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# The Counter-Intuitive Nature of Memoryless Distributions

Our intuitions about probability aren’t always right

Consider flipping a coin. If you have flipped the coin 100 times and gotten 100 tails in a row, what is the probability that you will get a heads on the next flip? What if you have gotten 10,000 tails in a row first? 1,000,000? Obviously, the answer is still 50%. While the odds of getting into a particular state of coin flips varies (these examples would be highly unusual to see in reality), each coin flip event is wholly independent of those that have come before.

In statistics, a probability distribution which has this property is called memoryless, in that the distribution has no memory of previous states.

The formal definition is: given a random variable $X \sim \text{dist}$ where $\text{dist}$ is a memoryless distribution, $$\textbf{Pr}\left[X > s + t | X > s\right] = \textbf{Pr}\left[X > t\right]$$ That is to say, the probability that $X$ has a value that is $t$ larger than some $s$, given the knowledge that $X$ is already greater than $s$, is the same as the probability that $X$ is greater than $t$ alone. The knowledge that $X$ already exceeds some value is irrelevant to the probability.

Let’s give a more specific example: flipping a coin until we get a heads. Say that we have already flipped the coin $s = 100$ times with nary a heads in sight. What is the probability that we’ll flip $t = 2$ more times and still not get a heads? That’s easy enough to calculate, the probability of getting two tails in a row is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$. Notice that the fact that we have a history of 100 tails in a row doesn’t even factor into the equation? That’s because the repeated flipping of a coin follows a memoryless distribution (a geometric distribution, to be precise).

Compare this to a uniform distribution. Say that we are waiting in a line, and the amount of time that we need to wait is uniformly distributed on an interval of $0$ - $10$ minutes. This means that the average wait time will be around $\frac{10 - 0}{2} = 5$ minutes. Let’s say that we have already been waiting for 2 minutes. What is the expected amount of time that we have left to wait?

A uniform distribution is not memoryless, so our wait time will change given this new information. The expected wait time after already waiting for 2 minutes is 4 minutes (not the 3 minutes you may have expected), because $\frac{10 - 2}{2} = 4$. This makes perfect, intuitive sense. As we wait in line, the expected amount of additional waiting time decreases.

Where things get very strange is when you start using a memoryless distribution to model things like wait time. One common example of this is the exponential distribution, which intuitively represents a continuous analog to the geometric distribution that we saw above with respect to coin flips.

An exponential distribution is used to model a process where events occur continuously and independently at a constant rate. Specifically, it has a density function of $$f(t) = \lambda e^{- \lambda t}$$ where $\lambda$ represents the rate.

Skipping a lot of math, the average (or expected) value of a random variable following an exponential distribution is, $$\mathbf{E}[X] = \frac{1}{\lambda}$$

So, let’s rexamine our wait time example above, except this time we’ll use $W \sim \text{exp}(\frac{1}{5})$ as our random variable representing the wait time. This means that the average wait time is $$\mathbf{E}[W] = \frac{1}{\frac{1}{5}} = 5 \text{ minutes}$$

Let’s pose the same question again. Given that I have already waited for 2 minutes, what is the expected amount of time that I still must wait?

Why, 5 minutes of course! The distribution is memoryless. So the amount of time that we have left to wait is independent of the amount of time that we have already waited. Which is a very counter-intuitive result.

To understand why this is the case, we need to go back to the definition. We are considering events to occur at a constant rate, so we might imagine that the service that we are in line for finishes serving the current person within a consistent amount of time. Okay, that makes sense. However the sneaky thing is that in this exponential model the next person to be served is selected at random from all of the people waiting. On each event completion, a random person is selected to be served next.

This means that the probability that you are selected has nothing to do with how long you have been waiting. And so the average amount of time that you need to wait is strictly a function of how many other people are waiting and how frequently a random person is picked (the effects of which are combined together and modeled using $\lambda$ in our equation). As a result, your average remaining wait time is constant, irrespective of how long you’ve been waiting already. This is the same reasoning that we saw initially when considering the probability of getting a heads on a given coin flip.

Naturally, this is a very bad model for a queue of people waiting for some service, at the very least from a psychological standpoint, but it does crop up a lot in queueing theory because of how convenient it is to work with memoryless distributions (and a very helpful relationship between exponential distributions and Poisson Processes–but that’s a topic for another day).

Thus, it’s worth at least being aware of memoryless distributions and some of their seemingly odd properties. And, even more generally, it’s worth ensuring that you understand the properties of any probability distribution that you are using to model a problem–beyond simply knowing the right equations to use–because our intuitive understanding of probability often falls short, and can lead us to either incorrect answers, or at least result in us doing a lot more work than we really need to.


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# International Geomagnetic Reference Field

$V(r,\phi,\theta) = a \sum_{\ell=1}^L\sum_{m=-\ell}^\ell \left(\frac{a}{r}\right)^{\ell+1} \left(g_\ell^m\cos m\phi + h_\ell^m\sin m\phi\right) P_\ell^m\left(\cos\theta\right)$
where $r$ is radial distance from the Earth's center, $L$ is the maximum degree of the expansion, $\phi$ is East longitude, $\theta$ is colatitude (the polar angle), $a$ is the Earth's radius, $g_\ell^m$ and $h_\ell^m$ are Gauss coefficients, and $P_\ell^m\left(\cos\theta\right)$ are the Schmidt normalized associated Legendre functions of degree $l$ and order $m$.

format: plain


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Consistent hashing is a strategy most notably used by distributed databases for determining to which slot a key belongs. Its main advantage is that if a new slot needs to be added, only K/n objects need to be moved (K=number of all keys, n=the number of slots). And this means adding and removing slots is relatively inexpensive.

## In Practice

Let’s say you have implemented a very successful online telephone book. Your users enter a person’s name and get their telephone number. You already have millions of person-phone mappings and more are added every day. Your DB server will soon not be able to handle the number of requests, thus you decided to partition the database.

• First, you have to choose a key.

Read and write requests to the DB will always be done for a specific person’s name. This means your key should be calculated from a person’s name.

key = some_hash_function(person.name)

When choosing a hash function, it is good to choose one, which distributes hashes uniformly.

Let’s say that our some_hash_function creates integer values between 0 and 1023.

• Next, you need to choose the number of partitions or slots.

Let’s say we will have 4 partitions.

4 was chosen completely at random. Considering our example it would have been enough to split the DB in half instead of into quarters, but it is easier to illustrate consistent hashing with more slots.

• Now you need to map the slots into the same hash-space as the keys.

We need to give each of our 4 partitions a hash value and this hash value must also be between 0 and 1023 just like the results of some_hash_function.

We can use any method to define the hashes of our 4 partitions. Let’s say we have chosen the following values:

  Partition0: 100
Partition1: 400
Partition2: 700
Partition3: 900

• In the last step, we define the way the keys map to slots.

Here is what we have until now:

Our hash space extends from 0 to 1023. All the keys (=person’s names) and all the slots (=DB partitions) have their position in the same hash space.

To avoid having to deal with the start and the end, the hash space is usually drawn as a circle. Like this:

Every key belongs to the first slot to the right on the circle. In our case all keys 0 - 100 belong to Partition0, 101 - 400 to Partition1, 401 - 700 to Partition2, 701 - 900 to Partition3. And the keys, which are bigger than 900, where do they belong? Also to the next slot on the circle, which is Partiotion0.

## What have we gained

Adding and removing of servers is relatively inexpensive.

If we add a server at roughly hash 250, then only the 2 keys after P0 and before 250 need to be moved to the new server.

If we remove the server P3, then only its 1 key needs to be moved to P0.

Moving of keys is always done to the new server or from a removed server, never between existing servers.

We are also not storing any information in any kind of a global directory. Our partitions are completely independent of each other.

## Gotchas

### How to choose the hash function

Your hash function should distribute keys evenly over the whole hash space. But it should also be fast. There is no need to take a cryptographic hash function, as these are usually on the slower side.

Wikipedia holds a list of hash functions

### How to remediate an uneven distribution of keys

In the real world, the distribution of keys and slots is often not even with most keys stored in just a few slots, leaving other slots mostly empty. This is usually addressed with virtual slots. Each physical server is responsible for several virtual servers.

Each physical server gets multiple slots on the ring, these slots do not have to be together. This has 2 advantages:

• when a physical server fails, its keys are redistributed to many other slots, not just to its next neighbour
• with many slots, the number of keys per slot goes down and the distribution of keys among slots becomes more even

### How to remediate some physical servers having better performance than others

This problem as well is solved with virtual servers. The number of virtual slots for each physical server is determined based on the performance of the physical server.


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# Add current directory to UltiSnips search directory

I want to have some specific snippets for each file that I am editing. Like, I want to type "ppp" just for "proper point groups", or "iff" for "if and only if".

To be more specific, I want an additional unique set of snippets for each file I am editing. For example, when I write about group theory, I will want things like above to be shorthanded. But when I write about geometry, I will want "hom" stands for "\operatorname{\mathcal{Hom}}", etc.

So far I succeeded to add a absolute path to UltiSnips search directory, like the following in my VIMRC:

let g:UltiSnipsSnippetDirectories=['UltiSnips','C:\Users\xxx\Drafts\UltiSnips']

I tried to add an relative path like

let g:UltiSnipsSnippetDirectories=['UltiSnips','.\UltiSnips']

But UltiSnips just ignores it. Any ideas?

Or, can I add a modeline to get the current directory, and then set the variable g:UltiSnipsSnippetDirectories?

FYI.: I am on windows 10.

First of all :h UltiSnips-snippet-search-path tells us:

Snippet definition files are stored in snippet directories. A snippet directory must be a subdirectory of a directory defined in the 'runtimepath' option.

Which makes me think that you can not create snippets directories anywhere since that would mean constantly add your current path to runtimepath which doesn't sound like a good idea.

Thus I would advise the following solution: Use one directory containing your customs snippets, in this directory create snippets for "custom" filetype and add these files types to the files you want to edit (As both Vim and Ultisnips handle multiple filetypes that shouldn't cause a problem).

For example:

• Use your current snippets directory 'C:\Users\xxx\Drafts\UltiSnips'
• In it, put files named custom1.snippets containing the snippets you want to use in the files which will use the ppp snippets.
• Then you can use :UltiSnipsAddFiletypes (here is the doc) to select the snippets used in your file.

Edit (After the answer was accepted)

Another option could be the following: I didn't test it and it is based on my understanding of the doc so it may not work:

The same doc topic also says:

If you keep your snippets in a .vim subdirectory called "mycoolsnippets" and you want to make use of the default snippets that come with UltiSnips, add the following to your vimrc file.

let g:UltiSnipsSnippetDirectories=["UltiSnips", "mycoolsnippets"]

If you do not want to use the third party snippets that come with plugins, define the variable accordingly:

let g:UltiSnipsSnippetDirectories=["mycoolsnippets"]

You can also redefine the search path on a buffer by buffer basis by setting the variable b:UltiSnipsSnippetDirectories. This variable takes precedence over the global variable.

