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Homeomorphic Complements

1. Feb 23, 2006

AKG

Let A and B be compact connected subsets of the plane, homeomorphic to one another. Are their complements homeomorphic?

I think they are. If A and B are empty, there's nothing to prove. Otherwise, let C denote the complement of A, let D denote the complement of B. We look at the connected components of C and D. Each will have one unbounded component with one hole in it. The unbounded component of C should be homeomorphic to the unbounded component of D. The rest of the components of C are bounded and, I believe, contractible and/or simply connected (are the two equivalent in the plane?). I want to argue that since A and B are homeomorphic, C and D have an equal number of these bounded components. I also want to argue that any two bounded, open, connected, contractible and/or simply connected subsets of the plane are homeomorphic. Actually, it might even be that any two open contractible subsets of the plane are homeomorphic. If this is so, then in technical terms, I believe we can find a bijection f from the collection of connected components of C to the collection of connected components of D such that X is homeomorphic to f(X). Since each X is open, I would like to argue that this induces a homeomorphism from all of C to all of D.

Is the above good?

2. Feb 24, 2006

Hurkyl

Staff Emeritus
Is this a homework question, or something you want to use to prove a homework question?

3. Feb 24, 2006

AKG

Neither.......

4. Feb 24, 2006

matt grime

Then why is it in the homework section?

5. Feb 24, 2006

AKG

Oops, force of habit I guess. It should be moved, but it doesn't matter to me.

6. Feb 24, 2006

StatusX

I believe this theorem is true, although it does not generalize to R3, with alexander's horned sphere as a counterexample. Your argument isn't quite rigorous, and it will take some work for it to handle pathological examples similar to the horned sphere that may arise in 2D.

7. Feb 24, 2006

Palindrom

Any two open, simply connected subsets of the plane are homeomorphic. We know that thanks to the Riemann mapping theorem in complex analysis.

8. Feb 27, 2006

AKG

It seems the hard part to prove is the following:

Suppose A and B are connected compacts in the plane, homeomorphic to one another. If AC has a connected component that is simply connected, so does BC.

It seems obvious, but how would one go about proving it?


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# How can my character start a thieves' guild without being disruptive to the rest of the group?

Background: D&D 5e with a steam punk flair.

My protagonist is a changeling rogue (can assume the role of other people) with a small bit of magic (Arcane Trickster). My deity has requested that I seed a thieves guild to act as a network of spies that will slowly grow as the protagonist progresses in their adventure.

In this question, I would like to focus on how I could go about doing this without being disruptive to the main storyline, or others at the table. I want to do this for fun, but not detract from the fun others are having.

On Page 186-187 the PHB discusses "Between Adventures":

Between trips to dungeons and battles against ancient evils, adventurers need time to rest, recuperate, and prepare for their next adventure. Many adventurers also use this time to perform other tasks, such as crafting arms and armor, performing research, or spending their hard-earned gold.

In some cases, the passage of time is something that occurs with little fanfare or description. When starting a new adventure, the DM might simply declare that a certain amount of time has passed and allow you to describe in general terms what your character has been doing. At other times, the DM might want to keep track of just how much time is passing as events beyond your perception stay in motion.

While this doesn't explicitly talk about seeding a underworld super cult that you will one day use to take over the world, I think this would be the appropriate place for this activity so as not to be disruptive. What I don't know is how much time/money investment should go to this.

From "Building a Stronghold" (DMG, p. 128) - thanks @John:

A character can spend time between adventures building a stronghold. Before work can begin, the character must acquire a plot of land. If the estate lies within a kingdom or similar domain, the character will need a royal charter (a legal document granting permission to oversee the estate in the name of the crown), a land grant (a legal document bequeathing custody of the land to the character for as long as he or she remains loyal to the crown), or a deed (a legal document that serves as proof of ownership). Land can also be acquired by inheritance or other means. Royal charters and land grants are usually given by the crown as a reward for faithful service, although they can also be bought. Deeds can be bought or inherited. A small estate might sell for as little as 100 gp or as much as 1,000 gp. A large estate might cost 5,000 gp or more, if it can be bought at all.

Once the estate is secured, a character needs access to building materials and laborers. The Building a Stronghold table shows the cost of building the stronghold (including materials and labor) and the amount of time it takes, provided that the character is using downtime to oversee construction. Work can continue while the character is away, but each day the character is away adds 3 days to the construction time.

$$\\begin{array}{|l|r|r|} \hline \textbf{Stronghold} & \textbf{Cost} & \textbf{Time} \\ \hline \text{Abbey} & 50,000\,\text{gp} & 400\,\text{days} \\ \text{Guildhall, town or city} & 5,000\,\text{gp} & 60\,\text{days} \\ \text{Keep or small castle} & 50,000\,\text{gp} & 400\,\text{days} \\ \text{Noble estate with manor} & 25,000\,\text{gp} & 150\,\text{days} \\ \text{Outpost or fort} & 15,000\,\text{gp} & 100\,\text{days} \\ \text{Palace or large castle} & 500,000\,\text{gp} & 1,200\,\text{days} \\ \text{Temple} & 50,000\,\text{gp} & 400\,\text{days} \\ \text{Tower, fortified} & 15,000\,\text{gp} & 100\,\text{days} \\ \text{Trading post} & 5,000\,\text{gp} & 60\,\text{days} \\ \hline \end{array} \$$

How can I do this within the rules of D&D 5e without being disruptive to the existing game?

