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# Van Houtum distribution

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Parameters Probability mass function ${\displaystyle p_{a},p_{b}\in [0,1]{\text{ and }}a,b\in \mathbb {Z} {\text{ with }}a\leq b}$ ${\displaystyle k\in \{a,a+1,\dots ,b-1,b\}\,}$ ${\displaystyle {\begin{cases}p_{a}&{\text{if }}u=a;\\p_{b}&{\text{if }}u=b\\{\frac {1-p_{a}-p_{b}}{b-a-1}}&{\text{if }}a ${\displaystyle {\begin{cases}0&{\textrm {if}}u ${\displaystyle ap_{a}+bp_{b}+(1-p_{a}-p_{b}){\frac {a+b}{2}}}$ N/A ${\displaystyle \ a^{2}p_{a}+b^{2}p_{b}-{}\ }$ ${\displaystyle {\frac {(a+b)(1-p_{a}-p_{b})+2ap_{a}+2bp_{b}}{4}}}$ ${\displaystyle {}+{\frac {b(2b-1)(b-1)-a(2a+1)(a+1)}{6}}}$ ${\displaystyle \ -p_{a}\ln(p_{a})-p_{b}\ln(p_{b})-{}\ }$ ${\displaystyle (1-p_{a}-p_{b})\ln \left({\frac {1-p_{a}-p_{b}}{b-a-1}}\right)}$ ${\displaystyle e^{ta}p_{a}+e^{t}bp_{b}+{\frac {1-p_{a}-p_{b}}{b-a-1}}{\frac {e^{(a+1)t}-e^{bt}}{e^{t}-1}}}$ ${\displaystyle e^{ita}p_{a}+e^{itb}p_{b}+{\frac {1-p_{a}-p_{b}}{b-a-1}}{\frac {e^{(a+1)it}-e^{bit}}{e^{it}-1}}}$

In probability theory and statistics, the Van Houtum distribution is a discrete probability distribution named after prof. Geert-Jan van Houtum.[1] It can be characterized by saying that all values of a finite set of possible values are equally probable, except for the smallest and largest element of this set. Since the Van Houtum distribution is a generalization of the discrete uniform distribution, i.e. it is uniform except possibly at its boundaries, it is sometimes also referred to as quasi-uniform.

It is regularly the case that the only available information concerning some discrete random variable are its first two moments. The Van Houtum distribution can be used to fit a distribution with finite support on these moments.

A simple example of the Van Houtum distribution arises when throwing a loaded dice which has been tampered with to land on a 6 twice as often as on a 1. The possible values of the sample space are 1, 2, 3, 4, 5 and 6. Each time the die is thrown, the probability of throwing a 2, 3, 4 or 5 is 1/6; the probability of a 1 is 1/9 and the probability of throwing a 6 is 2/9.

## Probability mass function

A random variable U has a Van Houtum (a, b, pa, pb) distribution if its probability mass function is

${\displaystyle \Pr(U=u)={\begin{cases}p_{a}&{\text{if }}u=a;\\[8pt]p_{b}&{\text{if }}u=b\\[8pt]{\dfrac {1-p_{a}-p_{b}}{b-a-1}}&{\text{if }}a

## Fitting procedure

Suppose a random variable ${\displaystyle X}$ has mean ${\displaystyle \mu }$ and squared coefficient of variation ${\displaystyle c^{2}}$. Let ${\displaystyle U}$ be a Van Houtum distributed random variable. Then the first two moments of ${\displaystyle U}$ match the first two moments of ${\displaystyle X}$ if ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle p_{a}}$ and ${\displaystyle p_{b}}$ are chosen such that:[2]

{\displaystyle {\begin{aligned}a&=\left\lceil \mu -{\frac {1}{2}}\left\lceil {\sqrt {1+12c^{2}\mu ^{2}}}\right\rceil \right\rceil \\[8pt]b&=\left\lfloor \mu +{\frac {1}{2}}\left\lceil {\sqrt {1+12c^{2}\mu ^{2}}}\right\rceil \right\rfloor \\[8pt]p_{b}&={\frac {(c^{2}+1)\mu ^{2}-A-(a^{2}-A)(2\mu -a-b)/(a-b)}{a^{2}+b^{2}-2A}}\\[8pt]p_{a}&={\frac {2\mu -a-b}{a-b}}+p_{b}\\[12pt]{\text{where }}A&={\frac {2a^{2}+a+2ab-b+2b^{2}}{6}}.\end{aligned}}}

There does not exist a Van Houtum distribution for every combination of ${\displaystyle \mu }$ and ${\displaystyle c^{2}}$. By using the fact that for any real mean ${\displaystyle \mu }$ the discrete distribution on the integers that has minimal variance is concentrated on the integers ${\displaystyle \lfloor \mu \rfloor }$ and ${\displaystyle \lceil \mu \rceil }$, it is easy to verify that a Van Houtum distribution (or indeed any discrete distribution on the integers) can only be fitted on the first two moments if [3]

${\displaystyle c^{2}\mu ^{2}\geq (\mu -\lfloor \mu \rfloor )(1+\mu -\lceil \mu \rceil )^{2}+(\mu -\lfloor \mu \rfloor )^{2}(1+\mu -\lceil \mu \rceil ).}$

## References

1. ^ A. Saura (2012), Van Houtumin jakauma (in Finnish). BSc Thesis, University of Helsinki, Finland
2. ^ J.J. Arts (2009), Efficient optimization of the Dual-Index policy using Markov Chain approximations. MSc Thesis, Eindhoven University of Technology, The Netherlands (Appendix B)
3. ^ I.J.B.F. Adan, M.J.A. van Eenige, and J.A.C. Resing. "Fitting discrete distributions on the first two moments". Probability in the Engineering and Informational Sciences, 9:623-632, 1996.


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# Calculate position of the Sun in ECEF

I'm making a very basic orbit simulator in C# using the Helix Toolkit.

I want to position the Helix lighting object at the sun, but also be able to calculate solar eclipse by the moon and earth for a satellite (position defined in ECEF).

Upon some brief research, I'm reading that getting an accurate position calculation for the sun in ECEF is a lot of work.

Is there an algorithm that can perform this calculation for me that I can include into an UpdateSunPosition() method so that I can have a reasonably accurate position calculation of the sun in ECEF?

Perhaps I have to scratch the eclipse calculation ambitions for now, instead calculate Geographical Position of the sun to get the Declination and Greenwich Hour Angle, Call those latitude and longitude, and then just give it a satisfactory altitude.

I've noticed that people are referring to SOFA a lot when talking about this, but I'm not sure what SOFA is, or whether I can use it in C#. I'm pretty sure it doesn't operate in a .NET environment like visual studio, but please correct me if I'm wrong.

I appreciate any and all suggestions or insights anyone may be able to offer here.

SOFA is the Standards Of Fundamental Astronomy, approved by the International Astronomical Union (IAU) as being the canonical standard for calculating positions and time in astronomy. It is available in C and F77 versions as standard, and wrappers have been written in other languages, most notably Python as part of AstroPy. For your purpose, the thirdparty part of the website mentions the World Wide Astronomy (WWA) library which is C# so that would be worth a look for your application. I would recommend reading the SOFA Cookbooks, particularly the ones on Earth Attitude and Time Scales to get a better idea of how the (many) various time and co-ordinate systems fit together.

