, PostgreSQL, SQLite), similar to the R dplyr package. Phillip Cloud from the pandas core team has been actively working on Ibis with me for quite a long time.
What's next?
In an upcoming blog post, I will go into some more detail about the roadmap for building an Arrow-native multithreaded in-memory execution engine and how that's relevant to the architecture of pandas2.
Dask makes it easy to read a directory of CSV files by running pandas.read_csv in parallel and then running a groupby operation on the entire dataset. Truly, what Matt Rocklin and team have built is an excellent piece of kit.
One issue with the Dask model is that it's using pandas as a black box. dask.dataframe does not solve pandas's inherent performance and memory use problems, but it spreads them out across multiple processes and helps mitigate them by being careful to not work with too large pieces of data all at once, which can result in an unpleasant MemoryError.
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2021-05-15 12:00:49
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http://math.stackexchange.com/questions/628541/how-does-correlation-measure-the-strength-of-a-linear-relationship
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# How does correlation measure the strength of a linear relationship?
Let $X$ and $Y$ be random variables
The correlation between $X$ and $Y$ is:
$\frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}=\frac{E[(X-\mu_X)(Y-\mu_Y)]}{\sigma_X \sigma_Y}$
So, correlation is suppose to measure the strength of a linear relationship.
But, given the above formula, I don't understand how it is going to work.
Let's assume that $X$ and $Y$ are perfectly correlated i.e. 1. Then, based on the formula, all I know is $Cov(X,Y)$ will some big number. How is it that dividing by the product of the standard deviation of $X$ and $Y$ helps to measure the strength of its linear relationship?
I also tried to draw a graph of $Y=X$ to get a geometric understanding of the formula but that hasn't help yet. My guess is that the correlation is suppose to be the slope of the line Y=X but the way the formula is defined does not make it obvious to me.
-
Erm what...? The size of covariance partly reflect how the variables are. If X=Y, then X and Y perfectly correlated, if var (X) = 0.00000000000001, then what is covariance between X and Y and how is that a very big number? Dividing by standard deviation scales it. – Lost1 Jan 5 '14 at 22:23
@Lost1 If $X=Y$, then $Cov(X,Y)=Var(X)$. $Var(X)$ can be a big number, can't it? – mauna Jan 5 '14 at 22:46
I just gave an example when it is very small. The point is it can be anything.... You wrote 'it will be a big number'. This is complete bs. – Lost1 Jan 6 '14 at 0:21
However the correlation is always 1, regardless what the variance is. That is why you look at correlation. It is always between -1 and 1 and independent of scales. If you look at relationship between height and weight. Using weight in kg and lb will give different covariance against height in m or feet, but correlation is always the same independent of choice of units. – Lost1 Jan 6 '14 at 0:23
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2015-04-28 10:54:02
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https://rkenmi.com/posts/decode-number-of-ways?lang=en
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@rkenmi - Decode number of ways
# Decode number of ways
## Problem
A message containing letters from A-Z can be encoded into numbers using the following mapping:
'A' -> "1"
'B' -> "2"
...
'Z' -> "26"
To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:
"AAJF" with the grouping (1 1 10 6) "KJF" with the grouping (11 10 6) Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".
Given a string s containing only digits, return the number of ways to decode it.
The answer is guaranteed to fit in a 32-bit integer.
## Input
• 1 <= s.length <= 100
• s, the string input, contains only digits and may contain leading zero(s).
## Approach
This is a dynamic programming problem because the results of a previously decoded subproblem can contribute to solve the current subproblem.
I am personally not a big fan of this problem because the decoding part can be really confusing at the implementation level when trying to re-use previous results. A clean approach to decoding delivers the best results. The clearest way to solve this problem is to do the following:
1. Cover the first two base cases
• For subproblem of size 0, what is the number of ways to decode? - This is naturally 1
• For subproblem of size 1, what is the number of ways to decode? - If the first digit in s is 0 then this would be 0. For any other digit, we have a mapping, so we would have a 1.
2. Loop through each character in s, starting from index 2
• If the current digit, i.e. "ones" digit is greater than 1, we'll re-use the answer for the previous subproblem i-1. This is the typical case where the number of ways to decode doesn't actually increase or decrease because the digit itself just has one mapping that we can add in a straightforward manner.
• For example, adding 3 to 12 means going from [AB, L] to [ABC, LC], which is still 2 ways to decode.
• If we look at the previous digit also and combine together to form a "double digit", and it happens to be within the range of 10-26, then we have a special mapping for it, so we'll also add the 2nd previous subproblem i-2 to the current results. This is because we want to combine the two different ways to decode - one from the additional mapping between digits 10-26 (i.e. 11 = K) and also one for the single digits (5 = E).