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Thin lenses in contact . (b) Angle of incidence, i = 60°Angle of refraction, r = 41° Using Snell's law, Refractive index of water wr.to air, μwa = sin isin r = sin  60°sin 41° = 1.32 (c) Angle of incidence in water, i =45°Refractive index of glass wr.to water,  μgω = μgaμwa =sin isin r         1.511.32 = sin 45°sin r = 0.7071sin r∴ sin r = 1.32 × 0.70711.51 = 0.6181 ⇒    r = 38.2°, which is the required angle of refraction of glass. Why Is The Focal Length Of A Convex Lens Always Positive Quora. Ltd. Download books and chapters from book store. Dispersive power : The ratio of angular dispersion to the angle of deviation for the mean wavelength (yellow colour) is called dispersive power of the material or the prism or the lens. The angle of minimum deviation is measured to be 40°. The surface power for a convex surface on the front vertex of a lens is positive because n lens >n air.If the back surface of a lens is convex, it also has a positive power since by the cartesian sign convention the radius is negative. Therefore, we can conclude that the power of a lens is inversely proportional to the focal length of the lens. This means that you can calculate the power of a lens using radii of curvature of two surfaces and the refractive index of the lens material. The dioptre is the unit for a measure of the refractive index of a lens. However, when the candle is closer than 18 cm from the mirror, the image would be virtual and therefore cannot be collected on the screen. Is lens formula applicable only for convex lens? For a convex lens, the converging ability is defined by power and in a concave lens, the diverging ability. Find the power of a concave lens of focal length 4 m. Find the focal length of a concave lens of power – 4 D. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from the lens. Do you notice the connection between the focal length and the bending of the light ray? Lens Formula Power of a lens NCERT Questions → Facebook Whatsapp. Table shows the sign convention for the values of object distance, image distance and focal length. Another thing you should keep in mind is that for a converging lens the optical power is positive and for a diverging lens, it is negative. Find the power of a plano-convex lens, when the radius of a curved surface is 15 cm and v =1.5. The focal length, f, of a lens in air is given by the lensmaker's equation: = (−) [− + (−)], where n is the index of refraction of the lens material, and R 1 and R 2 are the radii of curvature of the two surfaces. Power of a Lens is one of the most interesting concepts in ray optics. Your email address will not be published. Hence, the image formed is erect, and of course virtual. Concave lenses. What is power of a lens? ∴ Image is formed at 6.7 cm at the back of the mirror ( because v is positive). Calculate the image position and magnification in each case. Required fields are marked *. Lenses may be convex, concave or a combination. Where f is the focal length of the lens used. Define 1 dioptre. © SI unit of power is Dioptre (D). CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, difference between electric field and magnetic field, CBSE Previous Year Question Papers Class 10 Science, CBSE Previous Year Question Papers Class 12 Physics, CBSE Previous Year Question Papers Class 12 Chemistry, CBSE Previous Year Question Papers Class 12 Biology, ICSE Previous Year Question Papers Class 10 Physics, ICSE Previous Year Question Papers Class 10 Chemistry, ICSE Previous Year Question Papers Class 10 Maths, ISC Previous Year Question Papers Class 12 Physics, ISC Previous Year Question Papers Class 12 Chemistry, ISC Previous Year Question Papers Class 12 Biology, $$\frac{n_{1}}{n_{2}}=\frac{sin r}{sin i}$$, $$\mu =\frac{sin\frac{A+\delta _{m}}{2}}{sin\frac{A}{2}}$$, $$\frac{1}{f}=(\mu -1)(\frac{1}{R_{1}}-\frac{1}{R_{2}})$$. The lens formula may be applied to convex lenses as well as concave lenses provided the ‘real is positive’ sign convention is followed. This is the currently selected item. An important application of using the power of lenses is in ‘Optometry”. Here, the point P on the right of the lens acts as a virtual object. How does the power of a convex lens vary, if the incident red light replaced by violet light? Practice: Image formation in spherical lenses. When the candle is moved closer to mirror, the screen would have to be moved farther and farther. Describe the nature and size of the image. The minimum order quantity of plano-convex elements ranges between 100 and 1000 units. Hence, the image formed is erect, and of course virtual. What Is Lens Formula Quora. Define 1 dioptre. Therefore, the power of this particular lens if 5D. The refracting angle of the prism is 60°. Thus, the screen should be placed at a distance of 54 cm in front of the mirror on the same side as that of the object. Practice: Power of lens. Practice: Thin lenses questions. Draw the ray diagram and find the position, size and nature of the image formed. https://www.zigya.com/share/UEhFTjEyMDU3NTYy. To find the power of a lens in Ray Optics, the following formula can be used. Your email address will not be published. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. For a thin lens, the power of the lens is approximately the sum of … Describe what happens as the needle is moved farther from the mirror. converges the incident rays towards the principal axis) which is relatively thick across the middle and thin at the lower and upper edges. Practice: Convex and concave lenses. Power of lens. What is the power of a convex lens with a focal length of 25 cm? At what distance from the mirror should a screen be placed in order to obtain a sharp image? Calculating the focal length of a lens requires knowing the distance from an object to the lens and the distance from the lens to the image. Using lens maker s formula 1f = (μ-1) 1R1-1R2 ∴ Power of lens, P = 1f = (μ-1) 1R1-1R2Refractive index of violet light is greater than refractive index of red light.Therefore, power of the lens will be increased when red light is replaced by violet light.


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# listings package - Can I include a backslash \ in language keyword (\begin) for syntax highlighting in LaTeX code?

I want to highlight a LaTeX code (including backslash \, e.g. \begin). Unfortunately, by default a LaTeX listings package does syntax highlighting only keywords WITHOUT back slash (e.g. begin).

-

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[scaled=0.82]{beramono}
\usepackage{listings,xcolor}
\begin{document}

\begin{lstlisting}[basicstyle=\small\ttfamily,language={[LaTeX]TeX},
texcsstyle=*\color{red}\bfseries,
keywordstyle=\color{blue}\bfseries,
morekeywords=alignat,moretexcs=intertext]
\begin{alignat*}{4}
y &= -4   &+ 3 &+4     &-7      \\
y &=      &+ 3 &       &-7      \\
\intertext{Therefore}
a &= b    &d   &= cccc &e  &= d \\
a &= bbbb &d   &= c    &e  &= d
\end{alignat*}
\end{lstlisting}

\end{document}

-
Herbert, thank you, an excellent answer, it is simple and effective code! – Vladimir S. May 9 '11 at 7:41
Is it simply the star prepended to the texcsstyle that does the job ? – Carel Nov 24 at 19:49


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# Understanding a Measure-valued (Bochner?) Integral

Let $$X$$ be a compact metric space, and let $$\mathcal P(X)$$ be the (compact, metrisable) space of Borel probability measures on $$X$$. Similarly, $$\mathcal P (\mathcal P (X))$$ is the space of Borel probability measures on $$\mathcal P (X)$$. I want to make sense of the integral $$\int_{\mathcal P(X)} \mu \,\mathrm d \mathcal \tau (\mu) \tag{INT}\label{INT}$$ where $$\tau \in \mathcal P (\mathcal P (X))$$. Is there an elementary definition of this integral? (See the end of the question for an elaboration of what I mean by "elementary".)

It seems one way to interpret \eqref{INT} is to view it as a Bochner integral, by considering $$\mathcal P(X)$$ as a (compact, convex) subset of the Banach space of finite signed Borel measures on $$X$$ with the total variation norm. One property of the Bochner integral is that, for any bounded operator $$T\colon \mathcal P(X) \to Y$$ where $$Y$$ is another Banach space, we have that $$\int_{\mathcal P (X)} T\mu\,\mathrm d\tau (\mu) = T \left( \int_{\mathcal P(X)} \mu \,\mathrm d \mathcal \tau (\mu) \right). \tag{*}\label{*}$$

In particular, suppose that $$Y = L(C(X),\mathbb R)$$, the space of bounded linear functionals on $$C(X)$$, and $$T$$ is the operator given by \mu \mapsto \left(\begin{align*} C(X) &\to \mathbb R \\ f &\mapsto \int_X f(x) \,\mathrm d \mu(x) \end{align*}\right). Then, we can parse \eqref{*} as $$\int_X f(x) \, \mathrm d \mu_0 (x) = \int_{\mathcal P(X)} \int_X f(x) \,\mathrm d \mu(x) \,\mathrm d \tau (\mu) \tag{+} \label{+}$$ for each $$f \in C(X)$$, where $$\mu_0$$ is just the value of \eqref{INT}.

Hopefully, everything I've said so far is correct. (Please do point out mistakes if there are any.) If so, can I take \eqref{+} as the definition of \eqref{INT}? That is, I would like to define \eqref{INT} as the $$\mu_0 \in \mathcal P(X)$$ that satisfies \eqref{+} for each $$f \in C(X)$$. Do I lose any important properties of the integral by using this definition?

I am looking for a definition of \eqref{INT} without too much additional machinery, beyond what you would see in a typical introductory measure theory or functional analysis course. I was hoping \eqref{+} would do, but I am also open to alternative suggestions.

Additionally, if anyone can recommend a nice, readable reference for this material, I'd appreciate that as well. An ideal reference would be one that's accessible to someone who only knows a little measure theory and functional analysis, but I'll take what I can get.