• Thank you sir! I'll look into this and hopefully be able to update my question if I run into something I don't understand. – a_here_and_now Oct 5 '18 at 18:27
• I'd personally recommend CSPICE, which is what NASA uses, but SOFA is good to. – barrycarter Oct 7 '18 at 18:15
• Since the OP was looking for a C# version and seemed to have some exposure to SOFA already, it seemed the logical choice. It also comes with the reasonable precision ephemerides built in: epv00 for Earth position (reversible for Sun position) and plan94 for the planets. Saves having to download extra kernels – astrosnapper Oct 7 '18 at 21:34
• Given that the CSPICE toolkit is provided in C, how would I use it in C#? I actually am quite new to programming, and realistically I have no exposure to SOFA, just heard of it when I would search for previous inquiries toward my objective. Which one would be easier to implement? I have no problem installing extra stuff as long as its reasonably simple to do. Also, any tutorials or advice/pointers in the right direction toward implementing this would surely go a really long way and save me a lot of time. Otherwise, looking forward to learning and struggling with this for a day or so! – a_here_and_now Oct 8 '18 at 5:08
• I'm doing this for my work and they're pretty set up in the .NET environment with C# right now, so as much as I wish I could just use python, it's a bit of a stretch for my case since I can implement it with the SOFA functions. I've been reading through the SOFA documentation, they have equivalent C# functions posted on github already, and this seems to have potential to work really well. Thank you all! – a_here_and_now Oct 10 '18 at 23:06

@astrosnapper's answer is probably the one you want. But if you'd like something you can implement yourself, have a look at Astronomical Algorithms in this answer

This is something you'll have to dig into a bit, but if you like to program, it may be exactly what you're looking for. The Gaisma website is one of my favorites on the internet - easy to use and presents a bunch of information in easy-to-understand graphics. Click around!

I believe that this site uses algorithms from the collection found at this NOAA site. Click around there as well. They provide Excel spreadsheets which contain the algorithms and other resources. The "main" resource is a collection of algorithms published in the book Astronomical Algorithms - Jean Meeus. Search the books title and you can find that there are many similarly titled books. I'd recommend going to a library if possible, because it's (in my opinion) always good to go to libraries. However parts of these can be found on-line. For example, a few pages shown from the book Astronomical Formulae for Calculators (1988) include an interesting table of contents.

• Wow these are great resources. Thank you very much! – a_here_and_now Oct 10 '18 at 23:05

IAU SOFA and SPICE are overkill for what you need, and would require writing a C to C# wrapper around them. An implementation of VSOP87 in C# is all you really need. This is what was used to create NASA's Five Millennium Catalog of Solar Eclipses. The linked code has functions for getEarth() and getMoon() which return the heliocentric XYZ coordinates, the sun will always be at 0,0,0. To get geocentric coordinates, just subtract the Earth's position from the Moon and Sun positions.

You'll also want to look into Besselian elements, these are elements that are pre-computed in order to generate all of the data you see on NASA's pages. These already include all of the computations for light time, precession, nutation, abberation, etc. There are references on the Wikipedia page, but the most practical one is Elements of Solar Eclipses by Jean Meeus. To get an idea of what they can do, I used this method to create my Eclipse Map App. You can use the Besselian Elements either from Meeus' book, or from NASA's catalog. You'll find the elements in NASA's catalog after clicking through to the details, e.g. here's the one for the Aug 21 2017 Eclipse.

The benefit of using Besselian Elements rather than direct computation is first, a lot of the work is already done (e.g. precession, nutation...). Second, they are more computationally efficient, things like eclipse duration for a given lat/lon are solved directly rather that through itteration where you have to recompute the positions, precess, nutation, light time, etc all over again for each time step.

Each eclipse page in NASA's catalog also has a JavaScript implementation of how to use Besselian Elements to display the local circumstances on the map they display. But you might find the code a bit hard to follow without a reference like Meeus' book to explain what's going on.

Note the name Xavier Jubier in the NASA source code, he also has his own site on solar eclipses, has a similar JavaScript implementation and can also be another source of Besselian Elements.

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# How do I evaluate the integral $\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$?

I have no idea how to start, it looks like integration by parts won't work.

$$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$

If someone could shed some light on this I'd be very thankful.

• are you looking for exact value or just convergence? – Anurag A Jun 19 '14 at 9:37
• With wolfram alpha I know that the answer is $\pi /8e$. I do not know how to come to a solution – user1043065 Jun 19 '14 at 9:41
• Convergence is easy to prove since the function is bounded by an integrable function – user1043065 Jun 19 '14 at 9:44
• I'm not so sure that it is bounded by an obvious integrable function. Perhaps I'm not thinking correctly. – Anurag A Jun 19 '14 at 9:45
• M yields $\large{\pi \over 8{\rm e}} \approx 0.1445$. – Felix Marin Jun 20 '14 at 2:35

It is well known that $\displaystyle\int_0^\infty\frac{\sin x}{x}\,dx$ converges. Then $$\int_0^{\infty}\frac{x^5\sin x}{(1+x^2)^3}\,dx=\int_0^{\infty}\Bigl(\frac{x^5}{(1+x^2)^3}-\frac1x\Bigr)\sin x\,dx+\int_0^{\infty}\frac{\sin(x)}{x}.$$ This implies that the integral converges, since the the first integral on the right hand side converges absolutely,

To compute the value use calculus of residues. Let $$f(z)=\frac{z^5\,e^{iz}}{(1+z^2)^3}.$$ $f$ is meromorphic on an open set contaning the upper half plane, with a pole of order $3$ at $z=i$. For $R>1$ let $C_R$ be the semicircle $z=R\,e^{it}$, $0\le t\le\pi$. Then $$\int_{-R}^R\frac{x^5e^{ix}}{(1+x^2)^3}\,dx+\int_{C_R}f(z)\,dz=2\,\pi\,i\,\text{Res}(f;i).$$ Now you have to:

1. Prove that $\lim_{R\to\infty}\int_{C_R}f(z)\,dz=0$
2. Compute the residue.
• $f$ and $x^5\sin x/(1+x^2)^3$ do not coincide on the reals. Is that no problem? – user1043065 Jun 19 '14 at 10:48
• That the integral along the semi circle goes to 0 follows directl from the Jordan Lemma. – Zaid Alyafeai Jun 19 '14 at 12:01
• I have used Lebesgue dominated convergence theorem for that, concluding that the function goes to zero $\lambda$ - almost everywhere. Would that also be correct? – user1043065 Jun 19 '14 at 12:05
• I have edited the answer. The value of the integral will be the imaginary part of $2\,\pi\,i\,\text{Res}(f;i)$. – Julián Aguirre Jun 20 '14 at 11:51

Consider the function $f(t)=e^{\large-\sqrt a|t|}$, then the Fourier transform of $f(t)$ is given by \begin{align} F(x)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-ix t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-\sqrt a|t|}e^{-ix t}\,dt\\ &=\int_{-\infty}^{0}e^{\sqrt at}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-\sqrt at}e^{-ix t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(\sqrt a-ix)t}}{\sqrt a-ix} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(\sqrt a+ix)t}}{\sqrt a+ix} \right|_{0}^{t=v}\\ &=\frac{1}{\sqrt a-ix}+\frac{1}{\sqrt a+ix}\\ &=\frac{2\sqrt a}{x^2+a}. \end{align} Next, the inverse Fourier transform of $F(x)$ is \begin{align} f(t)=\mathcal{F}^{-1}[F(x)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(x)e^{ix t}\,dx\\ e^{-\sqrt a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sqrt a}{x^2+a}e^{ix t}\,dx\\ \frac{\pi e^{-\sqrt a|t|}}{\sqrt a}&=\int_{-\infty}^{\infty}\frac{e^{ix t}}{x^2+a}\,dx,\tag1 \end{align} where $(1)$ can be rewritten as $$\int_{0}^{\infty}\frac{e^{ix t}}{x^2+a}\,dx=\frac{\pi e^{-\sqrt at}}{2\sqrt a}.\tag2$$ Now differentiate $(2)$ with respect to $a$ twice and with respect to $t$ five times, take the real part, and set $a=t=1$ yields \begin{align} \Re\left[\int_{0}^{\infty}\frac{\partial^2}{\partial a^2}\frac{\partial^5}{\partial t^5}\left(\frac{e^{ix t}}{x^2+a}\right)\,dx\right]_{t=1,\,a=1}&=\left.\frac{\partial^2}{\partial a^2}\frac{\partial^5}{\partial t^5}\left(\frac{\pi e^{-\sqrt at}}{2\sqrt a}\right)\right|_{t=1,\,a=1}\\ -2\int_{0}^{\infty}\frac{x^5\sin x}{(x^2+1)^3}\,dx&=-\frac{\pi}{4e}\\ \int_{0}^{\infty}\frac{x^5\sin x}{(x^2+1)^3}\,dx&=\large\color{blue}{\frac{\pi}{8e}}. \end{align}

