text:
The previous post can be found here, be sure to check it out so you can follow the process step by step. Since then, another two very significant features of the algorithm have been implemented and tested: node pair candidate selection and feasibility checks.

## Introduction

As previously described, in the ISO problem we are basically trying to create a mapping such that, every node from the first graph is matched to a node from the second graph. This searching for “feasible pairs” can be visualized by a tree, where each node is the candidate pair that we should examine. This can become much clearer if we take a look at the below figure.

In order to check if the graphs $G_1$, $G_2$ are isomorphic, we check every candidate pair of nodes and if it is feasible, we extend the mapping and go deeper into the tree of pairs. If it’s not feasible, we climb up and follow a different branch, until every node in $G_1$ is mapped to a node $G_2$. In our example, we start by examining node 0 from G1, with node 0 of G2. After some checks (details below), we decide that the nodes 0 and 0 are matching, so we go deeper to map the remaining nodes. The next pair is 1-3, which fails the feasibility check, so we have to examine a different branch as shown. The new branch is 1-2, which is feasible, so we continue on using the same logic until all the nodes are mapped.

## Candidate Pair Selection

Although in our example we use a random candidate pair of nodes, in the actual implementation we are able to target specific pairs that are more likely to be matched, hence boost the performance of the algorithm. The idea is that, in every step of the algorithm, given a candidate

$$u\in V_1$$

we compute the candidates

$$v\in V_2$$

where $V_1$ and $V_2$ are the nodes of $G_1$ and $G_2$ respectively. Now this is a puzzle that does not require a lot of specific knowledge on graphs or the algorithm itself. Keep up with me, and you will realize it yourself. First, let $M$ be the mapping so far, which includes all the “covered nodes” until this point. There are actually three different types of $u$ nodes that we might encounter.

### Case 1

Node $u$ has no neighbors (degree of $u$ equals to zero). It would be redundant to test as candidates for $u$, nodes from $G_2$ that have more than zero neighbors. That said, we eliminate most of the possible candidates and keep those that have the same degree as $u$ (in this case, zero). Pretty easy right?

### Case 2

Node $u$ has neighbors, but none of them belong to the mapping. This situation is illustrated in the following figure.

The grey lines indicate that the nodes of $G_1$ (left 1,2) are mapped to the nodes of $G_2$ (right 1,2). They are basically the mapping. Again, given $u$, we make the observation that candidates $v$ of u, should also have no neighbors in the mapping, and also have the same degree as $u$ (as in the figure). Notice how if we add a neighbor to $v$, or if we place one of its neighbors inside the mapping, there is no point examining the pair $u-v$ for matching.

### Case 3

Node $u$ has neighbors and some of them belong to the mapping. This scenario is also depicted in the below figure.

In this case, to obtain the candidates for $u$, we must look into the neighborhoods of nodes from $G_2$, which map back to the covered neighbors of $u$. In our example, $u$ has one covered neighbor (1), and 1 from $G_1$ maps to 1 from $G_2$, which has $v$ as neighbor. Also, for v to be considered as candidate, it should have the same degree as $u$, obviously. Notice how every node that is not in the neighborhood of 1 (in $G_2$) cannot be matched to $u$ without breaking the isomorphism.

## ISO Feasibility Rules

Let’s assume that given a node $u$, we obtained its candidate $v$ following the process described in the previous section. At this point, the Feasibility Rules are going to determine whether the mapping should be extended by the pair $u-v$ or if we should try another candidate. The feasibility of a pair $u-v$ is examined by consistency and cutting checks.

### Consistency rules

At, first I am going to present the mathematical expression of the consistency check. It may seem complicated at first, but it’s going to be made simple by using a visual illustration. Using the notation $nbh_i(u)$ for the neighborhood of u in graph $G_i$, the consistency rule is:

$$\forall\tilde{v}\in nbh_2(v)\cap M:(u, M^{-1}(\tilde{v}))\in E_1) \wedge \forall\tilde{u}\in nbh_1(u)\cap M:(u, M(\tilde{u}))\in E_2)$$

We are going to use the following simple figure to demystify the above equation.

The mapping is depicted as grey lines between the nodes that are already mapped, meaning that 1 maps to A and 2 to B. What is implied by the equation is that, for two nodes $u$ and $v$ to pass the consistency check, the neighbors of $u$ that belong in the mapping, should map to neighbors of $v$ (and backwards). This could be checked by code as simple as:

for neighbor in G1[u]:
if neighbor in mapping:
if mapping[neighbor] not in G2[v]:
return False
elif G1.number_of_edges(u, neighbor) != G2.number_of_edges(
v, mapping[neighbor]
):
return False

where the final two lines also check the number of edges between node $u$ and its neighbor $\tilde{u}$, which should be the same as those between $v$ and its neighbor which $\tilde{u}$ maps to. At a very high level, we could describe this check as a 1-look-ahead check.

### Cutting rules

We have previously discussed what $T_i$ and $\tilde{T_i}$ represent (see previous post). These sets are used in the cutting checks as follows: the number of neighbors of $u$ that belong to $T_1$, should be equal to the number of neighbors of $v$ that belong to $T_2$. Take a moment to observe the below figure.

Once again, node 1 maps to A and 2 to B. The red nodes (4,5,6) are basically $T_1$ and the yellow ones (C,D,E) are $T_2$. Notice that in order for $u-v$ to be feasible, $u$ should have the same number of neighbors, inside $T_1$, as $v$ in $T_2$. In every other case, the two graphs are not isomorphic, which can be verified visually. For this example, both nodes have 2 of their neighbors (4,6 and C,E) in $T_1$ and $T_2$ respectively. Careful! If we delete the $V-E$ edge and connect $V$ to $D$, the cutting condition is still satisfied. However, the feasibility is going to fail, by the consistency checks of the previous section. A simple code to apply the cutting check would be:

if len(T1.intersection(G1[u])) != len(T2.intersection(G2[v])) or len(
T1out.intersection(G1[u])
) != len(T2out.intersection(G2[v])):
return False

where T1out and T2out correspond to $\tilde{T_1}$ and $\tilde{T_2}$ respectively. And yes, we have to check for those as well, however we skipped them in the above explanation for simplicity.