## Solution
def numDecodings(self, s: str) -> int:
results = []
dp = [0] * (len(s) + 1)
dp[0] = 1
dp[1] = 1 if s[0] != "0" else 0
if s[0] == "0" and len(s) == 1:
return 0
for i in range(2, len(s) + 1):
one_digit = int(s[i-1])
two_digits = int(s[i-2:i])
if one_digit >= 1:
dp[i] += dp[i-1]
if 9 < two_digits < 27:
dp[i] += dp[i-2]
return dp[-1]
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2022-11-26 13:28:40
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https://questioncove.com/updates/524bc530e4b06f86e8212399
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OpenStudy (anonymous):
Use linear approximation, i.e. the tangent line, to approximate 5.3^2 as follows: Let f(x)=x^2 and find the equation of the tangent line to f(x) at x=5 . Using this, find your approximation for 5.3^2
4 years ago
OpenStudy (anonymous):
anyone have a clue?
4 years ago
OpenStudy (anonymous):
$f(x)\approx L(x)$Where $L(x) = f'(a)(x-a)+f(a)$
4 years ago
OpenStudy (anonymous):
Typically $$a$$ is some point that is close to $$x$$ and $$f(a)$$ is easier calculated than $$f(x)$$ is.
4 years ago
OpenStudy (anonymous):
In this case $$a=5$$ and $$x=5.3$$.
4 years ago
OpenStudy (anonymous):
So first we find $$f'(5)=2x|_{x=5}=2(5)=10$$
4 years ago
OpenStudy (anonymous):
Next we find $$f(5)=x^2|_{x=5}=(5)^2=25$$
4 years ago
OpenStudy (anonymous):
This means $f(x)\approx L(x) = 10(x-5)+25 = 10x-50+25=10x-25$
4 years ago
OpenStudy (anonymous):
So given that $L(x)= 10x-15$We can approximate $$f(5.3)$$:$L(5.3)=10(5.3)-25=53-25=28$
4 years ago
OpenStudy (anonymous):
In short $$f(5.3)\approx 28$$
4 years ago
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2017-12-17 02:17:59
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https://www.mysciencework.com/publication/show/infinite-dimensional-hamiltonian-description-class-dissipative-mechanical-systems-c5fd31df
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# Infinite-dimensional Hamiltonian description of a class of dissipative mechanical systems
Authors
Type
Preprint
Publication Date
Mar 06, 2011
Submission Date
Jun 16, 2009
Identifiers
arXiv ID: 0906.3062
Source
arXiv
License
Unknown
External links
## Abstract
In this paper an approach is proposed to represent a class of dissipative mechanical systems by corresponding infinite-dimensional Hamiltonian systems. This approach is based upon the following structure: for any non-conservative classical mechanical system and arbitrary initial conditions, there exists a conservative system; both systems share one and only one common phase curve; and, the value of the Hamiltonian of the conservative system is, up to an additive constant, equal to the total energy of the non-conservative system on the aforementioned phase curve, the constant depending on the initial conditions. We describe in detail this relationship calling the conservative system substitute conservative system. By considering the dissipative mechanical system as a special fluid in a domain $D$ of the phase space, viz. a collection of particles in this domain, we are prompted to develop this system as an infinite-dimensional Hamiltonian system of an ideal fluid. By comparing the description of the ideal fluid in Lagrangian coordinates, we can consider the Hamiltonian and the Lagrangian as the respective integrals of the Hamiltonian and the Lagrangian of the substitute conservative system over the initial value space and define a new Poisson bracket to express the equations of motion in Hamiltonian form. The advantage of the approach is that the value of the canonical momentum density $\pi$ is identical with that of the mechanical momentum $m\dot{q}$ and the value of canonical coordinate $q$ is identical with that of the coordinate of the dissipative mechanical system. Therefore we need not to decouple the Newtonian equations of motion into several one-dimensional ordinary differential equations.
Seen <100 times
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2018-11-18 14:28:27
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https://chemistry.stackexchange.com/questions/6917/how-can-you-tell-how-much-actually-reacted-in-an-acid-base-reaction
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# How can you tell how much actually reacted in an acid base reaction
If you facilitate a weak acid base reaction. i.e. $\ce{NaOH + H2SO4}$. All of it may not react and the solutions will not neutralize. How can you determine how much actually reacted?