• Forgot to mention that $a>0$ and $t>0$ in $(2)$. You may also refer to this link to obtain $(1)$ using double integral method (elementary way). – Tunk-Fey Jun 19 '14 at 11:59
• Thank you very much for the effort! I am in a course of complex analysis so I think I'll use the other answer. I was still checking it so I did not yet check the answer acceptance button. sorry for that but still thanks for the effort! – user1043065 Jun 19 '14 at 12:04
• @user1043065 It's OK, no hurt feeling from me. I just enjoy your problem, very nice. $\ddot\smile$ – Tunk-Fey Jun 19 '14 at 12:05

An alternative to the obvious canonical answer given by Julián Aguirre: it is easy to compute $$\int_{-\infty}^{\infty} \frac{e^{i b x}}{x^2+a}dx$$ for $a, b > 0$, because the integrand has simple poles only. Now differentiate with respect to $a$ twice and with respect to $b$ 5 times, and set $a = b = 1$. The theorem you need to justify the differentiation can be found in your textbook.

Note \begin{eqnarray} I&=&\int_0^\infty \frac{x[(x^2+1)-1]^2\sin x}{(1+x^2)^3}dx\\ &=&\int_0^\infty \frac{x\sin x}{1+x^2}dx-2\int_0^\infty \frac{x\sin x}{(1+x^2)^2}dx+\int_0^\infty \frac{x\sin x}{(1+x^2)^3}dx. \end{eqnarray} From $$\int_0^\infty \frac{\cos(ax)}{b^2+x^2}dx=\frac{\pi}{2b}e^{-ab}, a>0, b>0,$$ we have \begin{eqnarray} \int_0^\infty \frac{x\sin(ax)}{b^2+x^2}dx&=&=-\frac{d}{da}\int_0^\infty \frac{\cos(ax)}{b^2+x^2}dt=\frac{\pi}{2e^{-ab}},\\ \int_0^\infty \frac{x\sin x}{(b^2+x^2)^2}dx&=&-\frac{1}{2b}\frac{d}{db}\int_0^\infty \frac{x\sin(ax)}{b^2+x^2}dt=\frac{a\pi}{4be^{-ab}},\\ \end{eqnarray} and \begin{eqnarray} \frac{d^2}{db^2}\int_0^\infty \frac{x\sin (ax)}{b^2+x^2}dx&=&8b^2\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^3}dx-2\int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^2}dx. \end{eqnarray} The latter one implies \begin{eqnarray} \int_0^\infty \frac{x\sin(ax)}{(b^2+x^2)^3}dx=\frac{a(ab+1)\pi}{16b^3e^{-ab}}. \end{eqnarray} Thus $$I=\frac{\pi}{8e}.$$


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## anonymous one year ago An oil drop with a mass of 3.5*10^-15 kg accelerates downward at a rate of 2.50 m/s^2 when placed between 2 horizontal parallel plates that are 1.00 cm apart. Assuming the oil drop is negative and the top plate positive, how many excess electrons does the oil drop carry if the potential difference between the plates is 5.38*10^2 V?

• This Question is Open
1. IrishBoy123

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2. IrishBoy123

$$\large n_e = {q \over e^-}$$


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College Physics 2e

26.5Telescopes

College Physics 2e26.5 Telescopes

Learning Objectives

By the end of this section, you will be able to:

• Outline the invention of a telescope.
• Describe the working of a telescope.

Telescopes are meant for viewing distant objects, producing an image that is larger than the image that can be seen with the unaided eye. Telescopes gather far more light than the eye, allowing dim objects to be observed with greater magnification and better resolution. Although Galileo is often credited with inventing the telescope, he actually did not. What he did was more important. He constructed several early telescopes, was the first to study the heavens with them, and made monumental discoveries using them. Among these are the moons of Jupiter, the craters and mountains on the Moon, the details of sunspots, and the fact that the Milky Way is composed of vast numbers of individual stars.

Figure 26.23(a) shows a telescope made of two lenses, the convex objective and the concave eyepiece, the same construction used by Galileo. Such an arrangement produces an upright image and is used in spyglasses and opera glasses.

Figure 26.23 (a) Galileo made telescopes with a convex objective and a concave eyepiece. These produce an upright image and are used in spyglasses. (b) Most simple telescopes have two convex lenses. The objective forms a case 1 image that is the object for the eyepiece. The eyepiece forms a case 2 final image that is magnified.

The most common two-lens telescope, like the simple microscope, uses two convex lenses and is shown in Figure 26.23(b). The object is so far away from the telescope that it is essentially at infinity compared with the focal lengths of the lenses ($do≈∞do≈∞$). The first image is thus produced at $di=fodi=fo$, as shown in the figure. To prove this, note that

$1di=1fo−1do=1fo−1∞.1di=1fo−1do=1fo−1∞.$
26.25

Because $1/∞=01/∞=0$, this simplifies to

$1di=1fo,1di=1fo,$
26.26

which implies that $di=fodi=fo$, as claimed. It is true that for any distant object and any lens or mirror, the image is at the focal length.

The first image formed by a telescope objective as seen in Figure 26.23(b) will not be large compared with what you might see by looking at the object directly. For example, the spot formed by sunlight focused on a piece of paper by a magnifying glass is the image of the Sun, and it is small. The telescope eyepiece (like the microscope eyepiece) magnifies this first image. The distance between the eyepiece and the objective lens is made slightly less than the sum of their focal lengths so that the first image is closer to the eyepiece than its focal length. That is, $do′do′$ is less than $fefe$, and so the eyepiece forms a case 2 image that is large and to the left for easy viewing. If the angle subtended by an object as viewed by the unaided eye is $θθ$, and the angle subtended by the telescope image is $θ′θ′$, then the angular magnification $MM$ is defined to be their ratio. That is, $M=θ′/θM=θ′/θ$. It can be shown that the angular magnification of a telescope is related to the focal lengths of the objective and eyepiece; and is given by

$M=θ′θ=−fofe.M=θ′θ=−fofe.$
26.27

The minus sign indicates the image is inverted. To obtain the greatest angular magnification, it is best to have a long focal length objective and a short focal length eyepiece. The greater the angular magnification $MM$, the larger an object will appear when viewed through a telescope, making more details visible. Limits to observable details are imposed by many factors, including lens quality and atmospheric disturbance.

The image in most telescopes is inverted, which is unimportant for observing the stars but a real problem for other applications, such as telescopes on ships or telescopic gun sights. If an upright image is needed, Galileo’s arrangement in Figure 26.23(a) can be used. But a more common arrangement is to use a third convex lens as an eyepiece, increasing the distance between the first two and inverting the image once again as seen in Figure 26.24.

Figure 26.24 This arrangement of three lenses in a telescope produces an upright final image. The first two lenses are far enough apart that the second lens inverts the image of the first one more time. The third lens acts as a magnifier and keeps the image upright and in a location that is easy to view.

A telescope can also be made with a concave mirror as its first element or objective, since a concave mirror acts like a convex lens as seen in Figure 26.25. Flat mirrors are often employed in optical instruments to make them more compact or to send light to cameras and other sensing devices. There are many advantages to using mirrors rather than lenses for telescope objectives. Mirrors can be constructed much larger than lenses and can, thus, gather large amounts of light, as needed to view distant galaxies, for example. Large and relatively flat mirrors have very long focal lengths, so that great angular magnification is possible.

Figure 26.25 A two-element telescope composed of a mirror as the objective and a lens for the eyepiece is shown. This telescope forms an image in the same manner as the two-convex-lens telescope already discussed, but it does not suffer from chromatic aberrations. Such telescopes can gather more light, since larger mirrors than lenses can be constructed.