## Conclusion

At this point, we have successfully implemented and tested all the major components of the algorithm VF2++,

• Node Ordering
• $T_i/\tilde{T_i}$ Updating
• Feasibility Rules
• Candidate Selection

This means that, in the next post, hopefully, we are going to discuss our first, full and functional implementation of VF2++.


text:
# Visual Conditions¶

You can associate conditions to your visual objects so you control rendering and other visual behavior. When those conditions match or not, then your visual objects gain the property you associate with the condition.

Conditions are mapped in the view element id_view as attribute data-xp-cond-rules:

In this case we compare to settings and isSocialLogged. These variables would relate to the visual context. In case you want to compare with a form field, you would do: form_myform.myfiel. You may at your server-side write variables to the visual context with self._add_attr('isSocialLogged', False) from service view.

You can compare:

• ==
• !=
• >
• <
• >=
• <=

## Logical Operators¶

• OR : ||
• AND : &&

You can include parenthesis ( and ) to provide right logic.

## Visual Objects Conditions¶

After you define your view conditions, it is time to map those conditions to your objects.

Our current release forces you to include container visual object, where you place condition logic. In future releases, we will have condition support in all objects.

data-xp-type="function.render"

You can also:

data-show-faces="true"
data-width="200"
data-max-rows="1"

You can include a set of conditions to match. The first to match, we apply action and skip the rest. The only action currently supported os render, where you can place to true to show or false to hide.

format: plain


text:
# [Tex/LaTex] Font size of Figure Caption Header

captionsfontsize

My question is about changing the font size.

I have to insert two figures side by side. Coming to inserting the captions for the figures individually:

Figure 3: blah blah.....................Figure 4: blah blah...

I am able to change the font size of "blah blah". What I want to know is how to change the font size of "Figure 3" or "Figure 4", say the figure headers or something. I do not know what they are called to search on-line.

You can do this (and more!) with the caption package

\usepackage[font=small,labelfont=bf]{caption}

will make Figure in bold, and make the caption content small.

Referencing Section 2.3 of the documentation, the size options for font are as follows:

• scriptsize
• footnotesize
• small
• normalsize
• large
• Large

Here's a MWE to play with

\documentclass{article}

\usepackage[font=scriptsize]{caption}
\usepackage{mwe}

\begin{document}

\begin{figure}
\centering
\begin{minipage}{.4\textwidth}
\includegraphics[width=\textwidth]{example-image-a}
\caption{My caption here}
\end{minipage}%
\hfill
\begin{minipage}{.4\textwidth}
\includegraphics[width=\textwidth]{example-image-b}
\caption{My other caption here}
\end{minipage}
\end{figure}

\end{document}


text:
# solving this system of linear equation.

This topic is 1842 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I need a bit of help recognizing the pattern to figuring out the following problem:

so, here's the problem:
the numbers are arranged in how the tile's are layed out in memory, however, i want to be able to get a patch of tile's relative to how the white lines are layed out.

for example: column 1 is 6, 9, and 13. 2 is 3, 7, and 10, and 3 is 1, 4, and 8.
so my input would be something like this: the bottom tile's index in the world(in this case that is tile #6), and then i'd specify which tile i was last on, and finally i'd pass the column and row size to get the chunk of.

the goal is to quickly determine which column/row the last tile is on, increment first by column, then by row. so i'd iterate over tiles 6, 9, 13, 3, 7, 10, 1, 4, and 8.

i'm having trouble creating an algorithm which can extract the row/column from the index of the tile(and do the reverse).

edit: after a bit more analysis, i figured out an algorithm to suit my needs.

edit2: ok, i can get the column/row, and thought i could figure out how to get back easily. But i can't seem to figure out how to get back to the index.

This is what i'm doing to get the column/row:
//above example, the width and height can be taken as 3, and 5.
//note all divisions are integers, and round down.
void Get(int Index){
int x = Index%width;
int y = Index/width;
int column = x+(height/2-((height&1)^1))-(y/2);
int row = x+(y+1)/2;
}

Edited by slicer4ever

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Rearrange your column equation to solve for x, or y if you prefer. Do the same for the opposite axis in the row = equation. now subsititute on for the variable in the other. with X and Y you can repeat the first step with the other equation to get back to Index. But why not just use the axii of the tile map(up left and up right in your image above) as opposed to the world X and Y axii. The numbers and indices become much easier to work with.

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Burnt_Fyr, on 03 Jun 2013 - 08:53, said:
Rearrange your column equation to solve for x, or y if you prefer. Do the same for the opposite axis in the row = equation. now subsititute on for the variable in the other. with X and Y you can repeat the first step with the other equation to get back to Index.

I've tried this, but my math isn't coming up right somewhere, this is what i've done:
int x = Index%width;
int y = Index/width;
int column = x+(height/2-((height&1)^1))-(y/2);
int row = x+(y+1)/2;

width = 3, height = 5 = (5/2 = 2 as int)
Index(x, y) = column, row

Index: 0 (0, 0) = 2, 0
c = x+2-y/2;
2 = x+2-y/2
-2 =  -2
0 = x-y/2
+y/2 =+y/2
y/2 = x
*2   = *2
y = 2x

r = x+(y+1)/2
0 = x+(2x+1)/2
0  = x+x ?//not sure if my math here is correct, as it's integer math, the 1/2 would round out right?
0 = 2x
/2  = /2
0 = x
y = 2x = 2(0) = 0

Index: 3(0, 1) = 2, 1
c = x+2-y/2
2 = x+2-y/2
-2 =  -2
0 = x-y/2
+y/2 = +y/2
y/2 = x
*2  = *2
y = 2x

r = x+(y+1)/2
1 = x+(2x+1)/2
1 = x+x ?
1 = 2x
/2 = /2
0 = x
y = 2x = 2(0) = 0 != 1