• I'm not sure if this is what you're looking for, but you do not need a lot of machinery to set up the Bochner integral. These lecture notes of Marc Rieffel (which define Bochner integration in Chapter 4 -- everything before that is standard measure theory) assume nothing but the definition of a Banach space and some point-set topology: math.berkeley.edu/~rieffel/measinteg.html May 11, 2020 at 16:51
• @AidanBackus Thanks for the notes; I'll check them out. I agree that the Bochner integral does not require a lot of machinery; in fact, (as far as I understand) it is the most natural extension of the Lebesgue integral to Banach spaces. However, this question is motivated by research I'm doing in economics, and I think the audience I'm writing for there might take better to definitions that use objects they're already likely to know something about (e.g. standard Lebesgue integrals). May 11, 2020 at 17:26
• That's a fair point; I tried to think of a nice way to reduce the Bochner integral to the Lebesgue integral in general but I don't know how to do it without appealing to the Hanh-Banach theorem or something else that economists might not be comfortable with. May 11, 2020 at 17:31
• @AidanBackus That seems interesting! Please feel free to include it in your answer. May 11, 2020 at 17:33
• Sure, I outlined the argument. The hard part if you approach it this way is showing that the Bochner integral actually exists, but for an economics talk that could probably just be blackboxed :-) May 11, 2020 at 17:47

This is a really interesting question! I think $$(+)$$ is the only sensible definition of $$\mu_0$$. As you correctly observed, this follows once we've decided to define $$\mu_0$$ using Bochner integration.

But the Bochner integral is the unique linear map satisfying a few "obvious" criteria, like the $$L^1$$ triangle inequality and dominated convergence. See Bochner Integral: Axioms. If you want to define $$\mu_0$$ in any way that is vaguely useful, it had better satisfy those axioms.

Per the discussion in the comments, let me outline how one could express the Bochner integral in terms of the Lebesgue integral. Let $$X$$ be a measure space, $$B$$ a Banach space, $$B^*$$ its dual, and $$L^1(X \to B)$$ the space of Bochner-integrable functions $$X \to B$$.

By the Hanh-Banach theorem, for every $$x,y \in B$$, $$x=y$$ if and only if for every $$\varphi \in B^*$$, $$\varphi(x) = \varphi(y)$$. In other words, to specify a point $$x \in B$$, it suffices to specify $$\varphi(x)$$ for every linear functional $$\varphi$$ (though there is no guarantee that $$x$$ will exist, and this seems unfixable due to Enflo's theorem that there are Banach spaces which do not admit a Schauder basis, if $$x$$ exists this uniquely defines it; maybe $$\mathcal P(X)$$ actually has a Schauder basis, and then you can just use the Schauder basis rather than fiddle around with linear functionals like this).

As you observe, the Bochner integral commutes with bounded linear maps. In particular it commutes with bounded linear functionals, i.e. elements of $$B^*$$. So if $$f \in L^1(X \to B)$$, we have $$\varphi\left(\int_X f(x)~dx\right) = \int_X \varphi(f(x)) ~dx.$$ Here the integral $$\int_X \varphi(f(x))~dx$$ is just the plain jane Lebesgue integral. We have specified $$\varphi\int_Xf$$ for every $$\varphi$$, so we have specified $$\int_Xf$$, provided that $$\int_Xf$$ actually exists. (It does, by the usual construction of the Bochner integral in terms of ISF, but the point of this outline was to give a cheap definition that averts this modulo technical details.)

• Thanks for that! The argument you outline is roughly the idea behind Simon's proof of the theorem I state in my answer. He just uses a nice trick to keep track of all the $\phi$'s at the same time, and then leverages compactness. I am confused about one thing, though: do we really need Hahn-Banach to conclude that $B^*$ separates points? I can see how it follows from HB, but that seems like an extremely overpowered way to do it. May 12, 2020 at 16:34
• In the case you're interested in I think it can be avoided. Here B* is the double-dual of C(X), and I think C(X) should already separate points in P(X). May 12, 2020 at 17:37
• In general it cannot be avoided. The reason is that in certain models of set theory without the axiom of choice, there exist Banach spaces B such that B* is the trivial vector space! So we must use the axiom of choice somehow, and this manifests as the Hanh-Banach theorem. May 12, 2020 at 17:39
• Yes, the general case was what I was interested in, but I was happy to keep AC. The argument I had in mind was to construct separating linear functionals using the basis on $B$, but of course, the existence of such a basis depends on AC. This strikes me as simpler than relying on Hahn-Banach, but I suppose that is a matter of taste. May 12, 2020 at 17:41
• I think any argument that uses Hanh-Banach can be replaced by using AC to choose a basis and induct on its (transfinite) length, but yeah that does come down to taste. May 12, 2020 at 17:42

I think I have an answer, but there are some details I need to think about more carefully.

Yes, we can take (+) as the definition of (INT). This follows from a version of the Strong Krein-Milman Theorem, which I state below. (This is Theorem 9.1 in Simon; reference below.) We denote the continuous dual of a topological vector space $$X$$ by $$X^*$$, and the closed convex hull of a set $$A$$ by $$\mathrm{cch}(A)$$.

Let $$A$$ be a compact convex subset of a real locally convex (Hausdorff) vector space $$X$$ and let $$\mu \in \mathcal P(A)$$. Then, there is a unique point $$r(\mu)\in A$$, called the barycenter or resultant of $$\mu$$, so that for any $$\ell \in X^*$$, $$\ell (r (\mu)) = \int_A \ell(x)\,\mathrm d \mu(x).$$ The map $$r$$ is a continuous affine map of $$\mathcal P(A)$$ (with the weak-$$*$$ topology) onto A and is the unique such map with $$r(\delta_x) = x$$. More generally, if $$B \subset A$$ is closed, and $$\nu (A \setminus B) = 0$$, then $$r(\nu) \in \mathrm{cch}(B)$$ and $$r[\{ \nu : \nu(A\setminus B) = 0 \}] = \mathrm{cch}(B).$$

This means that the measure $$\mu_0$$ defined by (+) is well-defined, and guarantees that the integral (INT) is continuous in $$\tau$$.

Reference

Simon, B. (2011). Convexity: an analytic viewpoint (Vol. 187). Cambridge University Press.


text:
# Rocket Racer on Track#

A rocket racer toy of mass $$m = {{ params.m }} kg$$ is sent down a track from $$A$$ to $$B$$ from a height $$h = {{ params.h }} m$$ and a horizontal distance $$d = {{ params.d }} m$$ away from point B as shown in the diagram below. If the toy has a velocity of $${{ params.va }} m/s$$ down the incline at point $$A$$, and a velocity of $${{ params.vb }} m/s$$ at point $$B$$, what is the work done by friction as the toy moves from point A to point B?

## Part 1#

Please enter in a numeric value in $$J$$.


text:
# [tex-live] Bug#668562: Interoperability problem of frenchb.ldf and xeCJK.sty

Daniel Flipo daniel.flipo at free.fr
Tue May 1 21:36:28 CEST 2012

Hi Sun, hi all,

A bug report (see http://bugs.debian.org/668562) was made a fortnight
ago.  The origin of the problem is that xeCJK assigns "\XeTeXcharclass"
4 to 6 to various ASCII characters instead of creating its own classes
with "\newXeTeXintercharclass".

A minimal example of the bug follows:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass{article}
\usepackage{xeCJK}
\usepackage[french]{babel}
\begin{document}
anti\-atomique
\end{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Yet another test file that shows a problem (the origin is the same)
between polyglossia/french and xeCJK.sty:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass{article}
\usepackage{xeCJK}
\usepackage{polyglossia}
\setdefaultlanguage{french}
\begin{document}
% A spurious space is added before the 4 chars , . ] )
a,b.c x[z]y  u(v)w
\end{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The fix I suggested on April 19th was to define 3 named
"\newXeTeXintercharclass" and to use them instead of raw 4, 5, 6
classes, but I got no feedback from Sun.

Sun, please tell us if you are willing to fix this issue, thanks in advance.

I can fix it on frenchb.ldf side, but polyglossia or any package making
use of XeTeXinterchartokens' mecanism will have to add the same fix
also, IMHO it is not the right way to go.

Best wishes,
--
Daniel Flipo


text:
# Problem Of The Week #443 November 16th, 2020

Status
Not open for further replies.