So maybe (emphasis on maybe) you could:

• Create several snippets directories project1, project2 at the same level as the UltiSnips directory.
• Each of these directories would contain differents snippets files for different filetypes in different projects.
• Then based on the project your file belongs to you could let b:UltiSnipsSnippetDirectories=["project1"]

I think that this way you wouldn't need to play with the filetypes nor the runtimepath You simply would need to modify b:UltiSnipsSnippetDirectories which should be done manually for each buffer because let is prohibited on modelines.

Note That as I said previously I didn't test that so I may be missing something.

• Er... proper point group is not a filetype, is just a phrase that I want to abbreviate using UltiSnips. Then, can I use modeline to get the current directory and set the value for g:UltiSnipsSnippetDirectories. How do I do that? – taper Nov 14 '16 at 11:33
• @taper From UltiSnips-snippet-search-path Snippet definition files are stored in snippet directories. A snippet directory must be a subdirectory of a directory defined in the 'runtimepath' option.. I don't think it is possible to use any project directory. That's why I suggested that you define filetypes grouping you custom snippets and that you set the desired filetype to the files you edit. – statox Nov 14 '16 at 12:13
• I get your point. That's a clever way to circumvent that restriction. Thanks! – taper Nov 14 '16 at 12:24
• @taper Nice, I'll edit my answer to make it clearer – statox Nov 14 '16 at 12:25
• let is prohibited in modeline. So, to make the second suggestion work one has to extend the built-in modeline by plugins like let-modeline, or use a local .vimrc file. Both are, actually, a bit troublesome. :P – taper Nov 14 '16 at 13:36


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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 deleted 1 characters in body

By means of the Hopf fibration $\pi: S^3 \rightarrow S^2$, U. Pinkall showed that every compact Riemannian surface of genus one can be conformally embedded in $S^3$.\

More precisely: Let $p$ be a closed curve on $S^2$ of length $L$. Lifting it to $S^3$ yields a torus isometric to $R^2 / \Gamma$, with $\Gamma$ generated by $(2\pi, 0)$ and $(A/2, L/2)$, where $A$ is the area enclosed by $p$. (This is Proposition 1 in Pinkall's paper.) \

Now it's claimed that if you lift a great-circle you should get the Clifford-torus. The above proposition then yields that the Clifford-torus is isometric to $R^2 / \Gamma _1$, with $\Gamma _1$ generated by $(2\pi, 0)$ and $(\pi, \pi)$. The usual definition of the Clifford torus is $R^2 / \Gamma _c$ with $\Gamma _c$ generated by $(2\pi, 0 )$ and $(0, 2 \pi)$. \

Who know's how this fits together?

1

# Hopf Tori in $S^3$

By means of the Hopf fibration $\pi: S^3 \rightarrow S^2$, U. Pinkall showed that every compact Riemannian surface of genus one can be conformally embedded in $S^3$. \ More precisely: Let $p$ be a closed curve on $S^2$ of length $L$. Lifting it to $S^3$ yields a torus isometric to $R^2 / \Gamma$, with $\Gamma$ generated by $(2\pi, 0)$ and $(A/2, L/2)$, where $A$ is the area enclosed by $p$. (This is Proposition 1 in Pinkall's paper.) \ Now it's claimed that if you lift a great-circle you should get the Clifford-torus. The above proposition then yields that the Clifford-torus is isometric to $R^2 / \Gamma _1$, with $\Gamma _1$ generated by $(2\pi, 0)$ and $(\pi, \pi)$. The usual definition of the Clifford torus is $R^2 / \Gamma _c$ with $\Gamma _c$ generated by $(2\pi, 0 )$ and $(0, 2 \pi)$. \ Who know's how this fits together?


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Question

# If a,b and c represent the lengths of sides of a triangle, then the possible integral value of ab+c+bc+a+ca+b is

Open in App
Solution

## We know, 0<a<b+c⇒ab+c<1(Triangle Inequality) ∴ab+c+bc+a+ca+b<3 In the case of a=b=c, we have ab+c=12 ⇒32≤ab+c+bc+a+ca+b<3 ∴ Integral values of ab+c+bc+a+ca+b=2

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# Timbab

Member Since 17 Jun 2012
Offline Last Active Jul 03 2012 12:52 PM

### In Topic: .vsh / .psh - HLSL/ASM Shaders - SWGalaxies - Totally lost

22 June 2012 - 07:09 AM

Hey, thanks for the replies. I was gonna mention that pdf, but you found it already.

Good that you mention the compiler, I do have it, two actually, a new version (9 something) and one the one from 2004 (Used in some of the newer shaders in the game), and I can compile the files, but I don't get the files I want, for example if I copy the exact code clean from the 2d_blur.psh with out the added compiler bytes and compile, I get this (It looks screwed, cause it's in hex):

I'll attach 2 pics so you can see it in hex.
ÿÿþÿ-CTABÿÿxDHX"hs0«userConstants««ps_2_0Microsoft (R) HLSL Shader Compiler 9.29.952.3111Q @@@€@€° B€ä°ä €ä°
ä €
ä €ä€ ä°€ä€U ä°€ä€ª ä°B€ä€ä B€ä€ä B€ä€ä B€ä€ä €ä€	ä €ä€ä ä€€ä€
ä ä€€ä€ä ä€€ä€ä ä€€ä€ÿÿ
http://img854.images...05/62536327.png

FORM?PSHPFORM?0000PSRC?//hlsl ps_2_0

#include "pixel_program/include/functions.inc"

sampler s0 : register(s0);

float4 main
(
in float2 textureCoordinates  : TEXCOORD0
)
: COLOR
{
float4 result = 0.0f;
float2 textureCoordinateOffsets = 0.0f;
for (int i = 0; i < 16; ++i, textureCoordinateOffsets += userConstants[16].xy)
result += tex2D(s0, textureCoordinates + textureCoordinateOffsets) * userConstants[i];

return result;
}
PEXE?????-CTAB??xDHXhs0?userConstants??ps_2_0Microsoft (R) D3DX9 Shader Compiler 4.09.00.1126Q?ApAQ? A0A@APAQ??@?@AAQ?@@@?@?@????	?????????????????U??????????????????????????U???????????????????????B
?	????B	?????B?????B?????B?????B?????B?????B?????B?????B?????B?????
?
??	??	?	????
?????
??	???????????????????
??????????????????????????????????????????U??????????????????????????U???B?????B?????B?????B?????B?????????????????????????????????????????????
http://img9.imagesha...19/12675741.png

In cmd I'd open up fxc.exe and use this command: fxc /T ps_2_0 /Fo 2d_blur.psh 2d_blur.cpp, does anyone know what I'm doing wrong? I might have to add the form in manually, but regardless, the code won't even show up.

The .vsh I can edit as far as I know, fully, as it compiles on the fly. I've already tried some stuff with minor successes, but nothing major yet, I'm sure I could and can modify the .psh files once I can compile it in the right way. But I also talked to someone who manually managed to get new .psh files working ages ago, but he can't remember correctly how he did it, so it IS possible.

@Reedbeta, what do you limit to 'only shaders'? Right now I can edit their functions.inc, it also has 'module' .inc files, and well also the .psh(Soon hopefully) and vsh files. Outside of that, I found out I might be able to make multi passes work by editing the .eft (Effect files that link to the psh/vsh), I do know some things in game, like water, use multiple passes. I've attached all the shaders in the .zip above though, in case you want to check out what I mean.

Couldn't I use the player position/character as a focal point, somehow?

For instance, I found that part of the fog distance to the player that's used for the view distance (Though I found out too that in the end I can't modify it, at least not the normal fog for it), is calculated with this:

float calculateFog(float4 vertexPosition_o)
{
float4 position_w = mul(vertexPosition_o, objectWorldMatrix);
float3 viewer_w = cameraPosition_w - position_w;
float  viewerDistanceSquared = lengthSquared(viewer_w);
return 1.0 / exp(viewerDistanceSquared * fog.w);
}`

Couldn't you do something similar with a blur/DoF like effect? It doesn't have to be 100% DoF, a faux one would do too.

Look at this for example, a modder achieved the following a few years ago, he never said how he did it exactly, he just said this:

"It also allows for camera blurring when focal on "centric" objects. "

"Nothing. Didn't touch sandstorms or fog. The only exception is that the draw distance now has a parameter in the Vertex/Pixel shader files that can force Depth of Field on some cards - but that won't affect sandstorms or natural in-game fog."

http://forum.modsour...ttach=925;image
http://forum.modsour...ttach=929;image

Just as a reference, here is the original game with view distances 0%, 50% and 100%.
http://img528.images...eenshot0447.jpg
http://img526.images...eenshot0448.jpg
http://img194.images...eenshot0449.jpg

Any ideas would be fantastic, thanks in advance!

Edit: Got the compiling and importing fixed now.

Curious, is it hard converting .fx files into .psh and .vsh? I've been trying some stuff, but no real luck so far.


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Precalculus (10th Edition)

$f(x)= (x+1)^2+2$
If the vertex of a graph is at $(h, k)$ then the vertex form for the quadratic function is $f(x)=a(x-h)^2+k$. The vertex is at $(-1,2)$. hence, the tentative equation of the function is: $f(x)=a[x-(-1)]^2+2\\ f(x)=a(x+1)^2+2.$ $(1,6)$ is on the graph so its $x$ and $y$ coordinates satisfy the equation of the quadratic function. Hence plugging in the values of $x$ and $y$ of this point into the tentative equation gives $6=a\cdot(1+1)^2+2\\ 6=a(4)+2\\ 4=a(4)\\ 1=a$ Therefore, $f(x)=(x+1)^2+2$.


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# Math Help - Calculus Trouble

1. ## Calculus Trouble

I am currently having trouble with these calculus practice problems. I would appreciate any help in solving these. Thank you.

1. Find the volume of a solid whose base is enclosed by the circle x^2 + y^2=1, and whose cross sections taken perpendicular to the base are equilateral triangles with bases in the xy-plane.

2. Let R be the region enclosed by the graphs of y=ln(x^2 + 1) and y=cos(x). Setup, using integrals, but do not evaluate the following:
a. The area of R.
b. The base of a solid is region R. Each cross section of the solid perpendicular to the x-axis is an isosceles right triangle. Express the volume of the solid.

2. Originally Posted by babolat29
I am currently having trouble with these calculus practice problems. I would appreciate any help in solving these. Thank you.

1. Find the volume of a solid whose base is enclosed by the circle x^2 + y^2=1, and whose cross sections taken perpendicular to the base are equilateral triangles with bases in the xy-plane.
The volume V of a solid between a and b is the integral of its cross-sectional area A(x) from a to b, or $\int_{a}^{b} A(x) dx$
The area A of an equilateral triangle is $\frac{\sqrt{3}s^2}{4}$ where $s$ is the length of a side. Since the trianglular cross-sections have their base in a circle with radius 1, the side of the triangular cross-section at any x is twice the distance from the x-axis to a semi-circle of radius 1, or $s(x) = 2\sqrt{1 - x^2}$. Therefore $A(x) = \frac{\sqrt{3}}{4}(2\sqrt{1 - x^2})^2 = \frac{\sqrt{3}}{4}4(1-x^2) = \sqrt{3}(1-x^2).$
Since your solid is enclosed in a circle with radius 1, you're going to integrate the area from -1 to 1, or $\int_{-1}^{1} A(x) dx = \int_{-1}^{1}\sqrt{3}(1-x^2)dx$ which you should know how to do.