Burnt_Fyr, on 03 Jun 2013 - 08:53, said:
But why not just use the axii of the tile map(up left and up right in your image above) as opposed to the world X and Y axii. The numbers and indices become much easier to work with.

in this particular instance, it would make more sense to arrange the tiles that way, but when it comes to rendering, and figuring out what tile's been clicked, or what tile an object is on, it's easier for me to arrange the tiles in world x/y instead of the grid's relative x/y

edit: ok, so here's a couple of wolfam's demonstrating the solution, but i'm not sure how wolfam figures out the integer solution.

this is the example equation I used for calculating the column/row:
http://www.wolframalpha.com/input/?i=%28x%2Bfloor%28h%2F2%29-floor%28y%2F2%29+%3D+c%2C+x%2Bfloor%28%28y%2B1%29%2F2%29+%3D+r%29+where+x%3D+0+and+y+%3D+2+and+h%3D5

and plugging in c and r back into the function, wolfam can figure out the correct solution, but i can't figure out how to solve this problem:
http://www.wolframalpha.com/input/?i=solve+%28x%2Bfloor%28h%2F2%29-floor%28y%2F2%29+%3D+c%2C+x%2Bfloor%28%28y%2B1%29%2F2%29+%3D+r%29+where+c%3D+1+and+r+%3D+1+and+h%3D5 Edited by slicer4ever

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i've yet to be able to solve my issue of going back, so i'm bumping this thread looking for a bit of help on achieving the correct result.

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It's too early for me to work through your math, escpecially when taken out of context of what your trying to do. So instead I'm going back to my suggestion earlier, and attempt to address the concerns you had above.

Using a grid of [x,y] for the tiles, the indices are easier to work with, as I said before.

If you want that angled look, during rendering, just apply a -45 degree rotation to the camera, and this gives the typical rts top down view that you are after. When it comes to picking a tile, you create the pickray, and then transform it back into the world via the projection and view matricies. This puts the pickray back into the right space, so that the coordinates where it intersects the tile map are correct relative to that map. For determining what tile a unit is on, simply grab the coordinates of the unit, as these are already in the same space of the tile map.

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i've already figured out the equations:
column = x-y/2;
row = x+(y+1)/2;

x = row/2+(colum+((row&1)?((column>0)?1:0):((column<0)?-1:0)))/2
y = row-column

Edited by slicer4ever

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text:
## Algebra 1

$9t^{4} - 3t^{2} + 11t + 12$
By simplifying the brackets we get = $9t^{4} + 5t + 8 - 3t^{2} + 6t + 4$ Then we add like terms together which are terms that have the same variables. = $9t^{4} - 3t^{2} + 6t + 5t + 4 + 8$ = $9t^{4} - 3t^{2} + 11t + 12$


text:
# 2.E: The Classical Wave Equation (Exercises)

These are homework exercises to accompany Chapter 2 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.

## Q2.1A

Find the general solutions to the following differential equations:

1. $$\dfrac{d^{2}y}{dx^{2}} - 4y = 0$$
2. $$\dfrac{d^{2}y}{dx^{2}} - 3\dfrac{dy}{dx} - 54y = 0$$
3. $$\dfrac{d^{2}y}{dx^{2}} + 9y = 0$$
Solution
1. To solve, we realize that the form of the differential equation is that of a quadratic function: $$ar^{2} + br + c = 0$$ where $$a = 1$$, $$b = 0$$, and $$c = - 4$$. We plug these into the quadratic formula $r = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\nonumber$ from which we obtain that $$r = \pm 2$$. The general solution has form $$Ae^{xr_1} + Be^{xr_2}$$. Thus, the solution is $y = Ae^{2x}+Be^{-2x}\nonumber.$
2. In a similar manner to part a, we use the quadratic formula with $$a = 1$$, $$b = -3$$, and $$c = -54$$ and obtain $$y = Ae^{-6x}+Be^{9x}$$.
3. We also use the quadratic formula with $$a = 1$$, $$b =$$, and $$c = 9$$. The difference is that we obtain imaginary roots. The solution to the quadratic equation is $$\pm 3i$$. We can substitute into the general solution and obtain $$y = Ae^{i3x}+Be^{-i3x}$$. Alternatively, we can use Euler's formula to write it as $y = A \cos(3x) + B \sin(3x) \nonumber .$

## Q2.1B

Find the general solutions to the following differential equations:

1. $$\dfrac{d^{2}y}{dx^{2}} - 16y = 0$$
2. $$\dfrac{d^{2}y}{dx^{2}} - 6\dfrac{dy}{dx} + 27y = 0$$
3. $$\dfrac{d^{2}y}{dx^{2}} + 100y = 0$$
Solution
1. To find the solution we use the characteristic equation: $$ar^{2} + br + c = 0$$ where $$a = 1$$, $$b = 0$$, and $$c = - 4$$. We use the quadratic formula to find that $$r = \pm 4$$. The general solution to differential equations of this form are $$Ae^{xr_1} + Be^{xr_2}$$. Thus, the solution is $$y = Ae^{4x}+Be^{-4x}$$.
2. Use same steps as part a but with $$a = 1$$, $$b = -6$$, and $$c = 27$$ to find that $$y = Ae^{9x}+Be^{-3x}$$.
3. Similar as above but with $$a = 1$$, $$b =0$$, and $$c = 100$$. We find that we have imaginary roots since $$r = \pm 10$$. We plug back into general solution to get $y = Ae^{i10x}+Be^{-i10x}\nonumber$ or use Euler's formula to find that $y = A \cos(10x) + B \sin(10x) \nonumber$Q2.1

## Q2.1C

Find the general solutions to the following differential equations:

1. $$\dfrac{dy}{dx} - 4\sin(x)y = 0$$
2. $$\dfrac{d^{2}y}{dx^{2}} - 5\dfrac{dy}{dx}+6y = 0$$
3. $$\dfrac{d^{2}y}{dx^{2}} = 0$$
Solution

a. Begin by moving the $$4\sin(x)y$$ to the right side.