#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

-----

For any natural number $n$, ($n\ge 3$), let $f(n)$ denote the number of non-congruent integer-sided triangles with perimeter $n$ (e.g., $f(3)=1,\,f(4)=0,\,f(7)=2$). Show that

a. $f(1999)>f(1996)$,
b. $f(2000)=f(1997)$.

-----

#### anemone

##### MHB POTW Director
Staff member
Congratulations to kaliprasad for his correct solution!

You can find the suggested solution as follows:
(a) Let $a,\,b,\,c$ be the sides of a triangle with $a+b+c=1996$ and each being a positive integer. Then $a+1,\,b+1,\,c+1$ are also sides of a triangle with perimeter 1999 because $b+c>a$ implies $(b+1)+(c+1)>a+1$ and so on. Moreover (999,999,1) form the sides of a triangle with perimeter 1999, which is not obtainable in the form $a+1,\,b+1,\,c+1$ where $a,\,b,\,c$ are the integers and the sides of a triangle with $a+b+c=1996$. We conclude that $f(1999)>f(1996)$.

(b) As in the case (a), we conclude that $f(2000)>f(1997)$. On the other hand, if $x,\,y,\,z$ are the integer sides of a triangle with $x+y+z=2000$, and say $x\ge y\ge z\ge 1$, then we cannot have $z=1$ for otherwise we would get $x+y=1999$, forcing $x,\,y$ to have opposite parity so that $x-y\ge 1=z$, violating triangle inequality for $x,\,y,\,z$. Hence $x\ge y \ge z >1$. This implies that $x-1\ge y-1 \ge z-1 >0$. We already have $y+z>x$. If $x\ge y+z-1$, then we see that $y+z-1\le x<y+z$, showing that $y+z-1=x$. Hence, we obtain $2000=x+y+z=2x+1$ which is impossible. We conclude that $x<y+z-1$. This shows that $x-1<(y-1)+(z-1)$ and hence $x-1,\,y-1,\,z-1$ are the sides of a triangle with perimeter 1997. This gives $f(2000)\le f(1997)$. Thus we obtain the desired result.

Status
Not open for further replies.


text:
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        the derivation of second order reaction rate constant expression when there are two different reactants are reacting.
7 years ago

Ashwin Sinha
520 Points
							Dear Neeta Gupta,
A second-order reaction depends on the concentrations of one second-order reactant, or two first-order reactants.
For a second order reaction, its reaction rate is given by:
$\ -\frac{d[A]}{dt} = 2k[A]^2$ or $\ -\frac{d[A]}{dt} = k[A][B]$ or $\ -\frac{d[A]}{dt} = 2k[B]^2$
In several popular kinetics books, the definition of the rate law for second-order reactions is $-\frac{d[A]}{dt} = k[A]^2$. Conflating the 2 inside the constant for the first, derivative, form will only make it required in the second, integrated form (presented below). The option of keeping the 2 out of the constant in the derivative form is considered more correct, as it is almost always used in peer-reviewed literature, tables of rate constants, and simulation software.[8]
The integrated second-order rate laws are respectively
$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$
or
$\frac{[A]}{[B]} = \frac{[A]_0}{[B]_0} e^{([A]_0 - [B]_0)kt}$
[A]0 and [B]0 must be different to obtain that integrated equation.
The half-life equation for a second-order reaction dependent on one second-order reactant is $\ t_ \frac{1}{2} = \frac{1}{k[A]_0}$. For a second-order reaction half-lives progressively double.
Another way to present the above rate laws is to take the log of both sides: $\ln{}r = \ln{}k + 2\ln\left[A\right]$
Examples of a Second-order reaction

$2\mbox{NO}_2(g) \rightarrow \; 2\mbox{NO}(g) + \mbox{O}_2(g)$

Plz. apptrove my answer by clicking ''Yes'' given below, if u loved it... Plz.. Plz... Plz...

7 years ago
E.Nandhini ketha
115 Points
							Chemical kinetics, also known as reaction kinetics, Chemical kinetics is the study of rates of chemical processes.

2 years ago
Think You Can Provide A Better Answer ?

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text:
# Find and compare the lift slopes for a straight wing and a swept wing with a half-chord line sweep of $$30^{\circ}$$.

A finite wing with an elliptical lift distribution has an aspect ratio of $$6$$. The airfoil of the wing has a lift slope of $$0.15$$ per degree. Find and compare the lift slopes for a straight wing and a swept wing with a half-chord line sweep of $$30^{\circ}$$.

Asked on 12th August 2021 in

High aspect ratio straight wing

Using Helmbold’s equation, for calculating the lift slope for straight wing, $a = \frac{a_{0}}{\sqrt{1+\left ( \frac{a_{0}}{\pi AR} \right )^{2}}+\left ( \frac{a_{0}}{\pi AR} \right )}$ Lift slope for the airfoil $$= a_{0} = 0.15\,/degree = 8.5944\,/radian$$.

Therefore,

$a = \frac{8.5944}{\sqrt{1+\left ( \frac{8.5944}{\pi \left ( 6 \right )} \right )^{2}}+\left ( \frac{8.5944}{\pi \left ( 6 \right )} \right )}=1.4569\,per\,radian$

Swept wing

Using Helmbold’s equation, for calculating the lift slope for a swept wing ,$a = \frac{a_{0}cos\Lambda }{\sqrt{1+\left ( \frac{a_{0}cos\Lambda}{\pi AR} \right )^{2}}+\left ( \frac{a_{0}cos\Lambda}{\pi AR} \right )}$Here, $$\Lambda = 30^{0}$$, is the sweep angle of the wing referenced to the half-chord line.  Therefore,  $a = \frac{8.5944cos30^{0}}{\sqrt{1+\left ( \frac{8.5944cos30^{0}}{\pi \left ( 6 \right )} \right )^{2}}+\left ( \frac{8.5944cos30^{0}}{\pi \left ( 6 \right )} \right )}=1.4436\,per\,radian$On comparing the lift slopes of straight wing and swept wing, we can see that there is a reduction in  lift slope for the swept wing.

Wing sweep is decreasing the lift slope, and it affects lift slope to a larger degree for a high aspect ratio wings than for a lower aspect ratio wings.


text:
Verbal Reasoning :: Puz Quiz 43
Home » Verbal Reasoning » Puz Quiz 43 » General Questions

free Online Test Puzzle Questions and Answers Quiz 43

Description: Free Online Test questions and answers on Puzzle with explanation for various competitive exams,entrance test. Solved examples with detailed answer test 43

Exercise

 "Hold a true friend with both hands." - (Proverb)
1 . Directions (1 - 5): Study the following information carefully and answer the questions given below:
V, U and T are seated in a circle facing the centre. A, B and C are also seated in the same circle but two of them are not facing the centre (facing opposite direction of the centre). V is second to the left of C. U is second to the right of A. B is third to the left of T. C is second to the right of T. A is seated next to V.

$Q.$ Which of the following are not facing the centre ?
 BA CA BC Cannot be determined
2 . Which of the following is T’s position with respect to B?
 Third to the right Second to the right Third to the left Third either to the right or the left
3 . Which of the following is V's position with respect to C?
 Second to the right Third to the left Fourth to the right Fourth to the left
4 . Which of the following is true regarding the seating arrangement?
 A, B and C are seated adjacent V, U and T are seated adjacent There are two persons whose seating arrangement cannot be ascertained Those not facing the centre are seated adjacent
5 . Which of the following is A's position with respect to U?
 Second to the left Second to the right Third to the right Cannot be determined
6 . Directions (6 - 10): Study the following information carefully and answer the questions given below:
In a Group of 5, each person has an exclusive and different preference (has/likes) for a pen, a watch and a car. Pen preferences are Parker, Lamy, Pointer, Lexi and Cello. Car preferences are Wagon R, Swift, Santro, Mica and City. Watch preferences are Timex, Titan, Fastrack, Samay and Citizen. Suman has Mica and Parker but does not prefer among watches -Titan or Fastrack. The one who has Swift, likes Fastrack. Mrudula has preference for City, Cello and Citizen. Amit has preference for Lamy and Timex, Veena prefers Wagon R and Lexi. Harsh's preference for a watch is not Titan.