• Welcome to Chemsitry.SE. There is a way to determine the extent of reaction of weak acids and weak bases using equilibrium calculations. However, your acid/base pair of choice is a strong acid $(\ce{H2SO4})$ and a strong base $(\ce{NaOH})$. Assuming you have stoichiometric equivalence between the two, this reaction will go to completion. – Ben Norris Nov 13 '13 at 12:03
• @BenNorris that would also be true with a weak acid and a strong base. One should also consider that $\ce{H2SO4}$ is not that strong on the second ionization. – mannaia Mar 21 '14 at 14:58
• @fp I think you are right. you should answer the question writing it is not a good praxis use comment to answer the question .. – G M Mar 22 '14 at 9:11
As noted by Ben in the comment, $\ce{NaOH}$ and $\ce{H_2SO_4}$ are strong base and acid, respectively, so they will react till one is completely used up. Those react in 1:2 molar ratio, so it would be easiest to determine if you have more moles of one or another, and it will simply react, using one mole of $\ce{H_2SO_2}$ per two moles of $\ce{NaOH}$, until there is only one left. If there is one left, you will have a non-neutral solution in the end.
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2020-01-18 10:06:04
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http://accesssurgery.mhmedical.com/content.aspx?bookid=427§ionid=40372728
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Chapter 84
The term pulmonary arteriovenous malformation (AVM) refers to lesions that have abnormal communications between the pulmonary arteries and pulmonary veins. Numerous other names have been used in the past to describe these lesions, such as pulmonary telangiectasias, aneurysms, fistulas, hemangiomas, and cavernous angiomas. These lesions can be congenital, usually as part of the hereditary hemorrhagic telangiectasia, also known as Rendu-Osler-Weber syndrome, or acquired from bronchiectasis, infections, hepatic cirrhosis, mitral stenosis, malignancies, or trauma. AVMs have been described based on number (single versus multiple), location (unilateral versus bilateral; parenchymal versus pleural), and size or type of drainage (simple versus complex).1,2
Clinical suspicion for the presence of pulmonary AVM should arise when there is presence of suggestive pulmonary nodules; family history of hereditary hemorrhagic telangiectasia; sequelae of right-to-left shunting such as hypoxemia, dyspnea, clubbing, cyanosis, and polycythemia; and systemic embolism such as cerebral stroke or cerebral abscess. Epistaxis can be reported in up to 85% of patients with hereditary hemorrhagic telangiectasia.1 A continuous bruit can be auscultated over the lesion. The triad of cyanosis, clubbing, and polycythemia is seen in 20% of patients. Approximately 90% of AVMs are unilateral, and 50–67% of patients have a single AVM.1,2
Workup of such patients should include a chest CT scan, which is the most sensitive test, evaluation of the shunt fraction, and pulmonary angiography to assess the feasibility of embolization. Approximately 25% of AVMs tend to enlarge up to a rate of 2 mm per year, and patients who are not treated have a stroke rate of 13% and a brain abscess rate of 11%. Complications are more common in patients who have AVMs greater that 2 cm or afferent vessels greater than 3 mm.1,2 At minimum, treatment should be offered to these patients, if not to all patients with angiographically accessible lesions.1
Embolization of pulmonary AVMs was first described in 1977.1,2 Pulmonary angiography is performed, and numerous techniques have been described, including coils, balloons, and sclerotic agents. Multiple AVMs can be embolized at a single session or a few weeks apart. Embolization is feasible in the many patients with angiographically accessible lesions, although treating patients with multiple feeding vessels can be challenging. Complications after balloon occlusion include balloon migration with distal embolization, balloon deflation, and pulmonary infarction. Long-term follow-up after embolization procedures is sparse. Recurrence after embolization has been reported.
Surgery is reserved for patients who cannot be embolized or who have failed embolizations. Surgical techniques used include thoracotomy and video-assisted thoracic surgery, and the extent of resection can range from fistulectomy and segmental resection to lobectomy and even pneumonectomy.3 While pneumonectomy has been used in the past, most lesions can be dealt today with lung-sparing techniques. Preoperative pulmonary function testing is done as indicated based on the degree of pulmonary resection.
When anatomic resections such as segmentectomy or lobectomy are ...
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## Subscription Options
### AccessSurgery Full Site: One-Year Subscription
Connect to the full suite of AccessSurgery content and resources including more than 160 instructional videos, 16,000+ high-quality images, interactive board review, 20+ textbooks, and more.
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2017-01-21 13:21:31
|
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https://www.toppr.com/ask/content/concept/inertial-gravitational-mass-208547/
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# Inertial and Gravitational Mass
## law
### Inertial and Gravitational mass
1) Inertial mass
This is defined by Newton's 2nd law- F = ma, which states that when a force F is applied to an object, it will accelerate proportionally, and that constant of proportion is the mass of that object. In very concrete terms, to determine the inertial mass, you apply a force of F Newtons to an object, measure the acceleration in m/s2, and F/a will give you the inertial mass m in kilograms.
2) Gravitational mass
This is defined by the force of gravitation, which states that there is a gravitational force between any pair of objects, which is given by
where G is the universal gravitational constant, and are the masses of the two objects, and r is the distance between them. This, in effect defines the gravitational mass of an object.
Gravitational mass is measured by comparing the force of gravity of an unknown mass to the force of gravity of a known mass. This is typically done by balance scale.