$\dfrac{dy}{dx} = 4\sin(x)y\nonumber$

Divide both sides by $$y$$ and multiply by $$dx$$

$\dfrac{1}{y}dy = 4\sin(x)dx\nonumber$

Integrate both sides

$ln(y) = -4\cos(x)+C\nonumber$

$y = Ce^{-4\cos(x)}\nonumber$

b. Begin by saying the solution has the form: $$y = e^{rx}$$ where $$r$$ is a constant. Plug in and factor out the $$e^{rx}$$ yields

$e^{rx}(r^2 - 5r + 6) = 0\nonumber$

Solve for the roots

$$r = 3$$ and $$r = 2$$

$y = Ae^{3x} + Be^{2x}\nonumber$

c. Simply take the integral twice to yield

$y = C_1x + C_2\nonumber$

## Q2.2A

Practice solving these first and second order homogeneous differential equations with given boundary conditions:

1. $$\dfrac{dy}{dx} = ay$$ with $$y(0) = 11$$
2. $$\dfrac{d^2y}{dt^2} = ay$$ with $$y(0) = 6$$ and $$y'(0) = 4$$
3. $$\dfrac{d^2y}{dt^2} + \dfrac{dy}{dt} - 42y = 0$$ with $$y(0) = 2$$ and $$y'(0) = 0$$
Solution

a. This first order differential equation can be solved by combining like variables and integrating. Simply multiply both sides by $$\dfrac{dx}{y}$$ to have the form

$\dfrac{dy}{y} = a\,dx\nonumber$

Integrating both sides,

$\int\dfrac{dy}{y} = \int adx \\ \ln|y| = ax +C \nonumber$

Solving for y we arrive at our general solution,

$y = Ce^{ax}\nonumber$

Using our boundary condition $$y(0) = 11$$ we can solve the particular solution for $$y$$

$y(0) = 11 = Ce^{a(0)} \\ 11 = Ce^0\nonumber$

Knowing $$e^0 = 1$$ we determine that $$C = 11$$ and we arrive at our final solution

$\boxed{y = 11e^{ax}}\nonumber$

b. This second order differential equation can be solved by making a typical guess for $$y$$ based on the flavor of the equation, checking accuracy then solving for the constants $$C_1$$ and $$C_2$$ Using the given boundary conditions.

A typical initial guess for this second order differential equation is $$y = e^{\pm\sqrt{a}t}$$. Test this guess by taking the first two derivatives of $$y$$ with respect to $$x$$ and compare to the given problem.

$y_{guess} = e^{\pm\sqrt{a}t}\nonumber$

Taking the first derivative of $$y_{guess}$$ with respect to $$x$$ we get

$\dfrac{dy_{guess}}{dx} = \pm\sqrt{a}e^{\pm\sqrt{a}t}\nonumber$

The second derivative of $$y_{guess}$$ with respect to $$x$$ is

$\dfrac{d^2y_{guess}}{dx^2} = ae^{\pm\sqrt{a}t}\nonumber$

To check how accurate this $$y_{guess}$$ substitute $$\dfrac{d^2y_{guess}}{dx^2}$$ and $$y_{guess}$$ into $$\dfrac{d^2y}{dt^2} = ay$$.

$ae^{\pm\sqrt{a}t} = ae^{\pm\sqrt{a}t}\nonumber$

since both sides of the equal sign are the same we know this was a great guess. Then our general solution will have the form

$y(t) = C_1e^{\sqrt{a}t} \ + \ C_2e^{-\sqrt{a}t}\nonumber$

Using our boundary conditions $$y(0) = 6$$ and $$y'(0) = 4$$ we can solve for $$C_1$$ and $$C_2$$

$y(0) = 6 = C_1e^{\sqrt{a}(0)} \ + \ C_2e^{-\sqrt{a}(0)} \\ 6 = C_1 + C_2 \\ \dfrac{dy(0)}{dx} = 4 = \sqrt{a}C_1e^{\sqrt{a}(0)} \ -\sqrt{a}C_2e^{-\sqrt{a}(0)} \\ 4 = \sqrt{a}C_1-\sqrt{a}C_2\nonumber$

Know we can use the two system of equations to solve for $$C_1$$ and $$C_2$$. Doing the algebra we find that

$C_1 = 3+\dfrac{2}{\sqrt{a}} \\ C_2 = 3-\dfrac{2}{\sqrt{a}}\nonumber$

Our particular solution then becomes

$\boxed{y(t) = (3+\dfrac{2}{\sqrt{a}})e^{\sqrt{a}t} \ + \ (3-\dfrac{2}{\sqrt{a}})e^{-\sqrt{a}t}}\nonumber$

c. To solve this second order differential equation we will use a similar method as part b) but we will add an extra step to determine the exponents of a good $$y_{guess}$$. We will determine the exponents or our exponentials by solving for the roots of this equation as if it was a quadratic equation in which

$\dfrac{d^2y}{dt^2} + \dfrac{dy}{dt} - 42y = 0\nonumber$

becomes

$r^2 + r - 42 = 0\nonumber$

Solving for the roots we find that

$(r+7)(r-6) \\ r = -7, \ 6\nonumber$

Knowing this, our exponents to our $$y_{guess}$$ become,

$y_{guess} = e^{-7t} + e^{6t}\nonumber$

Now we can follow the same process as part b). Take the first and second derivative of our guess:

$\dfrac{dy_{guess}}{dx} = -7e^{-7t}+6e^{6t}\nonumber$

Second derivative is

$\dfrac{d^2y_{guess}}{dx^2} = 49e^{-7t}+36e^{6t}\nonumber$

Substituting $$y_{guess}$$ and our first and second derivative into the original differential equation $$\dfrac{d^2y}{dt^2} + \dfrac{dy}{dt} - 42y = 0$$ we find that

$49e^{-7t}+36e^{6t} + -7e^{-7t}+6e^{6t} - 42e^{-7t} - 42e^{6t} = 0 \\ 0=0\nonumber$

Again, we were able to make a fantastic $$y_{guess}$$. We can then say that our general solution is