$Q.$ Which watch is Suman's preference?
 Titan Fastrack Samay Cannot be determined


text:
+0

# Quick question

+1
262
5
+736

What choose 2 is 120?

Mar 8, 2018

#1
+22188
+1

What choose 2 is 120?

$$\begin{array}{|rcll|} \hline \dbinom{n}{2} &=& 120 \quad & | \quad \dbinom{n}{2} = \dfrac{n}{2}\times \dfrac{n-1}{1} \\\\ \dfrac{n}{2}\times \dfrac{n-1}{1} &=& 120 \\\\ (n)(n-1) &=& 240 \\\\ n^2-n &=& 240 \\\\ n^2-n -240 &=& 0 \\\\ n &=& \dfrac{1\pm \sqrt{1-4(-240)} }{2} \\\\ &=& \dfrac{1\pm \sqrt{1+960} }{2} \\\\ &=& \dfrac{1\pm \sqrt{961} }{2} \\\\ &=& \dfrac{1\pm 31} {2} \quad & | \quad n > 0 \\\\ n &=& \dfrac{1+ 31} {2} \\\\ &=& \dfrac{32} {2} \\\\ &=& 16 \\ \hline \end{array}$$

$$\mathbf{\dbinom{16}{2} = 120}$$

Mar 8, 2018
#5
+736
+3

Thanks!

MIRB16  Mar 9, 2018
#2
+1

Very good!!.

Mar 8, 2018
edited by Guest  Mar 8, 2018
#3
0

CPhill: Would heureka's method work on large numbers such as: nC4 =3,921,225, where n = 100 ???.

Mar 8, 2018
#4
0

I don't think heureka's method will work on very large numbers such as: nC13=153,121,753,939,078,375

There is, however, a method that will work on any n and it goes as follows:

Take the above answer=153,121,753,939,078,375 x 13! =953,492,346,711,122,973,955,200,000

Now, take this huge last number and take the 13th root of it as follows:

953,492,346,711,122,973,955,200,000^(1/13) =118.941. round off this to 119. Then add 13 to it, or

119 + 13 = 132. Now take the average of [119 + 132] / 2 =125.5. Discard the fractional part and keep the integer part or 125, which is answer to n above.

Note: This method will work on ANY n, no matter the size.

Mar 8, 2018


text:
# Thoughts on a new biplane design?

Discussion in 'Hangar Flying' started by cluttonfred, Sep 3, 2018.

### Help Support HomeBuiltAirplanes Forum by donating:

1. Mar 10, 2019

### GeeZee

#### Active Member

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Sweet! But almost seems criminal to cover up that Verner. I’m definitely leaving mine hanging out in the breeze...

2. Mar 10, 2019

### cluttonfred

#### Well-Known Member

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I find that I am often attracted to the daintier, lighter biplanes rather than the powerful ones, which ought to make it easier to build a replica or just a look-alike as a U.S. LSA or European microlight. Those Verner engines have me thinking about radials, so here are few that I like.

DH Moth (Armstrong Siddeley Genet)

Kinner "Crackerbox" Airster (Lawrance?)

Vickers Vagabond (Blackburn Thrush)

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3. Mar 10, 2019

### jedi

#### Well-Known Member

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It may be time to update the Wikipedia Lil Bitts website with plans and ownership info. https://en.wikipedia.org/wiki/Raceair_Lil_Bitts

4. Mar 10, 2019

### GeeZee

#### Active Member

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Excellent idea and also to get it listed in Kitplanes buyers guide! The online version could be updated now.

5. Mar 10, 2019

### Dana

#### Super Moderator

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In the same class are the "other" Fisher planes (Fisher Flying Products, designed my Mike Fisher, no relation to Ed). There's the single seat FP404, also originally designed as an ultralight but too heavy, the 2 seat "Classic" version, and some others. My 404, with a half VW engine, was underpowered (it was designed for 503), but extremely economical (2 GPH).

6. Mar 10, 2019

### GeeZee

#### Active Member

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I like the 404 and have looked long and hard at that option. I’ve done a fair amount of woodworking so I believe I could build one ,no problem. This may sound crazy but I’m deliberately picking a fabrication method I don’t have any experience with. I want to learn to weld and cut/fit tubing. I helped a friend cover/ribstitch his kitfox so have an inkling of that process.
I’ve bought a TIG welder and have been running beads on scrap. I have recently discovered a local welding supplier that offers a 20hr “hobbyist” TIG class. I’m hoping 20 hrs of instruction will get me off to a good start.

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7. Mar 10, 2019

### cluttonfred

#### Well-Known Member

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You might want to try out a gas welding class to see if you like it, I did and was pleasantly surprised how much progress I made in couple of days. I wouldn't fly with my welds yet, but I can see that with practice I could do good work.

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8. Mar 10, 2019

### cluttonfred

#### Well-Known Member

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I would love to see a light, simple biplane along the lines of the strut-braced Fisher Classic with the side-by-side seating of the Javelin Wichawk and the cockpit arrangment (single stick, dual pedals) of the Evans VP-2.

Designed and built light with generous wing area it should fly well enough for low and slow fun on a modest 4-stroke engine (big VW, 5-cylinder Verner, A-65, etc.), think 65 hp J-3 Cub on a lazy summer day. The side-by-side seating is better for taking kids for a ride and it would make a great single-seater for the big and tall crowd. I'd go with enough stagger and/or sweep to keep the cockpit clear for easy access, maybe even include doors on one or both sides. Welded steel tube fuselage and struts, wood wings, all fabric-covered.

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9. Mar 10, 2019

### Toobuilder

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As usual, what is old is new again. Pitts S1C. Classic - check. Simple - check. Inexpensive- check. Reportedly the most pure form of personal flying yet invented by man - check. Proven - PLEASE!

Projects are available at VERY reasonable prices, and the dirt cheap O-290G will drag them around capably as long as you don't expect hard core acro. A nice 290 powered "C" just sold on the Biplane forum for well under $10k. Mind you this was a TURN KEY airplane and sold for less money than the raw materials. Many complete "projects" can be had for ~$3,500 (which is less than the cost of new flying wires ALONE).

So once again, the solution to cheap, classic, fun flying is out there in droves... Get off the computer and JUST DO IT!

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10. Mar 10, 2019

#### Well-Known Member

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11. Mar 10, 2019

#### Well-Known Member

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I am doing the same. I found a welding related makerspace where the owner gives all sorts of classes. I have a Gas welding set and the mini torch that the buttercup builders recommend, but my welds leave a lot to be desired. Albeit I am trying to learn on my own. I plan to also take the TIG class and see how I do. I may also take a gas class with this guy then make the decision on what I go with. The plus with gas is the tubes / welds normalize since they heat a large area and cool slowly. Tig is more concentrated and many say its best to heat the joint with a gas weld kit to get the same normalization. Yet many dont bother and seem to do just fine. Even Stinsons, Kitfox and I think Cubcrafters as well Mig weld which according to most is a no no...

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12. Mar 10, 2019

### Joe Fisher

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Last edited: Mar 10, 2019
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13. Mar 10, 2019

### Himat

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When there is a large number of old designs, why make yet another “old” design?

As Toobuilder states, there are working proved designs that fit the request. They can be had finished or part finished at a fraction of the cost of designing and building yet another.