Telescopes, like microscopes, can utilize a range of frequencies from the electromagnetic spectrum. Figure 26.26(a) shows the Australia Telescope Compact Array, which uses six 22-m antennas for mapping the southern skies using radio waves. Figure 26.26(b) shows the focusing of x rays on the Chandra X-ray Observatory—a satellite orbiting earth since 1999 and looking at high temperature events as exploding stars, quasars, and black holes. X rays, with much more energy and shorter wavelengths than RF and light, are mainly absorbed and not reflected when incident perpendicular to the medium. But they can be reflected when incident at small glancing angles, much like a rock will skip on a lake if thrown at a small angle. The mirrors for the Chandra consist of a long barrelled pathway and 4 pairs of mirrors to focus the rays at a point 10 meters away from the entrance. The mirrors are extremely smooth and consist of a glass ceramic base with a thin coating of metal (iridium). Four pairs of precision manufactured mirrors are exquisitely shaped and aligned so that x rays ricochet off the mirrors like bullets off a wall, focusing on a spot.

Figure 26.26 (a) The Australia Telescope Compact Array at Narrabri (500 km NW of Sydney). (credit: Ian Bailey) (b) The focusing of x rays on the Chandra Observatory, a satellite orbiting earth. X rays ricochet off 4 pairs of mirrors forming a barrelled pathway leading to the focus point. (credit: NASA)

A current exciting development is a collaborative effort involving 17 countries to construct a Square Kilometre Array (SKA) of telescopes capable of covering from 80 MHz to 2 GHz. The initial stage of the project is the construction of the Australian Square Kilometre Array Pathfinder in Western Australia (see Figure 26.27). The project will use cutting-edge technologies such as adaptive optics in which the lens or mirror is constructed from lots of carefully aligned tiny lenses and mirrors that can be manipulated using computers. A range of rapidly changing distortions can be minimized by deforming or tilting the tiny lenses and mirrors. The use of adaptive optics in vision correction is a current area of research.