3. For #2 See attachment


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1. ## Complex Number problem

Question: Express $z = \ \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}$ in the form x+iy where x and y are real numbers

Solution :
$z = \ \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}$

= $\frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}$ = ................= $\frac{\sqrt{104}i - \sqrt{72}}{24}$......Is this correct???

2. To do this problem you need to expand the denominator to the form $x+yi$ after this you need to mulitply the both numerator and denominator by the complex conjugate $x-yi$ . Is this what you did?

3. ## This is what i have done?

Originally Posted by pickslides
To do this problem you need to expand the denominator to the form $x+yi$ after this you need to mulitply the both numerator and denominator by the complex conjugate $x-yi$ . Is this what you did?
$\frac{\sqrt{2} + \sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}$

= $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}$

= $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{16}i + \sqrt{6}}$

= $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{12}- \sqrt{16}i}$

= $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{12}- \sqrt{16}i} \times \frac{\sqrt{12}+ \sqrt{16}i}{\sqrt{12}+ \sqrt{16}i}$

= $\frac{\sqrt{24} + \sqrt{32}i + \sqrt{72}i + \sqrt{96}i^2}{12 + \sqrt{192}i - \sqrt{192}i - 16 i^2}$

= $\frac{\sqrt{24} + \sqrt{104}i - \sqrt{96}}{12 + 16}$

= $\frac{ \sqrt{104}i - \sqrt{72}}{28}$.............Could any body please tell me what i have done wrong and how should i rectify it ???

4. $\frac{\sqrt{2} + \sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}$

= $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}$

= $\frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i}$

= $\frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i} \times \frac{2\sqrt{6} + \sqrt{8}i}{2\sqrt{6} + \sqrt{8}i}$

. . .

5. Originally Posted by zorro
$$= $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}$ = $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{16}i + \sqrt{6}}$ $- \sqrt{18}i + \sqrt{2}i \neq - \sqrt{16}i$ $- \sqrt{18}i + \sqrt{2}i = - \sqrt{9\times 2}i + \sqrt{2}i =- 3\sqrt{ 2}i + \sqrt{2}i =-2\sqrt{ 2}i$ also $\sqrt{6}+\sqrt{6} = 2\sqrt{6}\neq \sqrt{12}$ 6. Originally Posted by dedust = $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}$ = $\frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i}$$$

. . .
This can be simplified further $- \sqrt{8}i = - 2\sqrt{2}i$

7. ## Thanks Pickslide and Mark Goodstein

Thanks u for helping me
pickslides and Mark Goodstein

cheers

8. ## Is this correct?

Originally Posted by zorro
Thanks u for helping me
pickslides and Mark Goodstein

cheers

The result which i am getting is $- \frac{1}{2}$

9. ## Is the answer correct ?

Originally Posted by pickslides
This can be simplified further $- \sqrt{8}i = - 2\sqrt{2}i$


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# context.divkbaseb(k, b)¶

Generates an automaton that recognises the writing in base $b$ of the numbers divisible by the integer $k$.

Preconditions:

• $2 \le b$
• the labelset has at least $b$ generators

Postconditions:

• the automaton has $k$ states
• Result = quotkbaseb(k, b).project(0)

## Examples¶

In [1]:
import vcsn
c = vcsn.context('lal_char(0-9), b')
c.divkbaseb(3, 2)
Out[1]:
In [2]:
c.divkbaseb(2, 4)
Out[2]:

If $k$ and $b$ are coprime, then the result is known to be minimal.

In [3]:
div4base4 = c.divkbaseb(4, 4)
div4base4
Out[3]:
In [4]:
div4base4.minimize()
Out[4]:


text: Journal Home Page Cumulative Index List of all Volumes Complete Contentsof this Volume Previous Article Journal of Convex Analysis 27 (2020), No. 1, 179--204Copyright Heldermann Verlag 2020 Higher Order Problems in the Calculus of Variations: Du Bois-Reymond Condition and Regularity of Minimizers Julien Bernis Laboratoire de Mathématiques, Université de Brest, 29200 Brest, France julien.bernis@univ-brest.fr Piernicola Bettiol Laboratoire de Mathématiques, Université de Brest, 29200 Brest, France piernicola.bettiol@univ-brest.fr Carlo Mariconda Dip. di Matematica, Università di Padova, 35121 Padova, Italy carlo.mariconda@unipd.it [Abstract-pdf] This paper concerns an $N$-order problem in the calculus of variations of minimizing the functional $\smash{\int_{a}^{b}{\Lambda(t,x(t),\ldots,x^{(N)}(t))\mathrm{d}t}}$, in which the Lagrangian $\Lambda$ is a Borel measurable, non autonomous, and possibly extended valued function. Imposing some additional assumptions on the Lagrangian, such as an integrable boundedness of the partial proximal subgradients (up to the ($N\!-\!2$)-order variable), a growth condition (more general than superlinearity w.r.t. the last variable) and, when the Lagrangian is extended valued, the lower semicontinuity, we prove that the $N$-th derivative of a reference minimizer is essentially bounded. We also provide necessary optimality conditions in the Euler-Lagrange form and, for the first time for higher order problems, in the Erdmann-Du Bois-Reymond form. The latter can be also expressed in terms of a (generalized) convex subdifferential, and is valid even without requiring neither a particular growth condition nor convexity in any variable. Keywords: Calculus of variations, minimizer regularity, higher order problems, necessary conditions, Weierstrass inequality, Erdmann-Du Bois-Reymond condition. MSC: 49N60, 49K15. [ Fulltext-pdf  (688  KB)] for subscribers only.


text:
### Still have math questions?

Algebra
Question

(a) Find the $$x$$ -intercepts of the function $$y = x ^ { 2 } - 5 x - 14 .$$

(You may use either factoring or the quadratic formula.) (b) By putting $$x = 0$$ , find the $$y$$ -intercept. (c) Solve $$\frac { d y } { d x } = 0$$ and hence find the coordinates of the stationary point. (d) By examining the sign of $$\frac { d ^ { 2 } y } { d x ^ { 2 } }$$ , establish the nature of the stationary point. (e) Sketch a graph of the function, indicating all important features.


text:
# Probability Worksheets - Page 5

Probability Worksheets
• Page 5
41.
Find the odds in favor of getting exactly 1 head when three coins are tossed.
 a. 2 : 1 b. 5 : 3 c. 3 : 5 d. 1 : 2

#### Solution:

Sample space, S = {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

The odds in favor of getting exactly 1 head are 3 : 5
[Three outcomes are successes and 5 are failures.]

42.
A box contains 20 cards labeled 1 through 20. One card is drawn from it at random. Identify the event of drawing a card of an odd number.
 a. {11, 13, 15, 17, 19} b. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} c. {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} d. {1, 3, 5, 7, 9}

#### Solution:

Each outcome is a number of a card.

Let A be the event of drawing a card of an odd number.

A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

The event of drawing a card of an odd number is {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}.

43.
A box contains 20 cards labeled 1 through 20 on them. One card is drawn from it at random. Write the event of drawing a multiple of 5.
 a. {5, 10, 15, 20} b. {5, 10} c. {5, 15} d. {1, 5, 10, 20}

#### Solution:

Each outcome is a number on a card.

Let B be the event of drawing a card whose number is a multiple of 5.

B = {5, 10, 15, 20}

The event of drawing a card whose number is a multiple of 5 is B = {5, 10, 15, 20}

44.
Fifteen cards labeled A through O are placed in a bag. The bag is shaken and one card is drawn at random. Write the event of drawing a consonant.
 a. {A, B, C, D, E, F, G, H, I, J} b. {B, C, D, F, G, H, J, K, L, M, N} c. {B, C, D, F, G, H, I, J, K, L, M} d. {A, E, I, O}

#### Solution:

Let Q be the event of drawing a consonant.

Q = {B, C, D, F, G, H, J, K, L, M, N}

45.
What is the probability of getting a red card from a 52-card deck?
 a. $\frac{1}{26}$ b. $\frac{1}{2}$ c. $\frac{1}{13}$ d. $\frac{1}{4}$

#### Solution:

P(getting a red card) = 26 / 52 = 1 / 2
[There are 26 red cards.]

46.
If two dice are rolled, then what is the probability that the sum of the dots will be 8?
 a. $\frac{5}{36}$ b. $\frac{1}{9}$ c. $\frac{1}{6}$ d. $\frac{5}{6}$

#### Solution:

The number of elements in the sample space, S = 6 × 6 = 36.

Let E be the event of getting the sum of dots is 8.

E = {(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)}

The number of outcomes in E = 5

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 5 / 36

47.
Four cards are drawn from a bridge deck. What is the probability that all are diamonds?
 a. $\frac{4}{13}$ b. $\frac{11}{4165}$ c. $\frac{1}{4}$ d. $\frac{1}{13}$

#### Solution:

The sample space, S is the set of all possible 4 card hands in a 52-card deck.

The number of outcomes in S is 52C4 = 52! / 4!48! = 270725

Let E be the event of getting 4 diamonds from 13 in the deck.

The number of outcomes in E is 13C4 = 13! / 4!9! = 715

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 715 / 270725 = 11 / 4165

48.
Two dice are rolled. What is the probability that the sum of the dots showing will be less than 8?
 a. $\frac{1}{2}$ b. $\frac{7}{12}$ c. $\frac{13}{18}$ d. $\frac{5}{9}$

#### Solution:

The number of elements in sample space is 6 × 6 = 36.

Let E be the event of getting a sum less than 8 on the two dice.

E = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1)}

Number of outcomes in E = 21

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 21 / 36 = 7 / 12

49.
There are 12 boys and 10 girls. If 7 of them are selected at random, what is the probability that 4 are boys and 3 are girls?
 a. $\frac{{}_{7}{C}_{3}}{{}_{12}{C}_{4}}$ b. $\frac{{}_{12}{C}_{7}}{{}_{22}{C}_{7}}$ c. $\frac{\left({}_{12}{C}_{4}\right)\left({}_{10}{C}_{3}\right)}{{}_{22}{C}_{7}}$ d. $\frac{7}{22}$

#### Solution:

Number of boys = 12

Number of girls = 10

Total number of students = 22

The number of ways of selecting 7 from 22 is 22C7.

So, the number of outcomes in sample space is 22C7.

Let E be the event of selecting 4 boys from 12 and 3 girls from 10.