14. Mar 10, 2019

### GeeZee

#### Active Member

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I’ve got a little oxy set (the genrac brand in the blow molded carrying case) and took a gas welding sportair workshop when it came to town. The skill set for both Gas and TIG is somewhat similar so I’m thinking if I learn TIG I’ll be 3/4 the way to being able to gas weld. I know the debate about normalizing or not will go on forever but after my own research I don’t think it matters for what we do with tube aircraft frames. From what I’ve read true normalizing involves putting the welded assy into a heat treat oven to a certain temp then stepping it down over the course of about 8 hours. And after all that it only increases the ductility of the joint around 6% ( working from memory here so the numbers might not be quite correct). I guess the takeaway for me was that heating a cluster to cherry red then allowing it to cool over the course of several minutes does virtually nothing. As you said several certificated aircraft have been MIG welding tubes for a long time and have had no problems. I think MIG results in about a 6-8% more brittle weld than gas welding.

15. Mar 11, 2019

#### Well-Known Member

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If you have not seen the videos by tinmantech than have a look here: https://www.tinmantech.com/products/dvds/aircraft-aviation/

I rented the 4130 videos and they were very helpful. He recommends the meco midget torch and I found one on ebay and its quite nice. Would work well on the small tubes of the Lil Bitts. Here is link to torch on his site: https://www.tinmantech.com/products/welding/meco-torch-accessories/meco-midget-torch.php

Marc

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16. Mar 11, 2019

### GeeZee

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Thanks, those look very interesting. I’ll be renting them as well. I’ve heard of the midget and would consider one if I could also find a deal. Right now I’m still planning to TIG weld everything I can and use the torch to heat/bend tubing (like the lower longerons).

Edit: Well after checking out that site some more I think there’s definitely a place in my workshop for a meco midget! You can’t have too many tools, right?
Our sportair gas welding instructor found some aluminum flux and a couple of scraps of aluminum sheet so demonstrated aluminum welding for the class. It turned out beautifully but man you have to move quickly. He welded about a 12” long butt joint in about 3 seconds.

Last edited: Mar 11, 2019
17. Mar 11, 2019

### Little Scrapper

#### Well-Known MemberHBA Supporter

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Speaking from many years of experience.

Practice. There's just no way around it. Welding, regardless of type, is an art form. Every single day you should be welding even if it's only for 5 minutes. There's just no substitute, ya gotta suit up and lay bead.

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18. Mar 11, 2019

### cluttonfred

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Honestly, the pickings are slim in the size/weight/power range that interests me, meaning two seats, ~1000 lb gross, 60-80 hp. Fisher Classic is too light, Celebrity too heavy, Sherwood Ranger quite expensive, not many other options out there.

19. Mar 11, 2019

### Toobuilder

#### Well-Known Member

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There is a reason the pickings are slim. A 2 place Biplane with 80 HP is going to be a dog with a capital "D".

20. Mar 11, 2019

### cluttonfred

#### Well-Known Member

Joined:
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That depends on the kind of flying you do. If you’re used to a 65 hp J-3 Cub then an 80 hp biplane weighing substantially less than the Cub might seem pretty sporty.


text:
# Twenty

Twenty rabbits are put in 4 cells so that there are different number of rabbits in each cell contains at least 3 rabbits. What is the largest possible number of rabbits in one cell

Result

m =  8

#### Solution:

$m = 20 - (3+4+5) = 8$

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text:
# Marion and Thornton 14-16

ehrenfest
[SOLVED] Marion and Thornton 14-16

## Homework Statement

A photon is emitted at an angle \theta' by a star (system K') and then received at an angle \theta on Earth (system K). The angles are measured from a line between the start and Earth. The star is receding at speed v with respect to Earth. Find the relation between \theta and \theta'; this effect is called the aberration of light.

## The Attempt at a Solution

This is a special relativity question but I do not understand how you can answer it without GR. In special relativity, either theta=theta'=0 or theta is not 0 and the photon misses the earth.

Homework Helper
This is a special relativity question but I do not understand how you can answer it without GR. In special relativity, either theta=theta'=0 or theta is not 0 and the photon misses the earth.

GR is not needed because this phenomenon appears even when there is a constant relative velocity between the two rest frames. (In fact, aberration of starlight was known to 18th-Century astronomers, as the Wiki article shows, and was already understood classically. There is simply a modification to the equation that becomes necessary for relativistic velocities.)

The condition of the problem is that the star moves parallel to the x-axis (placing Earth at the origin), with the light ray reaching the Earth making an angle theta' to the star's line of motion in the star's reference frame. In the Earth's frame, what does that angle become?

I see the possible source of your confusion: the description for the light ray is poorly phrased... (In fact, this is not even the way this problem is generally posed. See, for instance, Taylor and Wheeler, 2nd ed., problem 3-9, pp. 80-81; D'Inverno, problem 3.4, p. 41; etc.)

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ehrenfest
So you are saying that for example, Earth is at (0,0,0) and the star is moving along the line (t,0,1)? I thought "receding" meant that the star was moving along the line (0,0,t).

Homework Helper
That's why I'm saying the problem is poorly stated. If the star were receding directly away from us, there would be no 'stellar aberration' at all (as you say, theta = theta' = 0). The only way the problem would make sense in that formulation would be to put the star's motion on a line not passing through the Earth and measure the angle relative to that line.

ehrenfest
Cool. Please confirm that this is what the question "means":

for example, Earth is at (0,0,0) and the star is moving along the line (t,0,1).

OK. Actually this is not homework. Since the problem statement does not make sense, I think I will just skip it. Thanks for trying to help.

Homework Helper
for example, Earth is at (0,0,0) and the star is moving along the line (t,0,1).

That would be one way of expressing the situation.

The version of the problem I've usually seen is this. A star is at rest at a point (x,0) [the distance is irrelevant, since we are concerned with a ray of light from it]. In the star's rest frame, a light ray travels to the origin of coordinates. The Earth is at the origin at the moment of reception, traveling with speed v along the y-axis. In the Earth's rest frame, what angle does the light ray makes to the x'-axis? (Thus, in what direction does the star appear to lie?)

Classically, Galilean velocity addition predicts that the star would appear to be at an angle given by tan(phi) = v/c "ahead" (in the direction of Earth's motion) of the x'-axis. In fact, this is pretty much what is observed for stellar observations from Earth, owing to the Earth's orbital motion plus the Sun's velocity relative to the star (so you get a sort of cycloidal apparent motion of the star, superimposed on the actual change in the star's direction due to differences in the Galactic orbits of Sun and star; as you can imagine, this complicates unraveling the dynamical behavior of Galactic stars a little...). The classical result means that the greatest possible aberration (angle) would be 45º. [The classical aberration is also used to explain the Poynting-Robertson effect, which imposes a 'drag' on orbiting dust particles in a planetary system, for instance.]

Using the relativistic addition relations, the prediction changes considerably, with the limit for aberration becoming 90º (that is, as v -> c , the star asymptotically approaches appearing straight ahead of you). Generalizing this to all directions, stars that are behind you in the stellar rest frame can appear to be ahead of you; as relative velocity increases, increasingly large fractions of the sky appear compressed into decreasingly smaller cones in the direction of 'flight'.

This is also related to the 'Terrell rotation' (concerning the visual appearance of objects in relative motion), a consequence of SR that was only recognized around 1963...

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Find the minimal polynomial of a particular matrix

Find the minimal polynomial of $A=\begin{pmatrix} 0 &0 & 0 &1\\ 1 & 0 & 0&0\\ 0 &1 &0 &0\\ 0 &0 & 1&0\end{pmatrix}$

Since this is a square matrix, the minimal polynmial will be such that $p(A)=0$. How do I find such a polynomial?