## definition
### equivalence principle in einstein's general relativity
Consider following three conditions to understanding the Equivalence of Inertial and Gravitational mass.
1. let's consider a person standing in spaceship resting on earth. His feet on the floor of the ship. Now we know the normal force is equal to an apparent weight of person which is mg, As acceleration a = 0 (spaceship at rest). Hence normal force = apparent weight = mg.
2. Man in a spaceship which is very far from the planet and he is floating in the ship, hence acceleration a = 0. An apparent weight of man becomes zero (weightlessness).
3. Man in a spaceship which is accelerating with acceleration a = g in space, now the apparent weight of man becomes mg. A spaceship is far away from planet there is no any g due to the planet. But man touching the floor and it gives the normal force to man which give the apparent weight = mg.
Hence the force due to gravity is same as that of acceleration. Hence there is an equivalence between inertial and gravitational mass.
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2022-08-20 05:17:20
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https://gamedev.stackexchange.com/questions/181270/what-is-the-correct-way-to-make-an-installed-build/181272
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# What is the correct way to make an Installed Build?
I am trying to use the Automation tool, following the documentation.
To create an Installed Build:
Run the Installed Build Script by invoking the AutomationTool with the following command line, replacing [PLATFORM] with either Win64 or Mac.
BuildGraph -target="Make Installed Build [PLATFORM]" -script=Engine/Build/InstalledEngineBuild.xml -clean
However, I am getting this error.
ERROR: Target 'Make Installed Build [Win64]' is not in graph
I am not sure what I should put as the platform.
I searched the config file, but don't see any specified platforms.
BuildGraph -target="Make Installed Build Win64" -script=Engine/Build/InstalledEngineBuild.xml -clean
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2022-10-05 12:59:32
|
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|
https://mathoverflow.net/questions/338212/is-there-a-morita-cocycle-for-the-mapping-class-group-modg-n-when-n-1
|
# Is there a Morita cocycle for the mapping class group Mod(g,n) when n > 1?
Write Mod(g,n) for the mapping class group of a genus-$$g$$ surface $$\Sigma$$ with $$n$$ boundary components. When $$n=0,1$$ we define the Torelli group $$T$$ to be the subgroup of Mod(g,n) which acts trivially on the homology $$H = H_1(\Sigma,\mathbf{Z})$$.
The Johnson homomorphism is a much-studied homomorphism from the Torelli group to $$\mathrm{Hom}(H,\wedge^2 H)$$ (when n=1) or a quotient of this (when n=0) whose kernel turns out to be the commutator subgroup of Torelli.
Morita showed in 1993 that the Johnson homomorphism extends to the whole group Mod(g,1), not as a homomorphism, but as a 1-cocycle in
$$H^1(\mathrm{Mod}(g,1), \mathrm{Hom}(H,\wedge^2 H))$$
where the action is given by the action of Mod(g,n) on $$H$$. (Thus the Morita cocycle restricts to a homomorphism on Torelli, as claimed.) It can be thought of as keeping track of the action of the mapping class on the quotient of $$\pi_1(\Sigma)$$ by the third term of its lower central series.
All of the above is well-known, or at least well-known to the people who know this kind of thing well. Now here's my question: is there a Morita cocycle on Mod(g,n) when n > 1?
Of course, such a cocycle would restrict to a Johnson homomorphism from the Torelli subgroup of Mod(g,n), and even this is subtle; but Church's paper "Orbits of curves under the Johnson kernel," gives a way to define a Torelli group and a Johnson homomorphism for Mod(g,n) which behaves well with respect to inclusion of subsurfaces. So a more specific version of my question would be: when n > 1, does Church's "Johnson homomorphism" extend to a "Morita cocycle" on all of Mod(g,n) which behaves well with respect to inclusion of subsurfaces?
The answer is "yes" -- in fact one can do better and get a class in
$$H^1(\text{Aut}(F_m), \text{Hom}(H, \wedge^2 H)),$$
where $$F_m$$ is the free group on $$m$$ generators and $$H$$ is the Abelianization of $$F_m$$, if I'm not mistaken. This gives the cocycle you want since there is an obvious map $$\text{Mod}_{g, n}\to \text{Aut}(\pi_1(\Sigma_{g, {n-1}}))\simeq F_{2g+n-2}$$ for $$n\geq 2$$, given by the conjugation action of $$\text{Mod}_{g,n}$$ on the point-pushing subgroup, namely $$\pi_1(\Sigma_{g, {n-1}})$$.