Number of outcomes in E = 12C4 × 10C3

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= (12C4)(10C3)22C7

50.
Two dice are rolled. What is the probability that both the dice show the same number of dots?
 a. $\frac{1}{6}$ b. 1 c. $\frac{1}{2}$ d. $\frac{1}{3}$

#### Solution:

The number of elements in sample space = 6 × 6 = 36

Let E be the event of getting the same number of dots on both the dice.

E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Number of outcomes in E = 6

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 6 / 36 = 1 / 6


text:
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Lesson 3: Decompose figures to find volume

# Decompose figures to find volume practice

Work through volume of irregular 3D figures problems that get increasingly more challenging.  The first items will have unit cubes.  Then, try some with only side-length labeled.

## Decompose figures with unit cubes

Problem 1A
Unit cube:
A unit cube is shown.
What is the volume of the following figure?
Front ViewBack View
unit cubes

Problem 1B
Unit cube:
A unit cube is shown.
What is the volume of the following figure?
Front ViewBack View
unit cubes

Want to practice more problems like this? Check out this exercise.

## Decompose figures with side-length labels

Problem 2B
The figure below is made of 2 rectangular prisms.
What is the volume of this figure?
A shape made up of two rectangular prisms. The first rectangular prism has a base that measures 2 inches length by 6 inches width, and has a height of 5 inches. The second rectangular prism sits behind the first rectangular prism. It has a base that measures 8 inches length by 1 inch width and has a height of 5 inches.
start text, c, u, b, i, c, space, i, n, end text

Problem 2A
The figure below is made of 2 rectangular prisms.
What is the volume of this figure?
A figure made up of 2 rectangular prisms. The first rectangular prism has the dimensions 10 centimeters length by 6 centimeters width by 3 centimeters height. The second rectangular prism has the dimensions 4 centimeters length by 6 centimeters width by 4 centimeters height.
start text, c, u, b, i, c, space, c, m, end text

Want to practice more problems like this? Check out this exercise.

## Want to join the conversation?

• I watched the videos for all of these but I still don't get problems 2B and 2A.
• No, it's easy it's just that you have to split them apart and then find the height for the first one.
(1 vote)
• This was not as easy as the two videos above it make it seem.
• I totally agree. In the videos he does all the special colors to draw and it kinda distracts you from the actual learning of it. In the videos it looks very simple and easy, but when you are actually doing it ends up being confusing and you have many questions. At least that's how it is for me. If I was to make the videos I would try to focus from the point of view of someone who does not get the math instead of from the point of view of someone who does get it showing you it. I think that the hints are more helpful because they are made for someone who is having a little trouble with the math. The bad part is though that if you use a hint that helps you it automatically says you got it wrong, but if you watch a video that makes you more confused you get energy points for watching it.
• Is there a faster way to do the section with decomposing figures with unit cubes?
• Yes there is a faster way, if the rectangular box-shaped pieces have a common dimension (that is, the same length, same width, or same height). For example, if the pieces have the same height, you could find the area of the base of the whole figure, by adding up areas of the rectangular bases of the pieces. Then you could multiply the area of the base of the entire figure by the height to finish the problem.
• Can you all start easy and if we is gitting it wright,it can start gitting harder?
• I watched both videos and pay attention and still don't get 2A and 2B.
• Basically, you have to split the shape into two rectangular prisms, then find the volume for each and solve.
• honestly all of these make no sence to me and im a 5th grader


text:
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# Is x odd?

Author Message
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Kudos [?]: 31 [0], given: 3

Is x odd? [#permalink]  15 Jul 2012, 03:30
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Difficulty:

95% (hard)

Question Stats:

19% (01:51) correct 81% (00:40) wrong based on 62 sessions
Is x odd?

(1) 2x - 1 is odd
(2) x^3 is odd
[Reveal] Spoiler: OA

Last edited by Bunuel on 15 Jul 2012, 03:41, edited 1 time in total.
Edited the question.
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Re: Is x odd? [#permalink]  15 Jul 2012, 03:51
2
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Expert's post
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This post was
BOOKMARKED
Is x odd?

(1) 2x - 1 is odd --> $$2x-1=odd$$ --> $$2x=even$$ --> $$x=\frac{even}{2}$$ --> $$x$$ is an integer. Now, $$x=\frac{even}{2}$$ could be even (consider $$x=\frac{even}{2}=\frac{4}{2}=2=even$$) as well as odd (consider $$x=\frac{even}{2}=\frac{2}{2}=1=odd$$). Not sufficient.

(2) x^3 is odd. If $$x=integer$$ then in order $$x^3=odd$$ to hold true, it must be odd (answer YES), but $$x$$ could also be some irrational number, for example $$x=\sqrt[3]{5}$$ (answer NO). Not sufficient.

(1)+(2) Since from (1) $$x=integer$$ then from (2) we have that $$x=odd$$. Sufficient.

Hope it's clear.
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Re: Is x odd? [#permalink]  16 Jul 2012, 10:18
@Bunuel:

I don't understand your approach for the first argument:

$$2x-1=odd$$

If the result muss be odd, so x must be even. It will be even if x is 2 or greater than 2. If x is odd the result won't be odd. If I say x = 1 so the result will be Zero. But Zero is neither even nor odd. So the satetement is sufficient ?
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Re: Is x odd? [#permalink]  16 Jul 2012, 12:06
Expert's post
Alexmsi wrote:
@Bunuel:

I don't understand your approach for the first argument:

$$2x-1=odd$$

If the result muss be odd, so x must be even. It will be even if x is 2 or greater than 2. If x is odd the result won't be odd. If I say x = 1 so the result will be Zero. But Zero is neither even nor odd. So the satetement is sufficient ?

Several things:

1. $$2x-1=odd$$ --> $$2x=odd+1=odd+odd=even$$ --> so $$2x=even$$ --> $$x=\frac{even}{2}=integer$$. Hence $$2x-1=odd$$ just means that $$x$$ is an integer (it can be even as well as odd).

2. If $$x=1$$the result wont be zero, it'l be 1, so odd: $$2*1-1=2-1=1=odd$$.

3. Zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form $$n=2k$$, where $$k$$ is an integer. So for $$k=0$$ --> $$n=2*0=0$$.

Hope it's clear.
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Re: Is x odd? [#permalink]  05 Sep 2015, 04:35
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# Rational equivalence hyperbola and union of two lines in $\mathbb{P}^2$

We say that two subschemes are rationally equivalent if their associated cycles are rationally equivalent. Two cycles are rationally equivalent if their difference is in $Rat(X)$. If $X$ is a scheme, we define $Z(X)$ the group generated by the reduced irreducible subschemes of $X$. I denote with $Rat(X) \subset Z(X)$ the subgroup generated by differences of the form $$\langle \Phi \cap (\{t_0\} \times X) \rangle - \langle \Phi \cap (\{t_1\} \times X) \rangle ,$$ where $t_0,t_1 \in \mathbb{P}^1$ and $\Phi$ is a subvariety of $\mathbb{P}^1 \times X$ not contained in any fiber ${t} \times X$ (I took this definition from Intersection theory in algebraic geometry of Harris). Now I have to prove rigorously that in $\mathbb{P}^2$ an hyperbola and the union of two lines are rationally equivalent. How can I do it? Thanks in advance.

• This is the same question as that one. Reposting a question under another username is bad étiquette. – Georges Elencwajg Dec 2 '13 at 12:49

Hint: Take the family of hyperbolae (I checked the plural of hyperbola and apparently hyperbolas works as well) $$p(t)=x^2-y^2-tz^2=0$$ for $t\in\mathbb{A}^1$. For $t\neq 0$, $p(t)$ is an actual hyperbola, but for $t=0$ we get the lines $x=y$ and $x=-y$.


text:
## Friday, March 13, 2009

### Fibonacci, a mathematician from Pisa

Leonardo Pisano (Leonardo Fibonacci, or simply Fibonacci meaning "filius nacci", the son of Nacci) was born in Pisa, Italy in about 1170 (the place where I live). He is well known for his Fibonacci Numbers.

Can you compute the Fibonacci numbers in optimal time?

#### 1 comment:

1. We can write the Fibonacci formula in matrix form

x(i+1) | 1 1 | x(i)
= | |
x(i) | 1 0 | x(i-1)

or y(i) = F y(i-1) and y(0) = [0 1], or simply
y(i) = F^n y(0). We must efficently calculate F^n.
This can be calculated using the Caley-Hamilton Theorem (A matrix is root of its characteristic polynomial). We obtain p(a) = -a(1-a) -1 and then F^2-F-I = 0 ==> F^2 = F+I. The divide et impera.


text:
The best way to prepare for your 5th Grade MAP Math test is to work through as many 5th Grade MAP Math practice questions as possible. Here are the top 10 5th Grade MAP Math practice questions to help you review the most important 5th Grade MAP Math concepts. These 5th Grade MAP Math practice questions are designed to cover mathematics concepts and topics that are found on the actual test. The questions have been fully updated to reflect the latest 2021 5th Grade MAP guidelines. Answers and full explanations are provided at the end of the post.

Start your 5th Grade MAP Math test prep journey right now with these sample 5th Grade MAP Math questions.

5th Grade MAP Math Practice Questions

1- In a bag, there are 65 cards. Of these cards, 13 cards are white. What fraction of the cards are white?

A. $$\frac{1}{5}$$

B. $$\frac{2}{7}$$

C. $$\frac{13}{52}$$

D. $$\frac{1}{4}$$

2- Which expression is equal to $$\frac{3}{11}$$ ?

A. $$3 – 11$$

B. $$3 ÷ 11$$

C. $$3 × 11$$

D. $$\frac{11}{3}$$

3- If $$A = 20$$, then which of the following equations are correct?

A. $$A + 20 = 40$$

B. $$A ÷ 20 = 40$$

C. $$20 × A = 40$$

D. $$A – 20 = 40$$

4- How long does a 425 miles trip take moving at 50 miles per hour (mph)?

A. 4 hours

B. 6 hours and 24 minutes

C. 8 hours and 24 minutes

D. 8 hours and 30 minutes

5- 50 is What percent of 40?

A. $$20 \%$$

B. $$25 \%$$

C. $$125 \%$$

D. $$150 \%$$

6- The perimeter of the trapezoid below is 52. What is its area? _________

7- The area of a circle is $$64π$$. What is the circumference of the circle?

A. $$8 π$$

B. $$16 π$$

C. $$32 π$$

D. $$64 π$$

8- A rope weighs 800 grams per meter of length. What is the weight in kilograms of 12.2 meters of this rope? (1 kilograms = 1000 grams)

A. 0.0976

B. 0.976

C. 9.76

D. 9760

9- Solve.
$$\frac{1}{2} + \frac{4}{5} – \frac{3}{10} = ?$$

A. $$\frac{9}{10}$$

B. $$\frac{2}{10}$$

C. 1

D. 2

10- How many $$\frac{1}{4}$$ cup servings are in a package of cheese that contains $$6 \frac{1}{2}$$ cups altogether?_________

Best 5th Grade MAP Math Prep Resource for 2021

1- A
There are 65 cards in the bag and 13 of them are white. Then, 13 out of 65 cards are white. You can write this as: $$\frac{13}{65}$$.
To simplify this fraction, divide both numerator and denominator by 13. Then:
$$\frac{13}{65}=\frac{1}{5}$$

2- B
$$\frac{3}{11}$$ means 3 is divided by 11. The fraction line simply means division or $$÷$$. Therefore, we can write $$\frac{3}{11}$$ as $$3 ÷ 11$$.