$y = C_1e^{-7t} + C_2e^{6t}\nonumber$

Using our boundary conditions $$y(0) = 2$$ and $$y'(0) = 0$$ we can solve for $$C_1$$ and $$C_2$$

$y(0) = 2 = C_1e^{-7(0)} \ + \ C_2e^{6(0)} \\ 2 = C_1 + C_2 \\ \dfrac{dy(0)}{dx} = 0 = -7C_1e^{-7(0)} \ +6C_2e^{6(0)} \\ 0 = -7C_1+6C_2\nonumber$

Now we can use the two system of equations to solve for $$C_1$$ and $$C_2$$. Doing the algebra we find that

$C_1 = \dfrac{12}{13} \\ C_2 = \dfrac{14}{13}\nonumber$

Our particular solution then becomes

$\boxed{y = \dfrac{12}{13}e^{-7t} + \dfrac{14}{13}e^{6t}}\nonumber$

## Q2.3A

Prove that $$x(t)$$ = $$\cos(\theta$$) oscillates with a frequency

$\nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \nonumber$

Prove that $$x(t)$$ = $$\cos(\theta$$) also has a period

$T = {2\pi}\sqrt{\dfrac{m}{k}} \nonumber$

where $$k$$ is the force constant and $$m$$ is mass of the body.

Solution

The angular frequency for a harmonic oscillator in units of radian/second is $\omega = \sqrt{\dfrac{k}{m}}\nonumber$

Angular frequency and frequency are related by: $\omega = 2{\pi}f\nonumber$

Substitute: $\omega = \sqrt{\dfrac{k}{m}}\nonumber$ for omega in: $\omega = 2{\pi}f\nonumber$

So: $\sqrt{\dfrac{k}{m}} = 2{\pi}f\nonumber$

Solve for f to find the frequency: $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\nonumber$

The period T is the inverse of the frequency so: $f = \dfrac{1}{T}\nonumber$

Substitute the frequency: $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\nonumber$

and solve for T: $\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} = \dfrac{1}{T}\nonumber$

So: $T = {2\pi}\sqrt{\dfrac{m}{k}}\nonumber$

## Q2.3B

Try to show that

$x(t)=\sin(\omega t)\nonumber$

oscillates with a frequency

$\nu = \omega/2\pi\nonumber$

Explain your reasoning. Can you give another function of x(t) that have the same frequency?

Both the functions of $$\sin(t)$$ and $$cos(t)$$ experience a complete cycle every $$2π$$ radians, so that they will oscillate at a frequency of 1/2π. Thus, the function $$x(t)=\sin(\omega t)$$ will go through $$\omega$$ complete cycles for every $$2π$$ radians, and thus it will have a frequency of

$\nu = \omega/2\pi\nonumber$

Another example of function $$x(t)$$ that will have the same frequency of $$\frac{ω}{2π}$$ can be

$x(t) = A\sin(\omega t) + B\cos(\omega t)\nonumber$

## Q2.3C

Which two functions oscillate with the same frequency?

1. $$x(t)=\cos( \omega t)$$
2. $$x(t)=\sin (2 \omega t)$$
3. $$x(t)=A\cos( \omega t)+B\sin( \omega t)$$
Solution
1. The function goes through $$\omega$$ cycles every 2$$\pi$$ radians so its frequency is $$v=\dfrac{\omega}{2\pi}$$. The 2 shortens the period to $$\pi$$. The function goes through $$\omega$$ cycles every $$\pi$$ radians so its frequency is $$v=\dfrac{\omega}{\pi}$$.
2. Therefore, functions A and C oscillate with the same frequency.
3. Both $$cos( \omega t)$$ and $$\sin( \omega t)$$ have a frequency of $$\dfrac{\omega}{2\pi}$$ so a linear combination of these functions will have the same frequency.

## Q2.3D

Prove that $$x(t) = \cos(\omega(t))$$ oscillates with a frequency

$\nu = \dfrac{\omega}{2\pi} \nonumber.$

Prove that $$x(t) = A \cos(\omega(t) + B \sin(\omega(t))$$ oscillates with the same frequency:

$\nu = \dfrac{\omega}{2\pi}. \nonumber$

Solution

Angular frequency which is in units of radian/second is

$\omega = \dfrac{2}{v\pi}\nonumber$

thus the frequency is $\nu = \dfrac{\omega}{2\pi}\nonumber$

In terms of $$x(t)$$ = A $$cos(\omega(t)$$ + B$$\sin(\omega(t)$$, both A and B will have the same angular frequency, thus

$\nu = \dfrac{\omega}{2\pi}\nonumber$

## Q2.4

Show that the differential equation:

$\dfrac{d^2y}{dx^2} + y(x) = 0\nonumber$

has a solution

$y(x)= \sin x + \cos x \nonumber$

Solution

$\dfrac{dy}{dx}= 2 \cos x - \sin x\nonumber$

$\dfrac{d^2y}{dx^2}= -2\sin x - \cos x\nonumber$

so

$-2 \sin x-\cos x +2 \sin x +\cos x =0\nonumber$

$\dfrac{d^2y}{dx^2} + y(x) = 0 \nonumber$

## Q2.7

For a classical harmonic oscillator, the displacement is given by

$\xi (t)=v_0 \sqrt{\dfrac{m}{k}} \sin \sqrt{\dfrac{k}{m}} t\ \nonumber$

where $$\xi=x-x_0$$. Derive an expression for the velocity as a function of time, and determine the times at which the velocity of the oscillator is zero.

Solution

We know that

$\dfrac{dx}{dt}=v\nonumber$

So

$\dfrac{d\xi}{dt}=v\nonumber$

$v=v_0{cos \left({\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}t\right)\ }\nonumber$

The times where the velocity is zero is given by setting the inside equal to the zeros of cosine.