A construction goes as follows. Let $$\mathbb{Z}[F_m]$$ be the group ring of $$F_m$$, and let $$\mathscr{I}$$ be the augmentation ideal. Then $$H \simeq \mathscr{I}/\mathscr{I}^2$$ canonically (via the map sending $$g$$ to $$g-1$$) and $$\mathscr{I}^2/\mathscr{I}^3\simeq H^{\otimes 2}$$ canonically (via the multiplication map). There is a short exact sequence of $$\text{Aut}(F_m)$$ modules $$0\to \mathscr{I}^2/\mathscr{I}^3\to \mathscr{I}/\mathscr{I}^3\to \mathscr{I}/\mathscr{I}^2\to 0,$$ which we can think of as an extension of $$H$$ by $$H^{\otimes 2}$$, and hence gives a class in
$$H^1(\text{Aut}(F_m), \text{Hom}(H, H^{\otimes 2})).$$
But in fact a direct computation of a crossed homomorphism representing this class shows that it lands in $$\text{Hom}(H, \text{Alt}^2(H)).$$
• Actually this will all be in a paper I'm writing with one of your students, among others! – Daniel Litt Aug 13 '19 at 0:00
• To fix the conjugation action, I am worried you need to fix a base point which is $\operatorname{Mod}(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something. – Will Sawin Aug 13 '19 at 0:57
• @WillSawin: This is a good point. One can fix the base-point dependence by taking the Inn-coinvariants of $Hom(H, \wedge^2 H)$, but this does lose a bit of information (and in fact I think the class one gets if one does this is pulled back from $\text{Mod}(g, n-1)$...) – Daniel Litt Aug 13 '19 at 1:36
• @JSE Don't you also have a paper on this stuff with one of your students? – Will Sawin Aug 13 '19 at 14:11
• We're working on it but for one of the theorems we want to prove I need this setup to work! And it's the same student, Wanlin Li! (Also Daniel Corey.) – JSE Aug 13 '19 at 15:24
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2020-02-21 13:47:25
|
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|
https://repo.scoap3.org/record/33752
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# Studies of Beauty Suppression via Nonprompt ${D}^{0}$ Mesons in Pb-Pb Collisions at $\sqrt{{s}_{\mathrm{NN}}}=5.02\text{}\text{}\mathrm{TeV}$
10 July 2019
Abstract: The transverse momentum spectra of ${D}^{0}$ mesons from $b$ hadron decays are measured at midrapidity ($|y|<1$) in $pp$ and Pb-Pb collisions at a nucleon-nucleon center of mass energy of 5.02 TeV with the CMS detector at the LHC. The ${D}^{0}$ mesons from $b$ hadron decays are distinguished from prompt ${D}^{0}$ mesons by their decay topologies. In Pb-Pb collisions, the $B\to {D}^{0}$ yield is found to be suppressed in the measured ${p}_{T}$ range from 2 to $100\text{}\text{}\mathrm{GeV}/c$ as compared to $pp$ collisions. The suppression is weaker than that of prompt ${D}^{0}$ mesons and charged hadrons for ${p}_{T}$ around $10\text{}\text{}\mathrm{GeV}/c$. While theoretical calculations incorporating partonic energy loss in the quark-gluon plasma can successfully describe the measured $B\to {D}^{0}$ suppression at higher ${p}_{T}$, the data show an indication of larger suppression than the model predictions in the range of $2<{p}_{T}<5\text{}\text{}\mathrm{GeV}/c$.
Published in: Physical Review Letters 123 (2019)
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2019-07-21 06:20:54
|
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|
https://tex.stackexchange.com/questions/312670/fill-polygon-with-image-crop-image-by-polygon
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# fill polygon with image / crop image by polygon
I would like to have an arbitrary polygon filled with an image (jpg, png, pdf, ...) for it to be included in a beamer frame.
I was thinking this may be possible using tikz, but I couldn't figure it out.
• Jun 2 '16 at 8:59
Crop jpeg into circular tikz node is valid for any kind of node (rectangle, circle or any already defined shape), but you can also define irregular polygons with a clip path:
\documentclass{beamer}
\usepackage{tikz}
\begin{document}
\begin{frame}{I'm watching you!}
\centering
\begin{tikzpicture}
\clip (-1,1)--++(-20:5cm)--++(75:4.5)--++(150:2cm)--++(200:3.5)--cycle;
\node at (2,2) {\includegraphics[width=6cm]{frog}};
\end{tikzpicture}
\end{frame}
\end{document}
• great! this is what I was looking for. Could you help me with the specification of the polygon nodes? Let's say I want a polygon with the corner coordinates (x,y) = (0cm,0cm) (1cm,0cm) (2cm,1cm) (0cm,1cm), what is the \clip line going to look like? Thanks Jun 2 '16 at 12:15
• @Bastian \clip (0,0) --(1,0)-- (2,1)--(0,1)--cycle; Jun 2 '16 at 13:26
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2021-09-27 20:07:27
|
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|
https://codegolf.stackexchange.com/questions/86463/interquartile-mean/86469
|
Interquartile Mean
Given (by any means) a sorted floating point dataset, return (by any means and within 1‰ of the correct value) the interquartile mean.