3- A
Plug in 20 for A in the equations. Only option A works.
$$A+20=40$$
$$20+20=40$$

4- D
50 miles : 1 hour
425 miles : $$425 ÷ 50 = 8.5$$ hours

5- C
Use percent formula:
$$part =\frac{percent}{100}× whole$$
$$50=\frac{percent}{100}× 40⇒ 50 = \frac{percent ×40}{100}⇒ 50 = \frac{percent ×4}{10}$$
multiply both sides by 10.
500 = percent $$×4$$, divide both sides by 4.
125 = percent

6- 104
First, find the missing side of the trapezoid. The perimeter of the trapezoid below is 52. Therefore, the missing side of the trapezoid (its height) is: $$52-12-18-14=52-44=8$$
Area of a trapezoid: $$A = \frac{1}{2} h (b_{1} + b_{2}) = \frac{1}{2} (8) (12 + 14) = 104$$

7- B
Use area and circumference of circle formula.
Area of a circle = $$πr^2 ⇒ 64π = πr^2 ⇒ r= 8$$
Circumference of a circle = $$2πr ⇒ C = 2×8×π ⇒ C =16π$$

8- C
1 meter of the rope = 800 grams 12.2 meter of the rope = $$12.2 × 800 =9,760$$grams = 9.76 kilograms

9- C
$$\frac{1}{2} + \frac{4}{5} – \frac{3}{10} = \frac{(5×1)+(2×4)-(1×3)}{10} =\frac{10}{10}=1$$

10- 26
To solve this problem, divide $$6\frac{1}{2}$$ by $$\frac{1}{4}$$.
$$6\frac{1}{2}÷\frac{1}{4}=\frac{13}{2}÷\frac{1}{4}=\frac{13}{2}×\frac{4}{1}=26$$


text:
# Fast Fourier Transform for Graph Laplacian?

In the case of a regularly-sampled scalar-valued signal $f$ on the real line, we can construct a discrete linear operator $A$ such that $A(f)$ approximates $\partial^2 f / \partial x^2$. One way to interpret this operator is via spectral decomposition of the corresponding matrix:

$$A = UVU^T.$$

If our operator $A$ has spectral accuracy, then $U^T$ is precisely the discrete Fourier transform matrix. Hence, we could compute the Fourier transform of $x$ by computing all the eigenvectors of $A$ and sticking them in the columns of $U$. Of course, we all know there's a quicker way to do it: use the fast Fourier transform (FFT).

What about in a more general setting? In particular, consider the graph Laplacian $L=UVU^T$ which for a weighted, undirected graph on $n$ vertices is an $n \times n$ matrix with the weights of incident vertices on the off-diagonal and (the additive inverse of) total incident weight on the diagonal.

Question:

Can we transform a signal on vertices into frequency space without computing the entire spectrum of $L$ (using something like the FFT)?

In particular I'm interested in the case where $L$ approximates the Laplace-Beltrami operator on some manifold $M$ -- here eigenvectors of $L$ approximate an orthogonal eigenbasis for square-integrable functions on $M$. However, pointers to nearby results (e.g., FFT for the combinatorial graph Laplacian) are appreciated.

-
From "Discrete Combinatorial Laplacian Operators for Digital Geometry Processing" by Hao Zhang: "the eigenvectors of the TL [Tutte Laplacian] represent the natural vibration modes of the mesh, while the corresponding eigenvalues capture its natural frequencies, resembling the scenario for [the] classical discrete Fourier Transform (DFT). However, the eigenvectors of the TL possess no analytical form in general and there are no fast methods, analogous to the Fast Fourier Transform, to compute the corresponding mesh signal transform." –  Steve Huntsman Apr 10 '10 at 3:22
This paper is available at, e.g. scholar.google.com/… –  Steve Huntsman Apr 10 '10 at 3:22
@SteveHuntsman It now says Your search did not match any articles. –  Ziyuan Lin May 22 at 14:14


text:
## What is the remainder? 2×4 – 4×2 + x – 3 x + 5 Your answer

Question

What is the remainder?
2×4 – 4×2 + x – 3
x + 5


text:
# Comparing radiation dose rates in soils and riverine sediment to track the dispersion of radioactive contamination in Fukushima coastal rivers

* Corresponding author
2 GEDI - Géochimie Des Impacts
LSCE - Laboratoire des Sciences du Climat et de l'Environnement [Gif-sur-Yvette] : DRF/LSCE
Abstract : The Fukushima Dai-ichi nuclear power plant (FDNPP) accident that occurred in March 2011 led to the formation of a 3000-km$^2$ radioactive pollution plume on soils located up to 70 km to the northwest of the damaged site. Forests and paddy fields are the dominant land uses in this mountainous region drained to the Pacific Ocean by several rivers that flow across densely inhabited coastal plains. It is then crucial to track the dispersion of radioactive material conveyed by those rivers to estimate the continental supply of radionuclides to the Ocean and to assess redistribution of radioactive sediment in those catchments. Radiations emitted by this contaminated material may indeed lead to an external exposure threat for local populations. As river discharge and sediment concentration data were not available during the first two years that followed the accident, alternative methods had to be developed to track this dispersion. We therefore organized field campaigns every six months and conducted local ground dose rate measurements to estimate whether fresh sediment drape deposits were more or less contaminated compared to local soils. Overall, our results showed that, in those regions exposed to violent typhoons and spring snowmelt, transfers of sediment are massive and episodic, and that they followed a seasonal cycle in 2011-2012. Then, in May 2013, contamination levels measured in sediment found in the upper parts of the catchments were almost systematically lower than the ones measured in nearby soils, whereas their contamination was higher in the coastal plains. This could have indicated a drying-up of the upstream sources of contamination. However, after the violent typhoons that occurred during summer in 2013, dose rates measured in fresh sediment deposits in November 2013 increased again systematically across the region. We thereby suggest that remobilization of contaminated sediment by typhoons and their storage in reservoirs and in coastal sections of the river channels now represent the most crucial issues to protect the local populations and manage the most contaminated catchments.
Document type :
Conference papers
Domain :

https://hal-cea.archives-ouvertes.fr/cea-02640917
Contributor : Olivier Evrard <>
Submitted on : Thursday, May 28, 2020 - 3:42:12 PM
Last modification on : Monday, September 7, 2020 - 4:22:03 PM

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EGU2014-1702.pdf
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• HAL Id : cea-02640917, version 1

### Citation

Olivier Evrard, Yuichi Onda, Hugo Lepage, Caroline Chartin, Irène Lefèvre, et al.. Comparing radiation dose rates in soils and riverine sediment to track the dispersion of radioactive contamination in Fukushima coastal rivers. EGU General Assembly 2014, Apr 2014, Vienne, Austria. pp.EGU2014-1702. ⟨cea-02640917⟩

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# Residual income valuation

Last updated

Residual income valuation (RIV; also, residual income model and residual income method, RIM) is an approach to equity valuation that formally accounts for the cost of equity capital. Here, "residual" means in excess of any opportunity costs measured relative to the book value of shareholders' equity; residual income (RI) is then the income generated by a firm after accounting for the true cost of capital. The approach is largely analogous to the EVA/MVA based approach, with similar logic and advantages. Residual Income valuation has its origins in Edwards & Bell (1961), Peasnell (1982), and Ohlson (1995). [1]

In finance, the cost of equity is the return a firm theoretically pays to its equity investors, i.e., shareholders, to compensate for the risk they undertake by investing their capital. Firms need to acquire capital from others to operate and grow. Individuals and organizations who are willing to provide their funds to others naturally desire to be rewarded. Just as landlords seek rents on their property, capital providers seek returns on their funds, which must be commensurate with the risk undertaken.

In microeconomic theory, the opportunity cost, or alternative cost, of making a particular choice is the value of the most valuable choice out of those that were not taken. In other words, opportunity that will require sacrifices.

In Economics and Accounting, the cost of capital is the cost of a company's funds, or, from an investor's point of view "the required rate of return on a portfolio company's existing securities". It is used to evaluate new projects of a company. It is the minimum return that investors expect for providing capital to the company, thus setting a benchmark that a new project has to meet.

## Concept

The underlying idea is that investors require a rate of return from their resources – i.e. equity – under the control of the firm's management, compensating them for their opportunity cost and accounting for the level of risk resulting. This rate of return is the cost of equity, and a formal equity cost must be subtracted from net income. Consequently, to create shareholder value, management must generate returns at least as great as this cost. Thus, although a company may report a profit on its income statement, it may actually be economically unprofitable; see Economic profit. It is thus possible that a value deemed positive using a traditional discounted cash flow (DCF) approach may be negative here. RI-based valuation is therefore a valuable complement to more traditional techniques.

In finance, return is a profit on an investment. It comprises any change in value of the investment, and/or cash flows which the investor receives from the investment, such as interest payments or dividends. It may be measured either in absolute terms or as a percentage of the amount invested. The latter is also called the holding period return.

In accounting, equity is the difference between the value of the assets and the value of the liabilities of something owned. It is governed by the following equation:

Shareholder value is a business term, sometimes phrased as shareholder value maximization or as the shareholder value model, which implies that the ultimate measure of a company's success is the extent to which it enriches shareholders. It became popular during the 1980s, and is particularly associated with former CEO of General Electric, Jack Welch.

## Calculation of residual income

The cost of equity is typically calculated using the CAPM, although other approaches such as APT are also used. The currency charge to be subtracted is then simply

In finance, the capital asset pricing model (CAPM) is a model used to determine a theoretically appropriate required rate of return of an asset, to make decisions about adding assets to a well-diversified portfolio.

In finance, arbitrage pricing theory (APT) is a general theory of asset pricing that holds that the expected return of a financial asset can be modeled as a linear function of various factors or theoretical market indices, where sensitivity to changes in each factor is represented by a factor-specific beta coefficient. The model-derived rate of return will then be used to price the asset correctly—the asset price should equal the expected end of period price discounted at the rate implied by the model. If the price diverges, arbitrage should bring it back into line. The theory was proposed by the economist Stephen Ross in 1976. The linear factor model structure of the APT is used as the basis for many of the commercial risk systems employed by asset managers.

Equity Charge = Equity Capital x Cost of Equity,

and

Residual income = Net Income − Equity Charge.