${\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}t=\dfrac{2n+1}{2}\pi\nonumber$

$t={\left(\dfrac{m}{k}\right)}^{1/2}\dfrac{2n+1}{2}\pi\nonumber$

The acceleration is

$\dfrac{dv}{dt}=a=-v_0{\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}{\sin \left[{\left(\dfrac{k}{m}\right)}^{\dfrac{1}{2}}t\right]\ }\nonumber$

If the \sin term is positive, the acceleration will be in the negative $$x$$ direction.

sine is positive from $$0$$ to $${\pi}$$, the acceleration will be in the negative $$x$$ direction for

$0<t\le {\left(\dfrac{m}{k}\right)}^{1/2}\pi\nonumber$

and in the positive direction from

${\left(\dfrac{m}{k}\right)}^{1/2}\pi<t\le {\left(\dfrac{m}{k}\right)}^{1/2}2\pi\nonumber$

### Q2.11

Verify that

$Y(x,t) = A \sin \left(\dfrac{2\pi }{\lambda}(x-vt) \right)\nonumber$

has a frequency $$\nu$$ = $$v$$/$$\lambda$$ and wavelength $$\lambda$$ traveling right with a velocity $$v$$.

Solution

All sine and cosine functions oscillate in a wave-like manner, so $$Y(x,t)$$ is a wave. We can write $$Y(x,t)$$ as

$Y(x,t) = A\sin(Bx - Ct)\nonumber$

where

$B = \dfrac{2\pi}{\lambda}\nonumber$

and

$C = \dfrac{2v\pi}{\lambda}\nonumber$

From these expressions we see that

Wavelength

$\lambda = \dfrac{2\pi}{B}\nonumber$

Frequency

$v = \dfrac{C}{2\pi} = \dfrac{v}{\lambda}\nonumber$

A standing wave has the equation $$y$$$$s$$ = $$A$$\sin($$k$$$$x$$$$s$$ ). The wave equation represented in this problem is related to the equation of a standing wave by $$x$$ = $$x$$$$s$$ + $$v$$$$t$$. The point $$x$$$$s$$ is arbitrary, and so we set the equation equal to 0. This gives $$x$$ = $$v$$$$t$$, so the wave is traveling right with velocity $$v$$.

## Q2.13

Explain (in words) how to expand the Hamiltonian into two dimensions and use it solve for the energy.

Solution

Begin by writing out the the Hamiltonian in two dimensions. This will include partial differential equations with respect to x and y. Then use that and the boundary conditions to use separation of variables to solve for the energy.

## Q2.13

Given that the Schrödinger equation for a two-dimensional box, with sides $$a$$ and $$b$$, is

$\dfrac{∂^2 Ψ}{∂x^2} + \dfrac{∂^2 Ψ}{∂y^2} +\dfrac{(8π^2mE) }{h^2}Ψ(x,y) = 0 \nonumber$

and it has the boundary conditions of

$$Ψ(0,y)= Ψ (a,y)=0$$ and $$Ψ(o,x)= Ψ(x,b)=0\] for all \(x$$ and $$y$$ values, show that

$E_{2,2}=\left(\dfrac{h^2}{2ma^2}\right)+\left(\dfrac{h^2}{2mb^2}\right).$

Solution

Let

$Ψ(x,y)= X(x)Y(y)\nonumber$

and use separation of variables.

$\dfrac{\partial^2 Ψ(x,y)}{\partial x^2} + \dfrac{\partial^2 Ψ(x,y)}{\partial y^2} +(8π^2mE / h^2) Ψ(x,y) = 0\nonumber$

$Y\partial^2 X/\partial x^2 +X\partial^2 Y/\partial y^2 +(8π^2mE / h^2) XY = 0\nonumber$

divide by XY to get:

$(1/X)\partial^2 X/\partial x^2 +(1/Y)\partial^2 Y/\partial y^2 +(8π^2mE / h^2) = 0\nonumber$

$(1/X)\partial^2 X/\partial x^2 +(1/Y)\partial^2 Y/\partial y^2 =-(8π^2mE / h^2)\nonumber$

Let

$(1/X)\partial^2 X/\partial x^2 =-M^2 and (1/Y)\partial^2 Y/\partial y^2 =-N^2\nonumber$

so that

$8π^2mE / h^2) =M^2+N^2\nonumber$

Apply the boundary conditions to get:

$X(x)=B \sin{Mx} = B \sin{\dfrac{n_x \pi x}{a}} for n_x =1,2,3...infty\nonumber$

and

$Y(y)=C\sin(Nx) = C\sin(n_yπx/b) for ny=1,2,3...\infty\nonumber$

since

$Ψ(x,y)=X(x)Y(y) \nonumber$

then

$$Ψ(x,y)=B \sin(\dfrac{n_xπx}{a}) C \sin(\frac{n_yπx}{b})\nonumber \] remember that $\dfrac{ 8\pi^2mE}{h2} =M^2+N^2\nonumber$ where \(M=n_xπ/a$$ and $$N=n_yπ/b$$

therefore

$\left(\dfrac{n_xπ}{a}\right)^2 + \left(\dfrac{n_yπ}{b}\right)^2](8π^2mE / h^2)=0 \nonumber$

so we get that

$E=\left[(\dfrac{n_xπ}{a})^2 + (\dfrac{n_yπ/b})^2\right](h^2/8\pi^2m)\nonumber$

now, you can plug in $$n_x=2$$ and $$n_y=2$$ to get

$E_{2,2}=(4h^2π^2/8mπ^2a^2)+(4h^2π^2/8mπ^2b^2) = (h^2/2ma^2)+(h^2/2mb^2)\nonumber$

### Q2.14

Explain, in words, how to expand the Schrödinger Equations into a 3 dimensional box.

Solution

Step 1: Set up the Hamiltonian for 3 dimensions

Step 2: set up the Schrödinger equation

Step 3: use separation of variables to solve it

Step 4: use solutions to solve for the energy.

## Q2.18

Solving for the differential equation for a pendulum gives us the following equation,

$\phi(x)= c_1\cos {\sqrt{\dfrac{g}{L}}} +c_2\sin {\sqrt{\dfrac{g}{L}}} \nonumber$

Assuming c1=2, c3= 5, g=7 and L=3, what is the position of the pendulum initially? Does this make sense in the real world. Why or why not? (We can ignore units for this problem).