One possible algorithm
1. Discard the lowest and highest quarters of the data points.
2. Calculate the average (sum divided by count) of the remaining data points.
Note: If the dataset size is not evenly split-able in four, you will have to weigh the datapoints that are shared by sub-sets. See Example evaluation 2 below.
Example evaluation 1
Given {1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38}
1. Data count is 12, so we remove lowest and highest 3 datapoints:
{1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38}
2. Average of the remaining 6 datapoints:
(5 + 6 + 6 + 7 + 7 + 8) / 6 = 6.5
Example evaluation 2
Given {1, 3, 5, 7, 9, 11, 13, 15, 17}
1. Count is 9, so each quarter has 2¼ datapoints:
{1, 2, (0.25×5), (0.75×5), 7, 9, 11, (0.75×13), (0.25×13), 15, 17}
2. Average of the remaining 4.5 datapoints:
(0.75×5 + 7 + 9 + 11 + 0.75×13) / 4.5 = 9
Scilab, 8 bytes
trimmean
See the documentation. By default, discard=50, so the IQM is computed.
EDIT: You know, this is a trivial built-in answer, so I’m marking it as CW.
• I guess this will be the winner. Well done. – Adám Jul 26 '16 at 19:30
Pyth, 11 10 bytes
.O><lQS*4Ql
.OsPtc4S*4
Test suite.
How it works
It quadruplicates the input list to ensure that the data count is divisible by 4.
It still needs sorting, because *4 applies to the whole list instead of to each individual element.
Then, it splits the list into four equal parts, then takes away the first and the last part.
The remaining list is flattened and the average is taken.
MATL, 12 11 bytes
4Y"G"6L)]Ym
Input is a horizontal vector, with the format
[1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38]
or
[1 3 4 5 6 6 7 7 8 8 9 38]
Try it online!
Explanation
4Y" % Input horizontal vector implicitly. Repeat each element 4 times (run-length
% decoding). The resulting array is still sorted.
G" % Push input, for each: repeat as many times as the input size
6L) % Remove first and last elements, by applying the index "2:end-1"
] % End for each
Ym % Compute mean. Display implicitly
• I don't get it. How does 6L) remove the first and last elements? When I do it, it pushes a bunch of complex numbers. – DJMcMayhem Jul 24 '16 at 17:11
• @DrGreenEggsandIronMan Complex numbers can be used for that in MATL. The imaginary unit stands for the end of the array, and if there are two of three numbers they define a range. So [2, -1+i] when used as an index means 2:end-1 – Luis Mendo Jul 24 '16 at 17:19
Snowman, 66 bytes
}vg","aS:10sB;aM4aRAsOal,4nD,aG0AaGalNdEAaL1AfL:nA;alaF,nDtSsP
Try it online!
Uses the same algorithm as @LeakyNun's answers.
} enable variables b, e, and g
vg read a line of input into b
","aS split on commas (in-place)
:10sB;aM convert each element in resulting array to number ("frombase(10)-map")
4aR repeat the array 4 times
AsO sort the array
al take the length and put it in e without consuming b (the array)
, swap b and e, then move e to g; now b=length g=array
4nD divide b by 4 (4 was stored in e, which is why the array was moved)
, move the array and length/4 back to their original positions
aG split the array into groups of length (length/4)
0AaG take all elements with index >0 (i.e. remove the first element)
al store the length of the new array in e again
NdE bring it up to b, decrement, and put it back
AaL take all elements with index <length-1 (i.e. remove last)
1AfL flatten the array 1 level deep
:nA; push a block that adds two numbers (to e)
al store the length of this new array in g
aF fold b over e (sum the numbers)
, move g (the length) into e
nD divide the sum by the length, resulting in the average
tSsP to-string and print
• This language looks awful. I love it. – Mego Jul 25 '16 at 9:11
Python 3, 50 bytes
lambda n:sum(sorted(n*4)[len(n):-len(n)])/len(n)/2
Ideone it!
How it works
It is a translation of my answer in Pyth.