## Valuation formula

Using the residual income approach, the value of a company's stock can be calculated as the sum of its book value and the present value of its expected future residual income, discounted at the cost of equity, ${\displaystyle r}$, resulting in the general formula:

In accounting, book value is the value of an asset according to its balance sheet account balance. For assets, the value is based on the original cost of the asset less any depreciation, amortization or impairment costs made against the asset. Traditionally, a company's book value is its total assets minus intangible assets and liabilities. However, in practice, depending on the source of the calculation, book value may variably include goodwill, intangible assets, or both. The value inherent in its workforce, part of the intellectual capital of a company, is always ignored. When intangible assets and goodwill are explicitly excluded, the metric is often specified to be "tangible book value".

In economics and finance, present value (PV), also known as present discounted value, is the value of an expected income stream determined as of the date of valuation. The present value is always less than or equal to the future value because money has interest-earning potential, a characteristic referred to as the time value of money, except during times of negative interest rates, when the present value will be more than the future value. Time value can be described with the simplified phrase, "A dollar today is worth more than a dollar tomorrow". Here, 'worth more' means that its value is greater. A dollar today is worth more than a dollar tomorrow because the dollar can be invested and earn a day's worth of interest, making the total accumulate to a value more than a dollar by tomorrow. Interest can be compared to rent. Just as rent is paid to a landlord by a tenant without the ownership of the asset being transferred, interest is paid to a lender by a borrower who gains access to the money for a time before paying it back. By letting the borrower have access to the money, the lender has sacrificed the exchange value of this money, and is compensated for it in the form of interest. The initial amount of the borrowed funds is less than the total amount of money paid to the lender.

${\displaystyle V_{0}=BV_{0}+\sum _{t=1}^{\infty }{RI_{t} \over (1+r)^{t}}}$

Here various adjustments to the balance sheet book value may be required; [2] see Clean surplus accounting.

In financial accounting, a balance sheet or statement of financial position or statement of financial condition is a summary of the financial balances of an individual or organization, whether it be a sole proprietorship, a business partnership, a corporation, private limited company or other organization such as Government or not-for-profit entity. Assets, liabilities and ownership equity are listed as of a specific date, such as the end of its financial year. A balance sheet is often described as a "snapshot of a company's financial condition". Of the four basic financial statements, the balance sheet is the only statement which applies to a single point in time of a business' calendar year.

The clean surplus accounting method provides elements of a forecasting model that yields price as a function of earnings, expected returns, and change in book value. The theory's primary use is to estimate the value of a company’s shares. The secondary use is to estimate the cost of capital, as an alternative to e.g. the CAPM. The "clean surplus" is calculated by not including transactions with shareholders when calculating returns; whereas standard accounting for financial statements requires that the change in book value equal earnings minus dividends.

Typically, the above formula will be applied such that the company is assumed to achieve maturity, or "constant growth". (Note that the value will remain identical: the adjustment is a "telescoping" device). Here, analysts commonly employ the Perpetuity Growth Model to calculate the corresponding terminal value [3] (although various, more formal approaches are also applied [4] ). Then, assuming long-run, "constant", growth ${\displaystyle g}$ from year ${\displaystyle m}$, the terminal value is

In finance, the terminal value of a security is the present value at a future point in time of all future cash flows when we expect stable growth rate forever. It is most often used in multi-stage discounted cash flow analysis, and allows for the limitation of cash flow projections to a several-year period. Forecasting results beyond such a period is impractical and exposes such projections to a variety of risks limiting their validity, primarily the great uncertainty involved in predicting industry and macroeconomic conditions beyond a few years.

${\displaystyle T_{m}={RI_{m} \over (r-g)}}$,

and the RI valuation would then be:

${\displaystyle V_{0}=BV_{0}+\sum _{t=1}^{m-1}{RI_{t} \over (1+r)^{t}}+{T_{m} \over (1+r)^{m-1}}}$.

## Comparison with other valuation methods

As can be seen, the residual income valuation formula is similar to the dividend discount model (DDM) (and to other discounted cash flow (DCF) valuation models), substituting future residual earnings for dividend (or free cash) payments (and the cost of equity for the weighted average cost of capital).

However, the RI-based approach is most appropriate when a firm is not paying dividends or exhibits an unpredictable dividend pattern, and / or when it has negative free cash flow many years out, but is expected to generate positive cash flow at some point in the future. Further, value is recognized earlier under the RI approach, since a large part of the stock's intrinsic value is recognized immediately – current book value per share – and residual income valuations are thus less sensitive to terminal value. [5]

At the same time, in addition to the accounting considerations mentioned above, the RI approach will not generally hold if there are expected changes in shares outstanding or if the firm plans to bring in "new" shareholders who derive a net benefit from their capital contributions. [6]

Although EVA is similar to residual income, there will be technical differences between EVA and RI, specifically Stern Stewart & Co, originators of EVA, recommend a fairly large number of adjustments to NOPAT before the methodology may be applied. [7] [8] See Economic value added § Comparison with other approaches.

## Notes

1. "8.10 Application IBM Step 1: Estimating Book Value". www.ftsmodules.com. Retrieved 2 October 2018.
2. Team, The AnalystNotes CFA. "December 2018 CFA Level 1: Study Session List". www.analystnotes.com. Retrieved 2 October 2018.
3. "Val22" (PDF). Retrieved 2018-10-02.
4. Martin, James R. "Management Accounting: Chapter 14". maaw.info. Retrieved 2 October 2018.
5. "EVA/Economic Profit Vs. Residual Income - AnalystForum". www.analystforum.com. Retrieved 2 October 2018.

## Related Research Articles

In finance, discounted cash flow (DCF) analysis is a method of valuing a project, company, or asset using the concepts of the time value of money. All future cash flows are estimated and discounted by using cost of capital to give their present values (PVs). The sum of all future cash flows, both incoming and outgoing, is the net present value (NPV), which is taken as the value of the cash flows in question.

In finance, the net present value (NPV) or net present worth (NPW) applies to a series of cash flows occurring at different times. The present value of a cash flow depends on the interval of time between now and the cash flow. It also depends on the discount rate. NPV accounts for the time value of money. It provides a method for evaluating and comparing capital projects or financial products with cash flows spread over time, as in loans, investments, payouts from insurance contracts plus many other applications.

The weighted average cost of capital (WACC) is the rate that a company is expected to pay on average to all its security holders to finance its assets. The WACC is commonly referred to as the firm's cost of capital. Importantly, it is dictated by the external market and not by management. The WACC represents the minimum return that a company must earn on an existing asset base to satisfy its creditors, owners, and other providers of capital, or they will invest elsewhere.

In corporate finance, as part of fundamental analysis, economic value added (EVA) is an estimate of a firm's economic profit, or the value created in excess of the required return of the company's shareholders. EVA is the net profit less the capital charge (\$) for raising the firm's capital. The idea is that value is created when the return on the firm's economic capital employed exceeds the cost of that capital. This amount can be determined by making adjustments to GAAP accounting. There are potentially over 160 adjustments but in practice only several key ones are made, depending on the company and its industry. EVA is a service mark of Stern Value Management.

In corporate finance, free cash flow (FCF) or free cash flow to firm (FCFF) is a way of looking at a business's cash flow to see what is available for distribution among all the securities holders of a corporate entity. This may be useful to parties such as equity holders, debt holders, preferred stock holders, and convertible security holders when they want to see how much cash can be extracted from a company without causing issues to its operations.

Rational pricing is the assumption in financial economics that asset prices will reflect the arbitrage-free price of the asset as any deviation from this price will be "arbitraged away". This assumption is useful in pricing fixed income securities, particularly bonds, and is fundamental to the pricing of derivative instruments.

Bond valuation is the determination of the fair price of a bond. As with any security or capital investment, the theoretical fair value of a bond is the present value of the stream of cash flows it is expected to generate. Hence, the value of a bond is obtained by discounting the bond's expected cash flows to the present using an appropriate discount rate.

In finance, intrinsic value refers to the value of a company, stock, currency or product determined through fundamental analysis without reference to its market value. It is also frequently called fundamental value. It is ordinarily calculated by summing the discounted future income generated by the asset to obtain the present value. It is worthy to note that this term may have different meanings for different assets.

The following outline is provided as an overview of and topical guide to finance:

The dividend discount model (DDM) is a method of valuing a company's stock price based on the theory that its stock is worth the sum of all of its future dividend payments, discounted back to their present value. In other words, it is used to value stocks based on the net present value of the future dividends. The equation most widely used is called the Gordon growth model (GGM). It is named after Myron J. Gordon of the University of Toronto, who originally published it along with Eli Shapiro in 1956 and made reference to it in 1959. Their work borrowed heavily from the theoretical and mathematical ideas found in John Burr Williams 1938 book "The Theory of Investment Value."

In finance, the T-model is a formula that states the returns earned by holders of a company's stock in terms of accounting variables obtainable from its financial statements. The T-model connects fundamentals with investment return, allowing an analyst to make projections of financial performance and turn those projections into a required return that can be used in investment selection. Mathematically the model is as follows:

The First Chicago Method or Venture Capital Method is a business valuation approach used by venture capital and private equity investors that combines elements of both a multiples-based valuation and a discounted cash flow (DCF) valuation approach.

A financial ratio or accounting ratio is a relative magnitude of two selected numerical values taken from an enterprise's financial statements. Often used in accounting, there are many standard ratios used to try to evaluate the overall financial condition of a corporation or other organization. Financial ratios may be used by managers within a firm, by current and potential shareholders (owners) of a firm, and by a firm's creditors. Financial analysts use financial ratios to compare the strengths and weaknesses in various companies. If shares in a company are traded in a financial market, the market price of the shares is used in certain financial ratios.

Dividend policy is concerned with financial policies regarding paying cash dividend in the present or paying an increased dividend at a later stage. Whether to issue dividends, and what amount, is determined mainly on the basis of the company's unappropriated profit and influenced by the company's long-term earning power. When cash surplus exists and is not needed by the firm, then management is expected to pay out some or all of those surplus earnings in the form of cash dividends or to repurchase the company's stock through a share buyback program.

Corporate finance is an area of finance that deals with sources of funding, the capital structure of corporations, the actions that managers take to increase the value of the firm to the shareholders, and the tools and analysis used to allocate financial resources. The primary goal of corporate finance is to maximize or increase shareholder value. Although it is in principle different from managerial finance which studies the financial management of all firms, rather than corporations alone, the main concepts in the study of corporate finance are applicable to the financial problems of all kinds of firms.

The sum of perpetuities method (SPM) is a way of valuing a business assuming that investors discount the future earnings of a firm regardless of whether earnings are paid as dividends or retained. SPM is an alternative to the Gordon growth model (GGM) and can be applied to business or stock valuation if the business is assumed to have constant earnings and/or dividend growth. The variables are:

In corporate finance, free cash flow to equity (FCFE) is a metric of how much cash can be distributed to the equity shareholders of the company as dividends or stock buybacks—after all expenses, reinvestments, and debt repayments are taken care of. Whereas dividends are the cash flows actually paid to shareholders, the FCFE is the cash flow simply available to shareholders. The FCFE is usually calculated as a part of DCF or LBO modelling and valuation. The FCFE is also called the levered free cash flow.