Figure 26.27 An artist’s impression of the Australian Square Kilometre Array Pathfinder in Western Australia is displayed. (credit: SPDO, XILOSTUDIOS)
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Model of HCl ( hydrochloric acid) Covalent Compounds. 8 0 obj Practice naming ionic compounds when given the formula If you're seeing this message, it means we're having trouble loading external resources on our website. Grade/level: 9th to 12th grade Age: 14-18 Main content: Naming Compound Other contents: Covalent and Ionic Formulas and Compounds Add to my workbooks (6) Download file pdf Embed in my website or blog Add to Google Classroom Add to Microsoft Teams Share through Whatsapp Choose from 500 different sets of grade 9 chemistry compounds naming binary flashcards on Quizlet. 10 0 obj What is the correct name for MoO 3 (think about how compounds have to be reduced to the simplest integer ratio) x�{�������! Compounds. ���+�_�t����YU�����2����w���I���PF��kb �؂��1�mkd�̯�&�?�x�#��>ȿap�&g,��S/�4Y�ͪ�Mv6��U=�����?�������ES9 Find out â¦ The chemical formula for salt is NaCl. �s������4�W:#��>M~շ�ڍ���S��)_֒�s��V��2ڄ�O��,��5BV��]>Vuق//�$5V��4�8��|���MF8��f��j��;,S���ˆu�}�Ԡ?�����6 <>>> About this quiz: All the questions on this quiz are based on information that can be found at Chemistry: Naming Compounds. Preview/Review Concepts W2 - Lesson 4 Science Grade 9 W2 - Lesson 4: Naming Chemical Compounds Salt is one of the most familiar chemical compounds. Let's start this quiz; it will give you the different functions of the International Union of Pure and Applied Chemistry (IUPAC) naming chemical compounds. q 16 0 0 1 0 0 cm/Image6 Do Q This could also be used to give students practice with compound practice. Home. Naming covalent compounds. 12 0 obj This Naming Compounds Spinner is a sure-fire way to help your students with chemical nomenclature! Edit. Naming Compounds in Chemistry. Types of Compounds Compounds Ionic Molecular. x��U�j�@��w9 SCIENCE 9 MODULE 1. Technology*Lesson*Plan*â*Rules*for*Naming*Compounds*(Grades*9=12)* * * *! <>>>/Length 29>> STA: Ch.5 30.ANS: C PTS: 1 DIF: L3 REF: p. 257 | p. 264 OBJ: 9.2.2 Apply the rules for naming and writing formulas for compounds with polyatomic ions. Element Project â Due Oct. 10 4. Molecular Compounds. 15 0 obj 2 0 obj You rocked this quiz! Back to Science for Kids 60% average accuracy. Sodium therefore comes first in the compound name. endobj 6 0 obj <>/Pattern<>/Font<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 720 540] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>> endobj Chemistry is the branch of science which deals specifically with the elements on the periodic table, as well as compounds, all of which are composed of atoms, molecules, and ions. d'- ����Ѕ-�����Ѵ�;��ʱ�@hvو���{�} When naming organic compounds, the IUPAC (International Union of Pure and Applied Chemistry) nomenclature (naming scheme) is used. How are the elements arranged on the Periodic Table? Displaying top 8 worksheets found for - Writing Formulas And Naming Compounds. 13 0 obj q 16 0 0 1 0 0 cm/Image6 Do Q This is to give consistency to the names. Grade 9 Elements And Compound Displaying top 8 worksheets found for - Grade 9 Elements And Compound . Naming Compounds â Part 2 â YouTube: This video explains how to use a chemical name to write the formula for that compound. Each compound has a unique name; Type 1: Compounds that contain a metal and a non-metal endobj %PDF-1.5 At this point we only need to learn how to name compounds consisting of two elements. Ninth Grade (Grade 9) Molecules and Compounds questions for your custom printable tests and worksheets. To ensure the best experience, please update your browser. <> endobj Save. Names and chemical symbols; Properties of elements in the same group; Names of compounds. Naming 1. stream Chemical bonds hold them together. A compound has a chemical formula; The Periodic Table. <> Grade 9 test 2019: Naming compounds and writing formulae DRAFT. We will consider three types: 1. The IUPAC system for naming compounds is very complex, but we do not need to learn all its rules. endobj It looks like your browser needs an update. answer choices . stream 11 0 obj stream Learn vocabulary, terms, and more with flashcards, games, and other study tools. It's obvious you've studied how to name ionic compounds and write the formulas from the names. 4 0 obj Ionic bonding powerpoint for grade 9 and below. <>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 720 540] /Contents 15 0 R/Group<>/Tabs/S/StructParents 1>> 5 0 obj <> 2. 9th grade. Start studying Naming-Chemistry-Grade 9. Grade 9 - Ionic Bonding - Free download as Powerpoint Presentation (.ppt / .pptx), PDF File (.pdf), Text File (.txt) or view presentation slides online. <> <> Ionic compounds are formed when a metal gives up its electrons to a non-metal. Non-metal goes second. Learn how to easily name ionic and covalent compounds. x��UMO�@�[�xG���~!$��T=�Rc>�ġI������IHH����2/�｝��w����e��)��NN���&RH)I)�ᕄ5�*M��N���p>��5�zW����H�H��Cw{i�)Mpv�V� ��#���-��e+Ȩ +���g@)� � Va ��[Fe�����ͣ4q!4; /�Pz x2"���e�p�����hc������T���5/�S�y���*[����Q��@,LbѢ�E�eW��x�C�L��M˝s[)\ �xk����l��-�G��#�)�A�4aʃ̈�Z-�p��-s�� �g��c�\fly�x��0I[���qc_���&�_��iV�*�*�h the element with "ide" at â¦ Naming compounds Learning Goals: Understand the conventions in naming chemical compounds. The next step is predicting whether â¦ endstream Naming Compounds Mrs. McCutchen's Class - Worksheet 5 - Naming Ionic Compounds Part 1 (Binary compounds with Group 1, 2 or 13 metals) Binary compounds: have only two kinds of elements, if there are three or more it is not a binary compound. endobj endobj Date:* 11/19/15=11/20/15* * Grade/Class/Subject:* Chemistry*(Grade*10=11)* stream When naming a covalent compound, the name of the first element is unchanged, the name of the second element ends with âideâ and if necessary, a prefix is used for the second element to indicate the number of atoms involved. 9 0 obj 3. Grade 9 Science: Unit 1: Atoms, Elements, and Compounds. %���� In the example of $$\text{NaCl}$$, sodium is a group 1 element on the left hand side of the table, while chlorine is in group 17 on the right of the table. This is part of my Naming Compounds Bundle AND my cost-saving, amazing Chemistry Bundle!The top layer asks the students 3 questions to help them identify which type of compound â¦ Naming Acids and Bases Acid names are based on the anion they form when dissolved in water; base names follow the rules for ionic, organic, or molecular compounds. In a compound, the element that is on the left of the Periodic Table, is used first when naming the compound. endobj 7 0 obj In this lesson, you will learn how to name and write formulas for various chemical compounds. <> endstream You might want to review the rules for naming ionic compounds to make certain you've mastered them. Grade 9, U1-L13-Molecular Compounds. Success Criteria: Given the chemical formula for a certain compound, you can write down its chemical name. In a hurry? 162 times. stream Ionic Compounds Without a Transition Metal. Naming Compounds More Practice With Naming Before naming a compound, you have to figure out what kind of compound it is. Organic Nomenclature and Functional Groups | Organic Chemistry Explained, Naming Alkanes | Organic Nomenclature and Functi…, Naming Alkyl Groups | Organic Nomenclature and F…, Functional Groups | Organic Nomenclature and Fun…, Grade 9 Academic Vocabulary | Knowsys Level 9 Guide, Naming compounds, Naming Chemical Compounds, Practice Tests | Organic Chemistry Explained, Organic Reaction Mechanisms and the Chemistry of…, Substitution and Elimination Reactions and the C…, Naming Binary Ionic Compounds - with transition metals, Science GCSE Prep - Biology, Chemistry, and Physics, Naming Ionic Compounds with Polyatomic Ions, Simple Binary Ionic Compounds - no transition metals, Naming Type III Binary Compounds (molecular compounds), PAP Chemistry chapter 5; naming binary compounds, Chemistry - Practice Naming/Writing Formulas of Compounds. The 21 compounds are shown here, either the name is given and the students need to write the formula, or else name the compound from the shown formula. Are either ionic or covalent. For example: Carbon with one oxygen atom â carbon monoxide (CO) 2. Review: Classification of Matter Definition of compound: pure substance two or more elements chemically combined. �9@�� v�.� Mrs Idone's Classroom. endobj OBJ: 9.2.1 Apply the rules for naming and writing formulas for binary ionic compounds. When naming a compound, what must the last part be? �G�N�e� ��wx돷����e����a>����ET-��A��I\B� ���+�Q�M6�!�,M8���������SI �iz�=yO��. endobj At this level we have to distinguish between two types of compounds, because the type of compound determines how it should be named. 1 0 obj <> Represented by a chemical formula. Mr. Idone's Grade 10 Applied Science Classroom. endstream z���w��p��.���30��B0V0Њ�҅���À�tgØ�L�MN��s��m�A�_�@T��5ů4���0��B�8n���TTs��2׷4���ہ�,LS� X&)5`��V���[���]�����*�� �P� })�Re��JyuKw����4�RU�j���T��d����R�Vլ���h�P� Some of the worksheets for this concept are Language handbook work, Atoms and molecules, Mixtures work answer key, Elements compounds and mixtures, Elements mixtures and compounds, Science 9th chemistry crossword name, Grammar practice workbook, 9th grade â¦ The chemical name for salt is sodium chloride. <> grade 9 science atoms elements and compounds questions and answers Oct 05, 2020 Posted By Rex Stout Publishing TEXT ID d666b40f Online PDF Ebook Epub Library in fixed ratios to form compounds ks3 science atoms and elements 01 join us login uk us india search join us login ks1 age 5 7 ks2 age 7 11 11 age 7 11 ks3 age 11 14 gcse 5 months ago. Q. Metal goes first and you get the name directly from the periodic table. It also enables every compound to have a unique name, which is not possible with â¦ 1. Chemistry. endobj Search this site. A pure substance made of two or more kinds of elements combined in fixed proportions. 14 0 obj 3 0 obj | 9.5.2 Apply Handout: Naming Compounds There are a lot of rules, so take care in reading the handout carefully. | 9.5.2 Apply the rules for naming chemical compounds by using a flowchart. <>>>/Length 29>> endobj If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Browse our pre-made printable worksheets library with a variety of activities and quizzes for all K-12 levels. Learn grade 9 chemistry compounds naming binary with free interactive flashcards. <> For webquest or practice, print a copy of this quiz at the Chemistry: Naming Compounds webquest print page. What information can we find on the Periodic Table? In this quiz, we aim to boost your knowledge on a chemistry-oriented level on some everyday items such as salt and aluminum foil, as well as your expertise on individual elements and their chemical symbol. kbolton_29678. This Ionic Compounds Worksheet is suitable for 9th Grade. � endobj endstream Examples of Molecular Compounds Gases: H2, O2, CO2, H2O, N2, ozone. <> endobj 0. Oh no! Vocabulary, terms, and more with flashcards, games, and study. Sets of grade 9 chemistry compounds naming binary with free interactive flashcards, the element that is on the Table. Is suitable for 9th grade, ozone students with chemical nomenclature compounds naming binary with free interactive.... 9 ) Molecules and compounds questions for your custom printable tests and worksheets Given chemical. Chemical nomenclature chemistry: naming compounds, H2O, N2, ozone goes first you! With compound practice 9th grade a metal gives up its electrons to non-metal. Same group ; names of compounds compounds naming binary with free interactive.... Group ; names of compounds, because the type of compound: pure substance two or more of. In fixed proportions tests and worksheets for binary ionic compounds to make certain you studied! A flowchart for various chemical compounds the conventions in naming chemical compounds by using a flowchart what the... Activities and quizzes for all K-12 levels based on information that can be found at:! Properties of elements in the same group ; names of compounds compound practice the arranged... Naming a compound, the element that is on the Periodic Table you 've studied how name... Rules for naming and writing formulas and naming compounds to name compounds consisting of or. If you 're behind a web filter, please make sure that domains. A web filter, please update your browser might want to Review the rules for naming compounds. More practice with naming Before naming a compound, the element that is on the Periodic Table: compounds... Choose from 500 different sets of grade 9 chemistry compounds naming binary flashcards on Quizlet pure! Or more kinds of elements in the same group ; names of compounds rules so... For binary ionic compounds Worksheet is suitable for 9th grade first when naming a compound, you will learn to... To use a chemical name to write the formulas from the Periodic Table part 2 â:... Could also be used to give students practice with naming Before naming a compound, what must the part... Of compound: pure substance two or more kinds of elements combined in fixed proportions of rules so! Should be named elements chemically combined are a lot of rules, take... 9Th grade please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked this we. Consisting of two elements for 9th grade at this point we only need learn! The left of the Periodic Table learn all its rules Covalent compounds metal gives up its electrons a... Compounds, because the type of compound determines how it should be named chemically combined this compounds... Of elements in the same group ; names of compounds to a non-metal other study.. Binary with free interactive flashcards, O2, CO2, H2O, N2, ozone, the that! Give students practice with naming Before naming a compound, the element that is on the left of Periodic. What information can we find on the Periodic Table other study tools it is: Classification of Matter of! | 9.5.2 Apply the rules for naming compounds â part 2 â YouTube: this video explains to! Chemical compounds a flowchart compounds There are a lot of rules, so care!, the element that is on the Periodic Table be used to give practice. Other study tools on Quizlet compounds more practice with compound practice the conventions naming... Can we find on the Periodic Table free interactive flashcards the type of compound it is name ionic and compounds... Its rules.kastatic.org and *.kasandbox.org are unblocked down its chemical name in fixed.! Covalent compounds gives up its electrons to a non-metal chemical symbols ; Properties of elements combined in proportions... Video explains how to name ionic and Covalent compounds â¦ Displaying top 8 worksheets found for writing. Activities and quizzes for all K-12 levels lesson, you can write down its chemical name write! Students with chemical nomenclature There are a lot of rules, so take care reading. Way to help your students with chemical nomenclature the IUPAC system for compounds! The formulas from the Periodic Table with flashcards, games, and other study tools examples of Molecular compounds:....Kastatic.Org and *.kasandbox.org are unblocked the formulas from the names, because the type of it.

marcy md 5191 parts


text:
# Altitudes

Geometry Level 2

True or False?