Jelly, 1413 12 bytes
x4ṫL‘$ḣLN$S÷LH
x4ṫLḊḣLN$S÷LH x4œs4ḊṖFS÷LH Try it online! Test suite. How it works It is a translation of my answer in Pyth. • I'm pretty sure this can be shortened, as I can do it 15 in APL. – Adám Jul 24 '16 at 14:15 • @Adám Please post your solution (so that I may copy haha) – Leaky Nun Jul 24 '16 at 14:16 • I want to give Marinus a chance... – Adám Jul 24 '16 at 14:20 • Here you go! – Adám May 18 '17 at 15:10 • Enough of a chance after more than 9 months, certainly – Luis Mendo May 18 '17 at 15:14 Pyke, 16 13 bytes 4*S4ftOsDsRl/ Try it here! • You broke my streak... – Leaky Nun Jul 24 '16 at 14:43 • I'm so sorry :( – Blue Jul 24 '16 at 14:45 Brachylog, 21 bytes :3jo@4brbcLl/N,L+:N*. Explanation This is basically @LeakyNun's Pyth answer algorithm. :3j Append 3 copies of the input to itself o@4 Sort and split in 4 lists of equal length brb Remove the head and the tail of the list of lists cL Concatenate the 2 sublists into a list L l/N, N is the inverse of the length of L L+:N*. Output is the product of N and the sum of the elements of L The only small trick there is in multiplying by the inverse of the length instead of dividing by the length, because division between 2 integers is integer division. Octave, 44 bytes @(x)mean(reshape(~~(1:4)'*x,[],4)(:,2:3)(:)) This defines an anonymous function. The input is a horizontal vector. Explanation The input horizontal vector is first matrix-multiplied (*) by a column vector of four ones (built with ~~(1:4)'). The result is a four-column matrix where each row is a copy of the input vector. This is then reshaped, while maintaining the linear order of the elements, into a 4-column matrix (reshape(...,[],4)). The center two columns are kept ((:,2:3)) and linearized into a single column ((:)), of which the mean is computed (mean(...)). • You can save 1 byte with the more readable [x;x;x;x] instead of ~~(1:4)'*x – Tom Carpenter Jul 26 '16 at 2:08 • @(x)mean([x;x;x;x](:)((b=numel(x))+1:3*b)) is also 2 bytes less. That was why I had come up with, but it's basically the same as your approach. – Tom Carpenter Jul 26 '16 at 2:12 • @TomCarpenter I don't think it's that similar. I think you should post it as a separate answer – Luis Mendo Jul 26 '16 at 9:14 J, 20 18 bytes 2 bytes thanks to @miles #-:@%~-@#+/@}.#}.4#] -@#(+/%#)@}.#}.4#] Usage >> f =: -@#(+/%#)@}.#}.4#] >> f 1 3 5 7 9 11 13 15 17 << 9 How it works It is a translation of my answer in Pyth. • – Adám Jul 24 '16 at 14:31 • @Adám Thanks, added. – Leaky Nun Jul 24 '16 at 14:32 • You can just directly take the average of the middle portion -@#(+/%#)@}.#}.4#] for 18 bytes. – miles Jul 24 '16 at 16:23 Actually, 2015 13 bytes ;l╗;+;+S╜@t╜τ@HΣ╜τ@/ ;l;τ;a;+;+StHΣ/ ;l;τ;aττStHΣ/ Try it online! How it works It is a translation of my answer in Pyth. • For once, an Actually answer which is readable (in Greek). – Adám Jul 24 '16 at 15:02 • @Adám Pyth uses ASCII. – Leaky Nun Jul 24 '16 at 15:03 Octave, 42bytes Another anonymous function for Octave. @(x)mean([x;x;x;x](:)((b=numel(x))+1:3*b)) You can try it online. Simply enter that command, and then do ans([1 2 4 5 6 9]) or whatever numbers are required. This one starts by creating from the input array one with 4 of each input element by first concatenating four copies vertically, and then flattening it vertically. This maintains the sort order. Then is extracts the range of elements from the length of the input array plus 1 up to three times the length of the input array. Because the new array is four times longer, this chops off the upper and lower quartiles. Finally the mean of the new array is returned. 05AB1E, 15 bytes €D€D¹gô¦¨˜DOsg/ Explanation €D€D # quadruple each element in list ¹gô # split into pieces the size of input ¦¨˜ # remove the first and last and flatten the middle 2 DOsg/ # sum and divide by length Try it online APL (Dyalog), 15 bytes IQM←(+/÷≢)≢↓-∘≢↓4∘/ Try it online! 4∘/ quadruplicate each element -∘≢↓ drop as many trailing elements as there are elements in the arguments ≢↓ drop as many leading elements as there are element in the argument () apply the following tacit function: +/ the sum ÷ divided by ≢ the tally JavaScript (ES6), 75 bytes a=>a.concat(a,a,a).sort(g=(x,y)=>x-y).slice(l=a.length,-l).reduce(g,0)/l/-2 Uses the obvious quadruplicate-and-sort approach, and I get to use reduce, which is nice. The only trickery here is to save 4 bytes by reusing the sort comparator to subtract all the array elements from zero, which gives me -2l times the answer I want. Golfscript, 28 29 bytes ~.4*$\,.@/1>2<{+}*{+}*'/'@2*
~.4*\$\,.@/1>2<{+}*{+}*\2*-1?*
Try it online!