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# TOTD 2018-04-28: “The Surprisingly Solid Mathematical Case of the Tin Foil Hat Gun Prepper”

Here is an article you should read and think about: “The Surprisingly Solid Mathematical Case of the Tin Foil Hat Gun Prepper”.  The author is a stormwater hydrologist—what he does for a living is study the behaviour of water as it moves through the Earth’s ecosystem and, in particular, extreme events such as floods.  It is he and his colleagues who draw the red lines on maps which determine whether you can get flood insurance at an affordable price and, in many cases, obtain a long-term mortgage on a property.  Those in his profession think deeply about “tail risks”: events which occur rarely but which have major, or even catastrophic, consequences when they happen.  Humans have evolved in an environment which has selected them to apply a number of heuristics that, in most cases, get a good enough answer without a complete understanding of the problem or an exhaustive analysis of the situation.  But evolution, biological or cultural, is poor at selecting for heuristics which apply for events which happen less frequently than the lifetime of most members of a species.  We use our intuition, and often we get the wrong answer.

There is some math in the cited paper, and in what follows, but nothing more complicated than multiplying numbers, which you can do with a calculator if you wish to work it out for yourself.  The only other thing you need to know are some very basic facts of probability and statistics: I’ve written an “Introduction to Probability and Statistics” which, while aimed at other applications, may help if you’re rusty and want to review the details.  All you really need to know is that if a series of events are independent of one another (the outcome of one doesn’t affect the others), and that they have a given individual probability, you get the probability of a series of events occurring one after another by multiplying their probabilities together.

If your eyes are beginning to glaze over, consider a simple case.  When you flip a fair coin, it’s equally likely to come up heads or tails—a statistician would say the probability of heads (or tails) is 0.5 (or ½).  What’s the probability of two heads (or tails) in a row?  You just multiply the probabilities and get 0.25 (¼).  The flips are independent because each flip has a probability of 0.5, regardless of the previous flips.  Even if you’ve flipped nine heads in a row (probability 0.00195, one in 512 times), the probability of the next flip being a head is still 0.5.

What does this have to do with floods, or prepping, or heuristics?  Just that we aren’t very good at understanding how these numbers behave when the probabilities are small and the consequences are dire.

You’ll often hear of a “hundred year flood”.  This is sloppy terminology which many people take to mean a flood that only happens once a century.  That’s wrong.  What it means is that it’s a flood, defined by the extent of flooding, which has a probability of 1/100 (0.01) of happening in a given year.  That is a very different thing.

Now let’s do that little bit of math I warned you about.  If your house is located within the red line of the “100 year flood”, then each year you’re rolling a 100 sided die and if it comes up zero, you lose everything (or almost).  Now, the chance of a flood in any given year doesn’t depend upon what happened in previous years so the probabilities are independent, and we can multiply them.  But now we’re going do to a little arithmetic and look at, not the probability of a flood, but the probability of no flood.  Well, that’s just one minus the probability of a flood (1−0.01 = 0.99).  Now, if we want to know the probability of going n years without a flood, we need only compute $$0.99^n$$.  Remember thirty year mortgages?  I do—I had one.  Let’s plug in 30 for n.  Tap-tap-tap on the calculator, or a little fiddling with the slide rule, and we get 0.74 or, of you like, around ¾ probability there will be no flood.  But that means there’s a 0.26 probability there will be a catastrophic flood before the mortgage is paid off.  Is it any wonder it’s hard to get a mortgage for a house in a “100 year” floodplain?

With this background and perspective, the author moves on to consider an event far more catastrophic than a flood (whose effects are local on a continental scale): a violent conflict over the government of a large territory.  Plugging in the numbers for the United States, he finds two such events in 340 years (the American Revolution and the War of Secession), and taking the mean lifespan of a person today as 78.7 years, calculates a probability of 37% that such an event will happen in their lifetime.  This is substantially higher than that of the flood, and will have consequences far more widespread and serious.

And the U.S. has been far more stable than most other countries in the world.

It’s fashionable to mock “preppers”, but taking the basic precautions against the collapse of our comfortable lives and the infrastructure which supports them is, based upon on a statistical calculation, no less reasonable than preparing for events such as a flood, hurricane, tornado, or earthquake.  And the well-equipped prepper is in a good position to ride out any of these events as well.

The zombie apocalypse is obviously pure fiction, but it has an allure to a few tongue-in-cheek preppers because of its functional completeness. If you are prepared for zombies, you are literally prepared for anything.

11+

## Author: John Walker

Founder of Ratburger.org, Autodesk, Inc., and Marinchip Systems. Author of The Hacker's Diet. Creator of www.fourmilab.ch.

## 18 thoughts on “TOTD 2018-04-28: “The Surprisingly Solid Mathematical Case of the Tin Foil Hat Gun Prepper””

1. Magnificent post, John.  I understand the fallacy of the frequentist argument, and yet I still fall prey to it on a routine basis.

Equally valuable is the bald assertions of the reasonableness of reasonable things — in the face of unreasonable opposition, more so.

7+
2. I remember using these kind of statistics back in the early 1980s to explain why the Soviets (not Russia – it was still the Soviet Union back then) found even a porous anti-missile system so threatening. My basic assumption was that a Soviet first-strike needed to destroy enough US missiles to ensure the subsequent counter-strike did not cripple the Soviet Union.

For my purposes I assumed the Soviet Union would survive without serious damage if no more than 50 US nukes survived a first strike. Given the number of warheads we had back then, it was amazing what an SDI system that even nailed 1% of incoming missiles would due to reducing the effects of a successful first strike taking out a US response.

I argued this meant that Reagan was right to offer the Soviets US ABM technology.  A porous ABM system would reduce the odds of either nation launching a first strike if both sides had it, but it suited the Soviets better to keep MAD in place because they assumed the uncertainty would eventually wear the US down.

7+
3. I was told I was going to be able to buy a parachute on the way down.

4+
4. 10 Cents:
I was told I was going to be able to buy a parachute on the way down.

A true libertarian is somebody who jumps off a cliff confident the market will supply a parachute.

10+
5. 10 Cents:
I was told I was going to be able to buy a parachute on the way down.

A true libertarian is somebody who jumps off a cliff confident the market will supply a parachute.

I believe in a fair country gravity should be outlawed. Think how many lives would be saved. #gravitysucks

4+
6. 10 Cents:
I was told I was going to be able to buy a parachute on the way down.

A true libertarian is somebody who jumps off a cliff confident the market will supply a parachute.

Now, you KNOW I like this sort of joke.

2+
7. Excellent! @johnwalker

I understood all of this article.  Why is this a big deal?  Well, when confronted with my first statistics class in graduate school , it required me to supplement my textbook with three borrowed library books on the subject just to get a B grade  (I did not take statistics in undergrad; hadn’t taken a math class in 20 years.  I used to sit at my study table with all four books opened to the same topic <smiling>)

Thus, I say again, Excellent!  So, well explained.

1+
8. I admit to being one of those who reads ‘prepper’ material (just always made sense to me to do so), although I’m not a prepper in any big sense.   However, that said, we are always prepared for a hurricane (prepared for up to at least 3 weeks of self-sufficiency).

Only problem, we forgot to find a solution for, is addressing the fact that we are no longer ‘young’, no longer quite so fit/strong/flexible as we used be.

But I hadn’t really thought about it, that is the ‘physical’ preparedness needed, until it was brought home to me last year when one of our large trees was pulled up during a hurricane and some of the huge branches ended up in the road.

Knowing it was my responsibility to move the tree/branches, I went over to ask a neighbor I saw with some thick rope if I could borrow it when he was done.  I had planned to tie it to the tree and its large branches to make it easier for me to pull it off  the road.  The neighbor, said yes, he would bring the rope over to my house in a bit.  I went inside to change clothes.

When I came back out just a short time later to see if he was on his way yet, I was surprised to see, the tree and all its big branches had been moved to the end of the block and onto the cul de sac circle.   Turns out he had gotten another neighbor to help him.  They tied it to his truck and towed it all away without saying a word to me.   I’m guessing I did not look to him quite up to being able to move the big tree myself, although it hadn’t occurred to me that I would not be able to; I’ve always been ‘able’.   (PS.  Thank you again, Jonathan)

In hindsight, not sure how I thought I was actually going to get the job done, but it never occurred to me for a second that I wouldn’t.

2+
9. Great article, with very straightforward math.

1+
10. 10 Cents:
I was told I was going to be able to buy a parachute on the way down.

A true libertarian is somebody who jumps off a cliff confident the market will supply a parachute.

I believe in a fair country gravity should be outlawed. Think how many lives would be saved. #gravitysucks

I agree 120.37%.  After all, gravity is just a social construct anyway.  And a racist and classist one, too, owing to height differences among various peoples, implying an entirely unequal vulnerability to dangerously rapid, uncontrolled depletions of gravitational potential energy (known to some as “falling”).

3+
11. 10 Cents:
I was told I was going to be able to buy a parachute on the way down.

A true libertarian is somebody who jumps off a cliff confident the market will supply a parachute.

I believe in a fair country gravity should be outlawed. Think how many lives would be saved. #gravitysucks

I agree 120.37%.  After all, gravity is just a social construct anyway.  And a racist and classist one, too, owing to height differences among various peoples, implying an entirely unequal vulnerability to dangerously rapid, uncontrolled depletions of gravitational potential energy (known to some as “falling”).

1+
12. My one quibble is that we frequently don’t know enough to say that a certain flood is a 100 year flood.  The joke is that we get a 100 year flood very five years.

1+
13. Off topic.

FYI, this is with my iPad in landscape. If I had it in portrait I would see even less of the notification. 🙂

1+
14. Richard Easton:
My one quibble is that we frequently don’t know enough to say that a certain flood is a 100 year flood.  The joke is that we get a 100 year flood very five years.

A 100 years doesn’t buy you as much time as it used to, Richard.

2+
15. Richard Easton:
My one quibble is that we frequently don’t know enough to say that a certain flood is a 100 year flood.  The joke is that we get a 100 year flood very five years.

A 100 years doesn’t buy you as much time as it used to, Richard.

2+
16. Garbage In, Garbage Out.   There are lots of places with poor or missing rainfall data, with the result that the estimates are based on non-representative information.   The guys that develop the flood maps have every incentive to get it right, and they keep getting better, but there are limits.

2+
17. Richard Easton:
My one quibble is that we frequently don’t know enough to say that a certain flood is a 100 year flood.  The joke is that we get a 100 year flood very five years.

A 100 years doesn’t buy you as much time as it used to, Richard.


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# Half Body Flow

In a previous article I talked about plane potential flows.  These flows are governed by Laplace’s equations which are partial differential equations.  As a result they can only be used for irrotational flows.  However, Laplace’s equations will allow you to find various basic velocity potentials as wells as stream functions.  In turn, using the method of superposition, these basic solutions can be combined to find more complex solutions.  For this article I will focus on the superposition of a source and a uniform flow, which in turn will allow us to determine the flow around a half body.