Solution

since we are starting at the initial value, we can assume $$t_0=0$$. Plugging $$t_0=0$$ into our equation yields.

$\phi(x)= c_1 \cos \left((0)\sqrt{\dfrac{g}{L}}\right) +c_2 \sin \left((0)\sqrt{\dfrac{g}{L}}\right) \nonumber$

$$\phi (0) = 2$$

In real life, this makes sense because the pendulum has to start with some potential energy (in our case $$\phi= 2$$) so that it can be transferred to kinetic energy and start oscillating with time.

## Q2.23

Consider a Particle of mass $$m$$ in a one-dimensional box of length $$a$$. Its average energy is given by

$\langle{E}\rangle = \dfrac{1}{2m}\langle p^2\rangle\nonumber$

Because

$\langle{p}\rangle\ = 0\nonumber$

$\langle p^2\rangle = \sigma^{2}_{p}\nonumber$

where $$\sigma_p$$ can be called the uncertainty in $$p$$. Using the Uncertainty Principle, show that the energy must be at least as large as $$\hbar/8ma^2$$ because $$\sigma_x$$, the uncertainty in $$x$$, cannot be larger than $$a$$.

Solution

From the given information we know that

$\dfrac{\hbar}{2\sigma_p} < \sigma_x \le a\nonumber$

Then

$\dfrac{\hbar}{2a}\le\sigma_p\nonumber$

and so

$$\dfrac{\hbar^2}{4a^2}\le\sigma^{2}_{p} \label{1}$$

We are given that $$\langle{p^2}\rangle = \sigma^{2}_{p}$$ so we write

$$\dfrac{\sigma^{2}_{p}}{2m} \ = \ \dfrac{\langle{p^2}\rangle}{2m} = \langle{E}\rangle \label{2}$$

Substituting Equation \ref{1} into Equation \ref{2} give

$\dfrac{\hbar}{8ma^2} \le \langle{E}\rangle\nonumber$

## Q2.33

Prove $$y(x, t) = A\cos[2π/λ(x - vt)]$$ is a wave traveling to the right with velocity $$v$$, wavelength $$λ$$, and period $$λ/v$$.

To prove y is a wave you can use the wave equation

$\dfrac{ ∂^2y}{∂t^2} = V^2*∂^2y/∂x^2\nonumber$

where V is the velocity of the wave

$\dfrac{∂y}{∂} = A^2πv/λ*\sin[2π/λ(x - vt)z\nonumber$

$\dfrac{∂^2y}{∂t^2} = -A^24π^2v^2/λ^2 \cos[2π/λ(x - vt)]\nonumber$

$\dfrac{∂y}{∂x} = -A^2π/λ \sin[2π/λ(x - vt)]\nonumber$

$\dfrac{∂^2y}{∂x^2} = -A^24π^2/λ^2 \cos[2π/λ(x - vt)]\nonumber$

$v^2 \dfrac{∂^2y}{∂x^2} = -A^24π^2/λ^2 \cos[2π/λ(x - vt)] V^2\nonumber$

Finally equating these

$\dfrac{∂^2y}{∂t^2} = v^2 \dfrac{∂^2y}{∂x} \nonumber$

gives

$-\dfrac{A^24π^2v^2}{λ^2} \cos[2π/λ(x - vt)] = -A^24π^2/λ^2 \cos[2π/λ(x - vt)] V^2\nonumber$

Leaving only $$v = V$$, so this is in fact a traveling wave with positive velocity v.

You can write y as $$Acos(Bx - Ct)$$

where $$B = 2π/λ$$ and $$C = 2πv/λ$$

so $$λ$$ is easily seen to be $$2π/B$$ and the period $$T = λ/v = 2π/C$$.


text:
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Found in: Page 134

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# A particle is “thermal” if it is in equilibrium with its surroundings – its average kinetic energy would be $\frac{\mathbf{\text{3}}}{\mathbf{\text{2}}}{{\mathbf{\text{k}}}}_{{\mathbf{\text{B}}}}{\mathbf{\text{T}}}$ . Show that the wavelength of a thermal particle is given by ${\mathbf{\text{λ=}}}\frac{\mathbf{\text{h}}}{\sqrt{{\mathbf{\text{3mk}}}_{\mathbf{\text{B}}}\mathbf{\text{T}}}}$

As a result, were able to demonstrate that :

$\text{λ=}\frac{\text{h}}{\sqrt{{\text{3mk}}_{\text{B}}\text{T}}}$

See the step by step solution

## Step 1:Given and unknowns.

Particle's average kinetic energy is$\text{KE}=\frac{\text{3}}{\text{2}}{\text{k}}_{\text{B}}\text{T}$

$\lambda \text{=}\frac{\text{h}}{\sqrt{{\text{3mk}}_{\text{B}}\text{T}}}$

## Step 2: Concept Introduction

The following equation can be used to describe the de Broglie wavelength:

${\mathbf{\text{p=}}}\frac{\mathbf{\text{h}}}{\mathbf{\text{λ}}}{\mathbf{.}}{\mathbf{........}}{\mathbf{\text{(1)}}}$

When it comes to momentum, it's best to think of it this way:

${\mathbf{\text{p}}}{\mathbf{=}}{\mathbf{\text{mv}}}{\mathbf{.}}{\mathbf{......}}{\mathbf{\text{(2)}}}$

The kinetic energy, on the other hand, can be expressed by the following equation:

${\mathbf{\text{KE}}}{\mathbf{=}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}{{\mathbf{\text{mv}}}}^{{\mathbf{\text{2}}}}{\mathbf{.}}{\mathbf{.......}}{\mathbf{\text{(3)}}}$

## Step 3:Expression for wavelength.

Get the expression for using Equations (1) and (2)$\lambda$ :

$\frac{\text{h}}{\text{λ}}\text{=mv}$

$\text{λ=}\frac{\text{h}}{\text{mv}}$

## Step 4: Expression for v.

Now use the average kinetic energy and Equation (3) to get the expression for the particle's speed because don't have one