Actually, 12 bytes
4α;l¼≈;±(Htæ
Try it online! (currently doesn't work because TIO is a few versions behind)
Explanation:
4α;l¼≈;±(Htæ
4α repeat each element 4 times
;l¼≈ length divided by 4, as integer
;± copy, unary negate
(Ht remove first and last quartiles
æ mean
Mathematica, 51 bytes
Mean@#[[(l=1+Length@#/4);;-l]]&@Sort@Join[#,#,#,#]&
Sorts four copies of the list (to prevent issues with list length not multiples of four), takes part "1 quarter the length of resulting list plus 1" through to the "1/4 length list + 1 from the end", takes their Mean.
Java 146 126 Bytes
Such java much verbose!
float m(float[]n){float r=0;int l=n.length,i=l/4;r-=(n[i])*(l%4)/4;r+=n[i*3]*(4-(l%4))/4;for(;i<l*3/4;r+=n[i],i++);return r/l*2;}
Older Ungolfed partially readable with test cases
/**
*
* @author rohan
*/
public Golf{
float m(float[]n){
//declarations
float r=0;
int x,i=0,l=n.length;
//sum the array
for(float m:n){r+=m;}
//remove the excess
for(;i<l/4;r-=n[i]+n[l-i-1],i++);
//weight the quartiles
r-=(n[l/4]+n[l*3/4])*(l%4)/4;
//return the sum/length but multiply by two since only half of the set is averaged
return r/l*2;
}
static void interQuartileMean(float... set){
System.out.println(new Golf().m(set));
}
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
//test cases pass with flying colours
interQuartileMean(1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38);
interQuartileMean(1, 3, 5, 7, 9, 11, 13, 15, 17);
}
}
Clojure, 82 81 bytes
Edit: 1 byte less by re-writing "didvide by 2 n" part.
#(let[n(count %)](*(/ n)0.5(apply +(subvec(vec(for[i % j(range 4)]i))n(* 3 n)))))
Previous:
#(let[n(count %)](/(apply +(subvec(vec(for[i % j(range 4)]i))n(* 3 n)))(* 2.0 n)))
Uses for to generate 4 repeated values, using float 2.0 not to have fractional results, the rest is just standard.
R, 17 11 bytes
mean(n,0.25)
Assuming n is the input vector in the standard R form n=c(1, 2, 3, ...).
This is in no way surprising since R can be considered “THE language for statistical computing” and has many statistical built-ins.
UPDATE. Saved 6 bytes thanks to rturnbull because trim is the first optional argument by default!
Test cases:
a <- c(1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38)
b <- c(1, 3, 5, 7, 9, 11, 13, 15, 17)
mean(a,trim=0.25) # Returns 6.5
mean(b,trim=0.25) # Returns 9
• Since trim is the default second argument, you don't need to name it; 0.25 can be shortened to .25 or 1/4. This saves you six bytes. – rturnbull Dec 26 '16 at 11:48
Excel, 17 bytes
=TRIMMEAN(A:A,.5)
Relaxed input format make this easy. Input one per row in Column A.
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2019-05-22 00:37:39
|
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https://zbmath.org/?q=an:0745.32004
|
# zbMATH — the first resource for mathematics
Equivalence of analytic and plurisubharmonic Phragmén-Lindelöf conditions. (English) Zbl 0745.32004
Several complex variables and complex geometry, Proc. Summer Res. Inst., Santa Cruz/CA (USA) 1989, Proc. Symp. Pure Math. 52, Part 3, 287-308 (1991).
[For the entire collection see Zbl 0732.00009.]
Let $$P(D)$$ be a partial differential operator with constant coefficients. It is known that some interesting properties of $$P(D)$$ can be characterized by whether or not satisfying certain estimates of Phragmén-Lindelöf type for plurisubharmonic functions of the form $$u=\log| f|$$, $$f$$ entire, on the algebraic variety $$V=\{z\in{\mathbb{C}}^ n:\;P(z)=0\}$$. In this paper, authors prove that for five Phragmén-Lindelöf conditions that have been appeared in the references the estimate holds for $$u=\log| f|$$ if and only if the estimates hold for all plurisubharmonic functions on the variety $$V$$. Here the idea of proving the equivalence of the analytic and plurisubharmonic versions of these Phragmén-Lindelöf principles is to write the plurisubharmonic function $$u$$ as an upper envelope of functions $$\log| f|$$, $$f$$ entire. From the above result, it follows that an arbitrary weakly plurisubharmonic function on the variety $$V$$ can nearly be written as such an upper envelope except on a small exceptional set.
##### MSC:
32U05 Plurisubharmonic functions and generalizations 31C10 Pluriharmonic and plurisubharmonic functions 35E20 General theory of PDEs and systems of PDEs with constant coefficients
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2021-01-15 14:52:56
|
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