The image above shows a combination of a uniform flow and a source.  When these two plane potential flows are superimposed an approximate representation of a flow around a half body will result.  As a result, the stream function can be seen below.

(Eq 1) $ψ=ψ_{uniform~flow}+ψ_{source}=Ur~sin~θ+\frac{m}{2π}θ$

In addition, the following equation represents the velocity potential.

(Eq 2) $Φ=Ur~cos~θ+\frac{m}{2π}~ln~r$

### Stagnation Point

In the above image a half body is shown.  As the flow approaches the stagnation point the fluid velocity for the uniform flow will cancel out.  Due to this, the fluid velocity will only be influenced by the source.  Hence, to determine the velocity of the source you will need to use the following equation.

(Eq 3) $v_r=\frac{m}{2πr}$

Hence when $x=-b$, than $U$ will be determined by the following equation.

(Eq 4) $U=\frac{m}{2πb}$

Next, in order to determine what the stream function will be at the stagnation point $ψ$, the flow will need to be evaluated at $r=b$ and $θ=π$.  As a result, the following equation is the stream function at the stagnation point.

(Eq 5) $ψ_{stagnation}=\frac{m}{2}$

Finally, taking in consideration that $m/2=πbU$ the equation of the streamline through the stagnation point will be the following.

(Eq 6)  $πbU=Ur~sin~θ+bUθ$

or

(Eq 7)  $r=\frac{b(π-θ)}{sin~θ}$

Where $θ$ is between $0$ and $2π$.

### Half Body

Superimposing a uniform flow and a source will allow us to determine the streamlines of a fluid moving around a half body.  In turn, the streamlines can be seen on the second picture of the image above.  It is called a half body because the body is open at the down stream end.  If you look at the image you will also notice that there are streamlines inside the the half body.  In addition, to those stream lines the source is also within the half body.  As a result, those streamlines are of no interest since we are only interested in the flow around the half body.  However, they need to be noted to show how the source is used help define the streamlines flowing around the half body.

When we are looking at the width of half body it asymptotically approached $2πb$.  As a result, the width of the half body is determined by the following equation.

(Eq 8)  $y=b(π-θ)$

For the above equation, the half-widths $θ→0$ and $θ→2π$ approach $±bπ$.

Next, since the stream function is known, the velocity component at any point can be determined.  As a result the velocity equations will be the following.

(Eq 9) $ν_r=\frac{1}{r}\frac{∂ψ}{∂θ}=U~cos~θ+\frac{m}{2πr}$

and

(Eq 10) $ν_θ=-\frac{∂ψ}{∂r}=-U~sin~θ$

Hence, the square of the velocity magnitude will be the following.

(Eq 11)  $v^2=ν_r^2+ν_θ^2=U^2+\frac{Um~cos~θ}{πr}+\left(\frac{m}{2πr}\right)^2$

In addition, since $b=m/2πU$ equation 11 will become the following.

(Eq 12) $v^2=U^2\left(1+2\frac{b}{r}~cos~θ+\frac{b^2}{r^2}\right)$

Now that the velocity is known, the pressure at any point can be determined by using the Bernoulli equation. The Bernoulli equation is used to determine the difference in the flow field between two points.  At the first point the pressure will be $p_o$ and the velocity will be $U$.  While, at the second point the pressure will be $p$ and the velocity $v$.  Hence,

(Eq 13)  $p_o+\frac{1}{2}ρU^2=p+\frac{1}{2}ρv^2$

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# Collision detection of two circular sectors in mixed polar and Cartesian coordinate

I am trying to solve the following collision detection problem. Suppose we have two circular sectors, each described in their own polar coordinate system with four values $r_1$, $r_2$, $d_1$ and $d_2$, where $r_1$ and $r_2$ ($r_2 \ge r_1$) are angles and $d_1$ and $d_2$ ($d_2 \ge d_1$) are two distances from the polar origin sandwiching the sector. The two sectors are placed on a 2-D Cartesian plane, each described with a coordinate $(x,y)$ (See illustration below).

Illustration:

The problem is to detect whether the two sectors collide. One trivial case is if the sum of the two $d_2$'s is greater than distance of two Cartesian points $\sqrt {(x_1 - x_2)^2 + (y_1 - y_2)^2}$, then the two shapes do not collide. I have yet to figure out how to solve rest of the problem (I thought of using the midline of $r_1$ and $r_2$, but can't exactly figure out how). Any ideas or suggestions?

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I have formatted your question, and am using variables like $x_1$ instead of $x1$ as in your figure, since that is more common. Please feel free to change it back if you want. – Paresh Dec 3 '12 at 6:02


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## Robust Vibration Control in Flexible Beam

This example shows how to robustly tune a controller for reducing vibrations in a flexible beam. This example is adapted from "Control System Design" by G. Goodwin, S. Graebe, and M. Salgado.

### Uncertain Model of Flexible Beam

Figure 1 depicts an active vibration control system for a flexible beam.

Figure 1: Active control of flexible beam

In this setup, a sensor measures the tip position $y\left(t\right)$ and the actuator is a piezoelectric patch delivering a force $u\left(t\right)$. We can model the transfer function from control input $u$ to tip position $y$ using finite-element analysis. Keeping only the first six modes, we obtain a plant model of the form

$G\left(s\right)=\sum _{i=1}^{6}\frac{{\alpha }_{i}}{{s}^{2}+2{\zeta }_{i}{\omega }_{i}s+{\omega }_{i}^{2}}$

with the following nominal values for the amplitudes :

$\alpha =9.72×1{0}^{-4},0.0122,0.0012,-0.0583,-0.0013,0.1199$

$\omega =18.95,118.76,332.54,651.66,1077.2,1609.2.$

The damping factors ${\zeta }_{i}$ are often poorly known and are assumed to range between 0.0002 and 0.02. Similarly, the natural frequencies are only approximately known and we assume 20% uncertainty on their location. To construct an uncertain model of the flexible beam, use the ureal object to specify the uncertainty range for the damping and natural frequencies. To simplify, we assume that all modes have the same damping factor $\zeta$.

% Damping factor zeta = ureal('zeta',0.002,'Range',[0.0002,0.02]); % Natural frequencies w1 = ureal('w1',18.95,'Percent',20); w2 = ureal('w2',118.76,'Percent',20); w3 = ureal('w3',332.54,'Percent',20); w4 = ureal('w4',651.66,'Percent',20); w5 = ureal('w5',1077.2,'Percent',20); w6 = ureal('w6',1609.2,'Percent',20);

Next combine these uncertain coefficients into the expression for $G\left(s\right)$.

alpha = [9.72e-4 0.0122 0.0012 -0.0583 -0.0013 0.1199]; G = tf(alpha(1),[1 2*zeta*w1 w1^2]) + tf(alpha(2),[1 2*zeta*w2 w2^2]) + ... tf(alpha(3),[1 2*zeta*w3 w3^2]) + tf(alpha(4),[1 2*zeta*w4 w4^2]) + ... tf(alpha(5),[1 2*zeta*w5 w5^2]) + tf(alpha(6),[1 2*zeta*w6 w6^2]); G.InputName = 'uG'; G.OutputName = 'y';

Visualize the impact of uncertainty on the transfer function from $u$ to $y$. The bode function automatically shows the responses for 20 randomly selected values of the uncertain parameters.

rng(0), bode(G,{1e0,1e4}), grid title('Uncertain beam model')

### Robust LQG Control

LQG control is a natural formulation for active vibration control. With systune, you are not limited to a full-order optimal LQG controller and can tune controllers of any order. Here for example, let's tune a 6th-order state-space controller (half the plant order).

C = tunableSS('C',6,1,1);

The LQG control setup is depicted in Figure 2. The signals $d$ and $n$ are the process and measurement noise, respectively.

Figure 2: LQG control structure

Build a closed-loop model of the block diagram in Figure 2.

C.InputName = 'yn'; C.OutputName = 'u'; S1 = sumblk('yn = y + n'); S2 = sumblk('uG = u + d'); CL0 = connect(G,C,S1,S2,{'d','n'},{'y','u'});

Note that CL0 depends on both the tunable controller C and the uncertain damping and natural frequencies.

CL0
CL0 = Generalized continuous-time state-space model with 2 outputs, 2 inputs, 18 states, and the following blocks: C: Parametric 1x1 state-space model, 6 states, 1 occurrences. w1: Uncertain real, nominal = 18.9, variability = [-20,20]%, 3 occurrences w2: Uncertain real, nominal = 119, variability = [-20,20]%, 3 occurrences w3: Uncertain real, nominal = 333, variability = [-20,20]%, 3 occurrences w4: Uncertain real, nominal = 652, variability = [-20,20]%, 3 occurrences w5: Uncertain real, nominal = 1.08e+03, variability = [-20,20]%, 3 occurrences w6: Uncertain real, nominal = 1.61e+03, variability = [-20,20]%, 3 occurrences zeta: Uncertain real, nominal = 0.002, range = [0.0002,0.02], 6 occurrences Type "ss(CL0)" to see the current value, "get(CL0)" to see all properties, and "CL0.Blocks" to interact with the blocks.

Use an LQG criterion as control objective. This tuning goal lets you specify the noise covariances and the weights on the performance variables.

R = TuningGoal.LQG({'d','n'},{'y','u'},diag([1,1e-10]),diag([1 1e-12]));

Now tune the controller C to minimize the LQG cost over the entire uncertainty range.

[CL,fSoft,~,Info] = systune(CL0,R);
Soft: [5.63e-05,Inf], Hard: [-Inf,Inf], Iterations = 80 Soft: [6.27e-05,0.000103], Hard: [-Inf,-Inf], Iterations = 75 Soft: [6.96e-05,7.38e-05], Hard: [-Inf,-Inf], Iterations = 76 Soft: [7.2e-05,7.2e-05], Hard: [-Inf,-Inf], Iterations = 34 Final: Soft = 7.2e-05, Hard = -Inf, Iterations = 265

### Validation

Compare the open- and closed-loop Bode responses from $d$ to $y$ for 20 randomly chosen values of the uncertain parameters. Note how the controller clips the first three peaks in the Bode response.

Tdy = getIOTransfer(CL,'d','y'); bode(G,Tdy,{1e0,1e4}) title('Transfer from disturbance to tip position') legend('Open loop','Closed loop')

Next plot the open- and closed-loop responses to an impulse disturbance $d$. For readability, the open-loop response is plotted only for the nominal plant.

impulse(getNominal(G),Tdy,5) title('Response to impulse disturbance d') legend('Open loop','Closed loop')

Finally, systune also provides insight into the worst-case combinations of damping and natural frequency values. This information is available in the output argument Info.

WCU = Info.wcPert
WCU = 3x1 struct array with fields: w1 w2 w3 w4 w5 w6 zeta

Use this data to plot the impulse response for the two worst-case scenarios.

impulse(usubs(Tdy,WCU),5) title('Worst-case response to impulse disturbance d')

Get trial now