If any 2 altitudes of a triangle are known, then it is possible to calculate the range of the third altitude.

×


text: Let us presume that there are two competing plants which starts production on the same day. Say that you have a cost function that gives you the total cost, C ( x ), of producing x items (shown in the figure below). Demand: Demand is generally the quantity of a good that an individual is willing and able to buy at a given price. Suppose that Joe ends up paying $1.25 per litre for milk. How would one calculate price function in this scenario? The marginal buyer is the consumer who will leave the market for a product first if the price was any higher. The marginal cost formula represents the incremental costs incurred when producing additional units of a good or service. Multiply both sides of this equation by price $$(P)$$: $$(P â MC) = 0.5P$$, or $$0.5P = MC$$, which yields: $$P = 2MC$$. That for any given the quantity of the good you're selling, that that point on the curve is actually showing the marginal benefit for that incremental unit. Economists use the term demand to refer to the amount of some good or service consumers are willing and able to purchase at each price. For businesses, this is also called the Marginal Revenue. Home; Uncategorized; how to calculate marginal benefit from demand curve; November 11, 2020 Formula to calculate marginal utility. Marginal cost, marginal revenue, and marginal profit all involve how much a function goes up (or down) as you go over 1 to the right â this is very similar to the way linear approximation works. Marginal Revenue Calculator How to Calculate Producer Surplus GDP per Capita Calculator GDP Deflator Calculator Money Multiplier Calculator Linear Demand Curve Calculator What is the Difference Between Demand Function and Inverse Demand Function? Viewed 549 times 1$\begingroup$This example is in a oligopoly market with two firms. The total gross benefit is therefore the sum of the marginal benefits from consuming successive units. Active 4 years, 7 months ago. Purchasing a car might improve the consumerâs mobility and efficiency. Ask Question Asked 5 years ago. Therefore, the demand curve is referred to as the marginal benefit curve. This example problem goes over the degree of comfort experienced at different levels of clean air. We can calculate the marginal net benefit of a decision by subtracting marginal cost from marginal benefit. Price and quantity demanded for most goods and services will be inversely related. Summary. DOWNLOAD IMAGE. Consumer surplus. The Demand Curve is a line that shows how many units of a good Inventory Inventory is a current asset account found on the balance sheet, consisting of all raw materials, work-in-progress, and finished goods that a company has accumulated. Marginal benefit is the benefit which we get through the consumption of one extra unit of any good or service. It is often deemed the most illiquid of all current assets - thus, it is excluded from the numerator in the quick ratio calculation. marginal revenue always lies behind the demand curve,and when demand increases marginal revenue also increases.demand curve is used to determine price of a commodity. So this is a marginal benefit for that first unit. The socially efficient quantity of lift tickets can be found by setting MSB = MSC: 80 â (1/4)Q = (1/4)Q. The demand curve reflects the marginal benefit consumers receive from purchasing an extra unit of the good (i.e., it reflects their marginal WTP for an extra unit). Why inverse? In the last video, we saw how you can actually view a demand curve as actually a marginal benefit curve. The marginal cost formula = (change in costs) / (change in quantity). The more goods you consume, the more the benefits you get. how to calculate marginal benefit from demand curve. But, the extra benefits you get from each additional unit of goods will decrease. In the following example, we will practice calculating marginal cost based on the learning curve. Marginal net benefit of the first drink is$13 ($20 â$7), the 2nd is $5 ($12 â $7), and the third is -$1 ($6 â$7). At any quantity demanded, the corresponding price depicted on the demand curve shows the willingness to pay of what Mankiw calls the "marginal buyer." The supply curve reflects the marginal cost to the firm of producing an extra unit. Let's call them plant A and B. You will not be graded on any changes you make to this graph. Of course, calculating marginal social benefit in the real world is far more complicated. The marginal benefit concept is essential in explaining the reasons behind the downward slope of the demand curve. There is a close relationship between any inverse demand function for a linear demand equation and the marginal revenue function. This is the marginal benefit for that second unit. Using Marginal Benefit to derive the Demand Curve * We can compare marginal benefit to the price this tells us the quantity that people are going to consume (quantity demanded). The marginal benefit is thus the change in the total benefit when an additional unit is consumed. Solved 1 The Table Presents Leff Stota Benefit From Drin. Use the graph input tool to help you answer the following questions. They are in an industry with a progress ratio of 80%. To determine the optimal quantity of a public good, it is necessary to first determine the demand for it. The formula for elasticity is (âQ/âP) × (P/Q). The variable costs included in the calculation are labor and materials, plus increases in â¦ The presence of an active car factory might also spur local commerce and generate jobs, and so on. Marginal utility quantifies the added satisfaction that a consumer garners from consuming additional units of goods or services. Hence the individual demand curve will be downward-sloping. How to Calculate AVC? Use the graph input tool to help you answer the following questions. The demand curve is essentially the âinverseâ of the marginal benefit curve. When we consider what factors will shift a demand curve, we need to make a distinction between the endogenous factors ... We will now go over several examples of calculating marginal benefit and marginal cost to show how to get this right on exam or homework questions. Using Marginal Benefit And Marginal Cost Curves To Find Net. When this is substituted into Equation \ref{3.5}, the result is: $$\dfrac{P â MC}{P} = 0.5$$. Demand is based on needs and wants, and while consumers can differentiate between a â¦ Since the whole quantity will be sold at the new lower price, the marginal revenue will be equal to the price of the extra unit sold minus the loss from selling all previous units at the new lower price . How does the marginal benefit explain the slope of the demand curve. The demand curve represents the marginal benefit curve because a consumer's willingness to pay for the next unit of a good is equal to the marginal... See full answer below. The demand curve shows the willingness to pay for an additional unit of the good or service, so it is equal to the marginal benefit of that good. The Marginal Revenue curve looks very similar to the Demand curve, just slightly steeper. Calculating marginal revenue from a linear demand curve The blue curve on the following graph represents the demand curve facing a firm that can set its own prices. How to calculate price function from demand curve when marginal cost is given? DOWNLOAD IMAGE. Utility is the satisfaction or benefit derived by consuming a product. If there are diminishing marginal returns, then peopleâs willingness to pay will also decline. * Marginal benefit of the good is the increase in benefit * Price of the good is the increase in â¦ Cost-benefit analysis: A systematic process for calculating and comparing the marginal benefits and marginal costs of a project or activity. You will not be graded on any changes you make to this graph. Therefore, at the competitive equilibrium, the price is where the marginal benefit equals the marginal cost. Relation to marginal revenue. Assume that a monopolist has a demand curve with the price elasticity of demand equal to negative two: $$E_d = -2$$. DOWNLOAD IMAGE. How To Calculate Marginal Benefit From Demand Curve DOWNLOAD IMAGE. Marginal Utility And Total Utility Video Khan Academy. The demand curve is given and also two firms' MC is given. Calculating marginal revenue from a linear demand curve The blue curve on the following graph represents the demand curve facing a firm that can set its own prices. Note that the inverse demand schedule can be considered as the marginal social benefit (MSB) curve. Demand, Willingness to Pay and Marginal Benefits The market demand curve for a good originates from what individuals are willing to pay (W2P) for the good. As long as the marginal net benefit is positive, we should increase our activity! Explain quantity demanded, and the law of demand; Identify a demand curve; Calculate consumer surplus given a Marginal Benefit curve and price; The Law of Demand. If the demand curve is linear, it is obvious that in order to sell an additional unit of x its price must fall. Marginal Benefit = ($112.50 â$50.00) / (15 â 5) Marginal Benefit = $6.25 per chocolate Since the next set of chocolates are priced higher ($6.25) than what the consumer in willing to pay ($5), he will not buy the next 10 chocolates. It can be thought of as the difference between the amount that the consumer was willing to pay and what he/she actually paid. is a measurement of the net benefit a consumer gains from consuming a certain amount of a good. Marginal Buyers. Maximum net benefits are found where the marginal benefit curve intersects the marginal cost curve at activity level D. Panel (b) shows that if the level of the activity is restricted to activity level E, net benefits are reduced from the light-green shaded triangle ABC in Panel (a) to the smaller area ABGF. To calculate the net benefit, of course, we have to subtract out the amount Joe actually pays for the amount he consumes . The âMarginal Benefitsâ are the extra benefit that a producer gets from producing one more unit of a good. This fact should be kept in mind when calculating elasticity. Economists use this marginal utility concept to determine how much of an item consumers are willing to purchase. By algebraic manipulation: Qd = 320 â 4P = 320 â 4*marginal social benefit, so MSB = 80 â (1/4)Q. The marginal utility they get will therefore influence their willingness to pay for something. This is just the tip of the iceberg as far as microeconomics is concerned. Total Benefit = Sum of Marginal Benefits. Referred to as the marginal Revenue function consumers are willing to pay will also decline scenario. An individual is willing and able how to calculate marginal benefit from demand curve buy at a given price out the Joe. Concept to determine the optimal quantity of a good that an individual is willing and able to at... A linear demand equation and the marginal social benefit in the real world is far more complicated for... Joe actually pays for the amount that the inverse demand function for a linear demand equation and marginal... Garners from consuming additional units of goods or services optimal quantity of a good and so on is. It is necessary to first determine the demand curve, just slightly steeper ) / ( in... A project or activity from marginal benefit curve utility they get will therefore influence their willingness pay! And marginal costs of a good we will practice calculating marginal social benefit in the last video, saw! Or service diminishing marginal returns, then peopleâs willingness to pay and he/she... From Drin net benefit of a project or activity gains from consuming additional units of goods decrease! Calculating marginal cost to the demand curve when marginal cost is given world is far more complicated changes you to... 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text:
How to solve an exponential ciphers?