• That's not a square matrix, it's $5 \times 4$. And it's not lower triangular either. – Robert Israel Nov 18 '15 at 20:28
• Hint: This is a permutation matrix – Omnomnomnom Nov 18 '15 at 20:35
• $\phantom{}$ $A^4 = I$. – Jack D'Aurizio Nov 18 '15 at 21:08
• @JackD'Aurizio thank you for the observation, can you explain how I can use that fact? – grayQuant Nov 18 '15 at 21:27
• @grayQuant: In addition to Jack's point, compute $A^n$ for $0\le n\le 3$ and observe that no two of them has a nonzero element in the same position -- so they are certainly linearly independent. – Henning Makholm Nov 18 '15 at 21:37

Note the eigenvalues of $A$ are $\pm 1,\pm i$, each of which has the multiplicity 1. So the minimal polynomial is just the characteristic polynomial $p(x)=x^4-1$.


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Question: Determine if the columns of the matrix form linearly iudependent set.8 3 2|

Question:

Question: Determine if the columns of the matrix form linearly iudependent set. 8 3 2|

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1. ## Linear Function

I would appreciate help with the following problem please.
What values of c in the function f : R to R given by f(x) =mx + c, is f linear.

2. ## Re: Linear Function

Originally Posted by mlg
I would appreciate help with the following problem please.
What values of c in the function f : R to R given by f(x) =mx + c, is f linear.
QUESTION: Do you know the definition of linear function?
If so there is a place to start.
If not then look it up.

3. ## Re: Linear Function

Any function of the form f(x)= ax+ c is linear for any numbers a and c.

4. ## Re: Linear Function

Originally Posted by HallsofIvy
Any function of the form f(x)= ax+ c is linear for any numbers a and c.
Actually there are two uses for the term linear function.
Moreover, this question was posted in the number theory forum which suggests the second meaning to me.

5. ## Re: Linear Function

Thank you Plato & Hallsoflvy for the quick reply.
I will now look up the definition of linear function.

6. ## Re: Linear Function

The definition of linear function is:
A linear function is a map f between two vector spaces that preserves vector addition and scalar multiplication:
f(x + y) = f(x) + f(y)
f(ax) = af(x)

How does this definition tie in with the value of c?

7. ## Re: Linear Function

Originally Posted by mlg
The definition of linear function is:
A linear function is a map f between two vector spaces that preserves vector addition and scalar multiplication:
f(x + y) = f(x) + f(y)
f(ax) = af(x)
How does this definition tie in with the value of c?
$f(x)=ax+c\\f(y)=ay+c\\f(x+y)=a(x+y)+c\\f(x)+f(y)= ~?$

8. ## Re: Linear Function

Thanks again Plato.

9. ## Re: Linear Function

To underscore this, I claim for a linear function $f: \Bbb R \to \Bbb R$ we must have:

$f(0) = 0$.

By LINEARITY:

$f(x + y) = f(x) + f(y)$. Taking $x = y = 0$, we get:

$f(0) = f(0+0) = f(0) + f(0)$ so that:

$0 = f(0) - f(0) = (f(0) + f(0)) - f(0) = f(0)) + (f(0) - f(0)) = f(0) + 0 = f(0)$.

Now if $f(x) = mx + c$, what is $f(0)$, and what does that tell you about what $c$ MUST be for $f(x) = mx + c$ to be linear?

10. ## Re: Linear Function

Thanks Deveno for your time and effort.

11. ## Re: Linear Function

f(x) = ax + c
f(y) = ay + c
f(x + y) = a(x + y) + c
f(x) + f(y) = ax + c + ay + c = a(x + y) + 2c.
But a(x + y) + c is not equal to a(x + y) +2c.
i.e. c is not equal to 2c, unless c = 0.
(The question mentioned values, but c = 0 is only one value).


text:
## Calculus (3rd Edition)

$$y'= -\frac{3}{x^2}\left(1+\frac{1}{x}\right)^2 .$$
Since $y=(1+\frac{1}{x})^3$, then by the chain rule $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $y'$ is given by $$y'=3\left(1+\frac{1}{x}\right)^2\left(1+\frac{1}{x}\right)'\\=3\left(1+\frac{1}{x}\right)^2\left(-\frac{1}{x^2}\right) = -\frac{3}{x^2}\left(1+\frac{1}{x}\right)^2 .$$


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# I/O

There are some useful I/O classes for parsing/reading/writing formulas and other data structures.

## Parsers

Parsers are used to parse formula strings into LogicNG formulas. Throughout the documentation, we used the default parser on formula factories by calling f.parse(...), but we can also generate parsers by hand or - most important - implement own parsers for our applications.

Parsers in LogicNG are generated by the parser framework ANTRL (LogicNG's only external dependency). You only define a grammar file and ANTLR generates the Java classes for the parser at compile time. LogicNG's parser grammars lie here. There are two parsers:

1. PropositionalParser for parsing propositional formulas without cardinality or pseudo-Boolean constraints. Its grammar can be found here
2. PseudoBooleanParser for parsing formulas with cardinality or pseudo-Boolean constraints.

Application Insight

You can easily implement you own parsers for you propositional language by looking at these grammar files and implementing your own parser inheriting from FormulaParser. Then you only need to write minimal code.

There are two formula readers implemented in LogicNG:

The formula reader can read propositional formulas with or without pseudo-Boolean constraints. Multiple lines in a file are interpreted as a conjunction of the formulas in each line.

## Writers

There are three writers for formulas in the org.logicng.io package:

• DIMACS writer for writing CNFs as DIMACS CNF files like described above.
• Formula writer for writing formulas in LogicNG syntax. The formula writer has the option splitAndMultiline which, if set to true, writes the different operands of a top-level conjunction in single lines. This can often improve readability, and the corresponding FormulaReader can read this format.
• Formula dot file writer for writing a graphical representation of formulas as a DOT file. Such a file can be used to generate a graphical representation e.g. with Graphviz. E.g. the graphics in the chapter on formulas were generated this way. The parameter alignLiterals allows to align all literals of a formula at the same level, meaning all terminal nodes are at the same level.

Other data structures have also their own writers.

### BDDs

Graphs of a BDD can be written as DOT file with the BDD dot file writer. The graphics of BDDs in this documentation were generated this way.

### Graphs

For graphs, there are two writers:


text:
Kirchhoff's Junction Rule Video Lessons

Concept

# Problem: In the picture on the below, what is the current in the wire above the junction?Does charge flow toward or away from the junction?

###### FREE Expert Solution

Current:

$\overline{){\mathbf{i}}{\mathbf{=}}\frac{\mathbf{V}}{\mathbf{R}}}$

Current through the 2Ω resistor:

I1 = 6/2 = 3A

91% (220 ratings)
###### Problem Details

In the picture on the below, what is the current in the wire above the junction?
Does charge flow toward or away from the junction?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

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text:
How do I prevent floating and jittering when on the ground and using spherical gravity?

TL;DR: My ship floats and jitters when it lands on a sphere.

To elaborate on the title: I'm experimenting with a 3D space game. I have created a sphere, which in turn has a larger sphere overlapping it, within which, gravity will take place. However, once my player's ship (currently using an arrowhead) lands on the planet, it doesn't actually make contact with the surface of the sphere, and then the ship begins to rotate erratically and jitter. Code below.

private void OnTriggerStay(Collider other)
{
if (other.CompareTag("SoI"))
{
inSoI = true;
}
}

This trigger checks to see if the ship is in the larger overlapping sphere. I have another identical block for OnTriggerExit which sets inSoI to false. I also have a OnCollisionEnter which sets another variable, landed to true, and a OnCollisionExit which sets it to false.

private void FixedUpdate()
{
if (inSoI && !landed)
{
gravityVector = (planet.transform.position - transform.position);
rb.AddForce(gravityVector.normalized * planetrb.mass * rb.mass / gravityVector.sqrMagnitude);
}
}

Simply, if in the larger sphere, then accelerate towards the planet. I tried disabling gravity if in contact with the sphere. It removed a lot of the jitter, however it still floats, and it still rotates side to side. The sphere by the way, is the standard Unity sphere, scaled 10000x in every axis. I don't need code (though it would help), any solution is fine, even if it means abandoning the project.