1. Apr 9, 2004

sachiyo

p=2633 , C=423 , e=29

plain text: EX (0423)
how to solve 423^29 mod 2633 without using calculator?

2. Apr 9, 2004

Hurkyl

Staff Emeritus
You only need 28 multiplies to do it in the most obvious fashion. *shrug* I bet you've learned a better way to do it if you read your notes / text.

3. Apr 10, 2004

sachiyo

There is some hints to do it by using Euler's theorem. But I still have no idea to do it.

4. Apr 20, 2004

Gokul43201

Staff Emeritus
it's painstaking

For standard RSA type encryption, the number N = 2633 (public key) is a product of 2 primes, p and q, which allows for decription using (p-1) and (q-1). However, in your choice, N is 2633 which itself is prime, and so does not allow for easy decription (also it is not the 29th Merseene Prime, or suchtypes). Anyway, there is a general approach to solve your problem...in this particular case, the numbers appear to conspire to giving large remainders with most powers I tried. I actually solved your problem without a calculator, and it took me about 1 hour, cause it's mostly "brute force" arithmetic. Here's how I did it :

Notation : a^e||p==x means that x is the residue when 'a' raised to the e'th power is divided by p.
Note : use (a^e)(b^f)||p==[(a^e||p)(b^f||p)] ||p and (a^e + b^f)||p==[(a^e||p) + (b^f||p)] ||p

423 = (3^2)*47, already a bad start ...stuck with a bg prime, 47.
So, 423^29 = (3^58)*(47^29)
3^8||2633=6561||2633==1295, which is large and unwieldy, 3^9, 3^10 & 3^11 also give large remainders so i tried 47*3^7||2633 ==102, a relatively small number, and close enough to 100 that it is easy to square and cube easily. Of course, using this, we will exhaust the 3's way before work through the 47's but that is a small price.

So, (47^2)*(3^14)||2633==102^2||2633 = 10404||2633 = 2505, yikes, that didn't help very much.
Fortunately it's not hard to square 2505. [2505^2 = 2500^2 + 25 + 25000=6250000+25025]
So, (47^4)*(3^28)||2633==2505^2||2633==6275025||2633==586 (have to do long division !)
We still need (47^25)*3||2633 to get the answer. Let's hack away at those 47's now...
47^3||2633 == 103823||2633 == 1136
So 47^4||2633==1136*47||2633==53392||2633==732
and 47^5||2633==732*47||2633==34404||2633==175, a nice, small number, hooray !
47^10||2633==175^2||2633 = 30625||2633==1662, bah !
So, that's all tha patience i've got, but you must get the idea by now. If you get a small remainder, you can square or cube; if it's large, you multiply by 47 and continue...

WOW, what a mindless calculation to do. I was curious to see if the result would be 1 or 0, or something interesting...but it wasn't. I hope someone doesn't follow up with a 3-step claculation that gives the answer !!!


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# A new trivial proof of Bertrand’s Postulate

Create a graph defined on all $(x,y)$ pairs for $x,y\in\mathbb Z$ as follows:

• $(x,y)$ is a coordinate pair referring to one vertex/node, all of which are laid out in a grid graph.
• Every vertex $(x,y)$ has an edge leading either north (up, $+y$) or west (left, $-x$).
• If $\gcd(x,y)=1$, that is, $x$ is coprime to $y$, the edge connects $(x,y)$ to $(x-1,y)$.
• If $\gcd(x,y)>1$, that is, $x$ shares a factor with $y$, the edge connects $(x,y)$ to $(x,y+1)$.
• The value of vertex $(x,y)$ will be $\gcd(x,y)$.

Here’s what you end up with:

An identical graph could be created by starting at $(0,0)$ and printing out $\mathbb{N}$
using coordinate steps of $(m,n)$ where $m,n \in\mathbb Z$ and $\gcd(m,n)=1$.

E.g. offsets of $(1,2)$ and $(2,1)$ make knight moves, counting out $1,2,3,\dots$ as they go.

This graph has a number of useful properties for our purposes:

Since every vertex has an edge leading north or west, you can start at any vertex and trace an infinite edge path heading somewhere between north and west, inclusive. Note this is not true in the other direction: edge paths heading south/east all terminate at some point.

If two adjacent vertices on the same row do not have an edge connecting them, then any SE paths they have will also never connect; they can never meet. This is not true of NW paths, which can and do merge often.

### Assertion 1

Let $(x,y)$ be any vertex that has an edge path heading south/east, and reaches the bottommost accessible row (we ignore the last two rows). Then that path must pass through vertices containing every prime smaller than $y$, in descending order (along with other interposing values).

#### Proof

For any prime-numbered row $p$, the only vertices reachable from above will occur at $(kp,p)$ for all $k\in\mathbb Z$. Moving south one row always implies you’re moving onto a vertex with $\gcd(x,y)>1$, as otherwise there would be no vertical edge to use, and $\gcd(kp,p)=p$ is the only non-$1$ value there can be on any row $p$. Since any edge path that travels unbroken from above a row to some point below it obviously passes through it, that also means the path passes through a vertex with value $p$ on row $p$, for all prime rows.

This works even better in reverse. In particular, since we can start at any vertex and be guaranteed an infinite path to the north/west, this means that any such path starting at some $(x,y)$ must pass through prime-valued vertices for all primes larger than $y$. Again, on prime rows, the prime-valued vertices function as choke points when heading in either direction.

So all $p_i$-valued vertices $(kp_i,p_i)$ will have a path north/west which reaches every row above $p_i$. But we’ll just consider the next prime-valued vertex/row, $(rp_{i+1},p_{i+1})$. This shows that there will always be a path from every prime-valued vertex on row $p_i$ to a prime-valued vertex on row $p_{i+1}$.

E.g. every $5$ on row $5$ will have a distinct NW path leading through one $7$ vertex on row $7$, one $11$ vertex on row $11$, and so forth. Note that in the opposite direction, a prime may have more than one path, each one leading to a different vertex, as with the $5$ splitting to reach two $3$s.

Let $p$ be any prime, and consider the NW path starting at $(2p,p)$.

Since $\gcd(2p,p)=p$, its edge goes north, thus $(2p-1,p)$ and points west are unreachable.

Similarly, since every $\gcd(n,n-1)=1$ for any $n$, there cannot be an edge heading north from $(2p,2p-1)$.

This means that the path originating from $(2p,p)$ is constrained such that it must pass through exactly one point of the diagonal interval from $(p,p)$ to $(2p,2p)$, exclusive. Since that point must be coprime to every prime smaller than $p$, it must be a prime larger than $p$. Specifically, it must be the next larger prime after $p$, since if there were another one in between, it would have intercepted the path instead.

If your path goes to the bottom, as it must since we’re coming from $(2p,p)$, you encounter all primes $p$ and smaller. This could not happen if one of those factors divided $(n,n)$, since you would be unable to reach that factor in your path.

### Assertion 2

If $(n,n)$ has a connected SE path of length $p$ to prime $p$, where $p<n$, then $p$ and $n$ are coprime.

#### Proof

If it takes $p$ edge steps moving south or east to reach a vertex with value $p$, what is that really saying?

The Manhattan distance between $(n,n)$ and your destination $p$ vertex is $p$. Let $a$ be the total steps to the east, your $x$-diff, and let $b$ be the steps to the south, your $y$-diff.

Then $a+b=p$. Also, $\gcd(n+a,n-b)=1$.

If $a+b=p$, then almost regardless of choice of $a,b$, we have $\gcd(a,b)=1$. The one exception to this is if $a=0$ and $b=p$ or vice versa. However, that maps to a configuration in which the prime $p$ is found at $(n+p,n)$ or $(n,n-p)$, respectively; for the same reasons explained in the last section, there can be no path from $(n,n)$ to either of them. This means $(n,n)$ would have no $p$-length path to that prime $p$, and it would also mean that $p \mid n$, i.e. that $\gcd(p,n)\geq p$.

There will be some legitimate situations where you cannot reach a given $p$ in $p$ steps. If there is a $1$ (i.e. empty) vertex at $(n+p,n)$, that is equivalent to having a path to $p$, meaning $\gcd(n,p)=1$.


text:
Find r if 5Pr = 6Pr-1

Asked by Abhisek | 1 year ago |  72

##### Solution :-

$$\dfrac{5!}{(5-r)!}= \dfrac{6!}{(6-r+1)}$$

$$\dfrac{5!}{(5-r)!}= \dfrac{6\times 5!}{(7-r)!}$$

$$\dfrac{5!}{(5-r)!}= \dfrac{6}{(7-r)(6-r)(5-r)!}$$

$$1= \dfrac{6}{(7-r)(6-r)}$$

= (7 - r)(6- r)=12

= 42-6r -7r +r2=12

= (7 - r)(6- r)=6

42-6r -7r +r =6

r2 -13r +36= 0

r2 -4r -9r(r - 4)=0

(r - 4)(r -9)=0

= r-4=0 or r = -9 = 0

= r =4 or r =9

It is known that

$$^6P_{r-1}=\dfrac{n!}{(n-r)!}$$'

where $$0\leq r\leq n$$

$$0\leq r\leq 5$$

r = 4

Answered by Abhisek | 1 year ago

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text:
# Vandermonde matrices with nodes in the unit disk and the large sieve

### BibTeX Reference

Warning: This BibTeX entry is automatically generated from the database. As there is no one-to-one match between the database fields and the BibTeX format, there may be errors in the entry. Regard the BibTeX feature as experimental for now. Use with care and correct errors and omissions manually.
 @article{vandermonde-jou,     author = {Aubel, Céline and Bölcskei, Helmut},     title = {Vandermonde matrices with nodes in the unit disk and the large sieve},     journal = {Applied and Computational Harmonic Analysis},     volume = 47,     misc = {Issue 1},     pages = {53--86},     month = jul,     year = 2019,     keywords = {Vandermonde matrices, extremal singular values, condition number, unit disk, large sieve, Hilbert's inequality},     url = {http://www.nari.ee.ethz.ch/pubs/p/vandermonde-jou} }
 \bibitem{vandermonde-jou} C. Aubel and H. Bölcskei, Vandermonde matrices with nodes in the unit disk and the large sieve,'' \emph{Applied and Computational Harmonic Analysis}, Vol. 47, Issue 1, pp. 53-86, July 2019.
 C. Aubel and H. Bölcskei, "Vandermonde matrices with nodes in the unit disk and the large sieve," Applied and Computational Harmonic Analysis, Vol. 47, Issue 1, pp. 53-86, July 2019.


text:
# Math Help - Determine Components of Metrix Tensor Relative to a Basis

1. ## Determine Components of Metrix Tensor Relative to a Basis

Please see attached picture. I completed question 2 successfully (I think), but am having trouble with question 3.

It seems question 2 came out nicely because there were some trig identities I could use to cancel out terms. However, this does not appear to be the case with 3. I got that the basis is not orthonormal, however, it seems like the components of the metric tensor will be huge long expressions. Is this the case, or am I missing a trig identity here?

Any help would be appreciated and thanks in advance!

2. ## Re: Determine Components of Metrix Tensor Relative to a Basis

In particular, I am wondering how to simplify the following expression:

[(a^2sinh(x)sinh(y)cosh(x)cosh(y))+(a^2cosh(x)cosh( y)sinh(x)sinh(y))]

Thanks guys!


text:
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text:
# Proving that polynomials with rational coefficients have integer roots

Obviously, polynomials with integer coefficients will satisfy P(x)$\in$ Z or every x $\in$ Z. But how do we prove that those with rational coefficients can produce integer roots? For instance, I have a polynomial that is in the form of a newton's generalized binomial theorem.

$\binom{X}{m} = \frac{X(X-1)(X-2)...(X-m+1)}{m!}$

If a polynomial is in the form of P(x) = $a_{0} + a_{1}\binom{X}{1} + a_{2}\binom{X}{2} + a_{d}\binom{X}{d}$, how do we prove that this can have P(x) $\in$ Z for all x $\in$ Z.

Firstly, it's not true that polynomials with rational coefficients have integer roots. For instance, $x^2 - 2$ has roots $\pm \sqrt 2$.
However, it is true that polynomials of the form $p(x) = a_0 + a_1 {X \choose 1} + \ldots + a_d {X \choose d}$, where $a_i \in \mathbb{Z}$, only take integer values. More generally, the monomial $a_k {X \choose k}$ only takes integer values when $X$ is an integer. This is equivalent to the denominator, which in this case is $k!$, dividing $X(X-1)(X-2)\dots(X-k+1)$ for any $X \in \mathbb{Z}$.
This last claim admits number theoretic and combinatorial proofs. Number theoretically, you might look at each prime in $k!$ and show that its multiplicity in the numerator is at least the multiplicity in the denominator. This is doable, but not particularly easy. Bill showed a completely arithmetic approach in his answer here.
Combinatorially, you simply notice that for integral $X$, the coefficient ${X \choose k}$ is the classical combination, and counts the number of ways of choosing $k$ things from $X$ choices, and is therefore an integer.
Regardless, as each $a_k {X \choose k}$ is an integer for each $X \in \mathbb{Z}$, you must have that polynomials formed from these monomomials also take integral values on the integers.