• The 10000x might be part of your problem. At that range from the origin, floating point positions have a resolution of about half a millimeter, which may be coarse enough to observe jerkiness or vibration at the scales you're using.This imprecision could easily be even greater if your sphere is itself a long way from the origin. Have you tried arranging your scene so the interaction happens close to the origin? – DMGregory Apr 17 '18 at 1:38
• Ah, thank you, this got my mind going. Turns out using an isosphere with a large size at 1x scale then using a mesh collider solved it. I made the isosphere in blender. Turns out is also has a few other pros. Thanks – Devintage Apr 17 '18 at 15:07

Can you describe, how did you set up the collision system? In which script are you checking your OnCollisionEnter and OnTriggerEnter? Which of your colliders are set to isTrigger? If the collider on your ship has the isTrigger checked, it does not fire the OnCollisionEnter on the inner sphere, hence, you never going to have your landed true.

If you have your the collder on your ship NOT set to isTrigger, then it is going to fire your OnCollisionEnter, when it has a RigidBody on and collides with another collider, which is not a trigger.

This should clear up things: https://docs.unity3d.com/Manual/CollidersOverview.html

I tried out a simple project, with a plane, and 10000x scaled up sphere, and I only got the jitter with the sphere, so it must be behind the unwanted behaviour of your ship.

• I decided to use a static collider for my planet and just use a public variable for the mass, this also means I'm not limited by the max mass a rigidbody can have. I did some debugging and noticed that landed was never set to true, with or without a rigidbody, and the player is non-kinematic so the collision detection should work. Any idea why this might be happening? – Devintage Apr 17 '18 at 15:17
• Is your the collider on your player set to isTrigger? – formatc2013 Apr 17 '18 at 17:27


text:
# Projectile motion of a baseball

hey, i'm having some difficulty with this problem.
A baseball player makes perfect contact with the ball, striking it 45 degrees above the horizontal at a point 1.3m above the ground. his homerun hit just clears the 3.0m wall 130m from home plate. with what velocity did the baseball player strike the ball?
Code:
  /
/                                                            |
[U]/ ) 45     [/U]                                                   |
|                                                         | 3.0m
| 1.3m                                                    |
|[U]                                                         |             [/U]
130m
this is pritty much what i have. i have no clue where to start. i've been trying to figure it out but no success... :(

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uh...i did...i did lyk..30 questions on projectile motion...i know how to do the questions. i'm just stuck on this question.

Last edited by a moderator:
Were you not given a value for time between the ball being hit and it clearing the wall?

(Although you may not need it for this question)

ok...i am pretty sure i can help you witht his one :)
if you have been learning projectile motion in class then you should know:
$$\Delta Y = (V_{0}sin\varphi)(t)-.5gt^{2}$$
where $$\varphi = 45 degrees (the angle of "launch"$$
also you should know:
$$\Delta X = (V_{0}cos\varphi)(t)$$

NOW here is what you will have to do:
solve the change in X function for t and plug that in for wherever "t" appears in the change in Y equation
set the change in Y to equal 1.7 m (the ball needs to at least hit that point to make it over the wall) and then solve that big nasty equation algebraically to get $$\ (V_{0}$$

***edit : sorry its my first time with the math code so ill write this out
phi = 45 cuz its the angle of launch (the writing was kinda hard to read)


text:
# Large Cardinals, Determinacy and Other TopicsThe Cabal Seminar, Volume IV

Volume 4

## $130.00 (C) Part of Lecture Notes in Logic Paul B. Larson, Alexander S. Kechris, Robert M. Solovay, Donald A. Martin, W. Hugh Woodin, Ilijas Farah, John R. Steel, Claude Dellacherie, Matthew Foreman, Andrés Eduardo Caicedo, Benedikt Löwe View all contributors • Date Published: November 2020 • availability: Available • format: Hardback • isbn: 9781107182998 ##$130.00 (C) Hardback

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• The proceedings of the Los Angeles Caltech-UCLA 'Cabal Seminar' were originally published in the 1970s and 1980s. Large Cardinals, Determinacy and Other Topics is the final volume in a series of four books collecting the seminal papers from the original volumes together with extensive unpublished material, new papers on related topics and discussion of research developments since the publication of the original volumes. This final volume contains Parts VII and VIII of the series. Part VII focuses on 'Extensions of AD, models with choice', while Part VIII ('Other topics') collects material important to the Cabal that does not fit neatly into one of its main themes. These four volumes will be a necessary part of the book collection of every set theorist.

• Includes updated/revised material from the original Cabal Seminars volume
• New, unpublished survey articles put the historical papers into context
• Now includes uniform and modern notation to make the book more accessible to the reader

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## Product details

• Date Published: November 2020
• format: Hardback
• isbn: 9781107182998
• length: 270 pages
• dimensions: 160 x 235 x 25 mm
• weight: 0.63kg
• contains: 2 tables
• availability: Available

Preface
Original numbering
Part VII. Extensions of AD, models with choice:
1. A brief history of determinacy Paul B. Larson
2. “AD plus uniformization” is equivalent to “half $AD_\mathbb{R}$” Alexander S. Kechris
3. The independence of DC from AD Robert M. Solovay
4. Games of countable length Donald A. Martin
5. Some consistency results in ZFC using AD W. Hugh Woodin
6. Subsets of $\aleph_1$ constructible from a real Alexander S. Kechris
7. AD and the uniqueness of the supercompact measures on $\wp_{\omega_1}(\lambda)$ W. Hugh Woodin
8. The extender algebra and $\sum_1^2$-absoluteness Ilijas Farah
Part VIII. Other topics:
9. On Vaught's conjecture John R. Steel
10. Capacities and analytic sets Claude Dellacherie
11. More saturated ideals Matthew Foreman
12. The fourteen Victoria Delfino problems and their status in the year 2020 Andrés Eduardo Caicedo and Benedikt Löwe
Bibliography.

• ## Editors

Alexander S. Kechris, California Institute of Technology
Alexander S. Kechris is Professor of Mathematics at the California Institute of Technology. He is the recipient of numerous honors, including the J. S. Guggenheim Memorial Foundation Fellowship and the Carol Karp Prize of the Association for Symbolic Logic. He is also a member of the Scientific Research Board of the American Institute of Mathematics.

Benedikt Löwe, Universiteit van Amsterdam
Benedikt Löwe is Universitair Hoofddocent at the Universiteit van Amsterdam, Professor of Mathematics at the Universität Hamburg and Fellow of Churchill College at the University of Cambridge. He is currently the president of the Deutsche Vereinigung für Mathematische Logik und für Grundlagenforschung der Exakten Wissenschaften (DVMLG) and the Secretary General of the Division for Logic, Methodology and Philosophy of Science and Technology (DLMPST).

John R. Steel, University of California, Berkeley
John R. Steel is Professor of Mathematics at the University of California, Berkeley. Prior to that, he was a professor in the mathematics department at the University of California, Los Angeles. He is a recipient of the Carol Karp Prize of the Association for Symbolic Logic and of a Humboldt Prize. Steel is also a former Fellow at the Wissenschaftskolleg zu Berlin and the Sloan Foundation.

## Contributors

Paul B. Larson, Alexander S. Kechris, Robert M. Solovay, Donald A. Martin, W. Hugh Woodin, Ilijas Farah, John R. Steel, Claude Dellacherie, Matthew Foreman, Andrés Eduardo Caicedo, Benedikt Löwe

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