$\text{KE=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

$\text{KE=}\frac{\text{3}}{\text{2}}{\text{k}}_{\text{B}}\text{T}$

$\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{=}\frac{\text{3}}{\text{2}}{\text{k}}_{\text{B}}\text{T}$

${\text{v}}^{\text{2}}\text{=}\frac{\text{2×}\frac{\text{3}}{\text{2}}{\text{k}}_{\text{B}}\text{T}}{\text{m}}$

$\text{v=}\sqrt{\frac{{\text{3k}}_{\text{B}}\text{T}}{\text{m}}}$

## Step 5: Derived expression for v.

The generated expression is then plugged into the expression of $\text{λ}$:

$\begin{array}{c}\text{λ=}\frac{\text{h}}{\text{mv}}\\ \text{=}\frac{\text{h}}{\text{m×}\left(\sqrt{\frac{{\text{3k}}_{\text{B}}\text{T}}{\text{m}}}\right)}\\ \text{=}\frac{\text{h}}{\sqrt{{\text{3mk}}_{\text{B}}\text{T}}}\end{array}$

As a result, were able to demonstrate that$\text{λ=}\frac{\text{h}}{\sqrt{{\text{3mk}}_{\text{B}}\text{T}}}$


text:
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## Which one of the following doesn’t generate same language as rest? (i) (a+b)*a(a+b)*a(a+b)* (ii) b * a b * a (a + b)* (iii) (a + b)* a b* a b* (iv) b * a (a + b)* a b*

All of regular expression are equal  and it genrate the language  at least two a's


text:
# Finding Impulse Response from Block Diagram

1. Apr 26, 2015

### Graphite

1. The problem statement, all variables and given/known data

Where $$v = y - bu$$
It's given that the transfer function is
$$h(t) = b u(t) + v(t) = b \delta(t) + e^{at}b \mu(t)$$
2. Relevant equations

3. The attempt at a solution
I can't seem to figure out how the impulse response above was found. I understand that the impulse response is y when $u = \delta(t)$, but I'm not sure where the exponential comes from. One of the steps in the problem is
$$y(t) = bu(t) + \int_{t_{0}}^{t} e^{a(t-\tau)} abu (\tau) d \tau$$
where the second term on the right is v. Going by the diagram, shouldn't v be equal to the following?
$$v = \int_{t_{0}}^{t} u(\tau) ab + v(\tau)a \; d\tau$$

Last edited: Apr 26, 2015
2. Apr 26, 2015

### rude man

Is v' = abu + av or abu - av or what? You need to put + or - signs in front of all your arrows.
The evolution of the exponential term will be apparent if you replace the integrator with a 1/s block and solve for V(s) or Y(s) and then invert back to the time domain.
You wrote "μ" in the exponential term, I think you meant "u".

3. Apr 26, 2015

### Graphite

It's v' = abu + av. The arrows are all positive.
So taking the Laplace transform of v' gives me
$$\dot{V} = abu + ay \\ s V(s) = ab + a V(s)$$
Since u is a unit impulse, the transform would just be 1, right? After that,
$$V(s) = \frac{ab}{s-a} \frac{1}{s}$$
Using partial fractions, I'm getting
$$\frac{b}{s-a} - \frac{b}{s} \rightarrow b e^{at} - b$$
I understand where the exponential is coming from, but I know I'm making a mistake somewhere above since I should be getting $be^{at} \mu$

I double checked also, and the it is mu that is being used, not u. I think my professor is just using different notation, where u is input and mu is the unit step function.

4. Apr 26, 2015

### rude man

No. Look at the diagram.
Further, your "given' output y is wrong. It's y = bδ(t) + ab eat u(t).
But I think you have the right idea, and yes, the laplace transform of δ(t) is 1.

BTW I use u(t) to mean the step function. This is standard notation. I've never seen anyone use μ(t) for that. Use x(t) for the input, not u(t).

Last edited: Apr 26, 2015
5. Apr 26, 2015

### Graphite

I'm now seeing where I made my mistake.

Thanks for the help.


text:
Why does NASA plan to put a meteoroid in Lunar orbit instead of Earth orbit?

NASA is working on an asteroid retrieval mission. A small asteroid will be moved into Lunar orbit where it will be visited by space walking astronauts. Since it will be only max seven meters in diameter, I prefer to call it a meteoroid. The one which exploded over Chelyabinsk last year was about 20 meters in diameter and hence about almost nine times more massive. The retrieved meteoroid will be too small to pose any danger to Earth. Also, I suppose it will have a lower speed relative to Earth, than would an average interplanetary asteroid. They will look for a low delta-v asteroid.

Wouldn't it be easier to visit it in low Earth orbit and maybe use the ISS to investigate it?

It would take vastly more $\Delta V$ to get it to a low-Earth orbit. The targets selected are close enough to Earth's orbit about the Sun that it only takes around $200\,\mathrm{m/s}$ to get it into a distant retrograde orbit about the Moon. To get the thing to a low Earth orbit would be around $3\,\mathrm{km/s}$. The tyranny of the rocket equation makes that infeasible.

• Very interesting! That sounds like a potentially very good explanation. But could you please explain a bit about what basically makes it so? The Moon orbits Earth at only 1 km/s, that could intuitively explain 1/3 of the difference in your numbers. But why would near Earth asteroids have lower than that delta-V relative to the Moon than to Earth? Because they could be pushed into trajectories which pass by the Moon several times, maybe? – LocalFluff Jan 28 '14 at 14:17
• Because the Moon has very nearly escaped from the Earth. Going the other way, it takes about $3.1\,\mathrm{km/s}$ to get from LEO (low-Earth orbit) to an orbit that just reaches the Moon. From LEO, it takes $3.2\,\mathrm{km/s}$ to escape Earth's gravity completely. Only $100\,\mathrm{m/s}$ more. So in terms of $\Delta V$, the Moon is much closer to the rest of the Solar System than it is to Earth. – Mark Adler Jan 28 '14 at 15:19
• @LocalFluff Gravity well of the Moon is shallower since it's only about 1.23% the mass of the Earth. It's also non-negligible that it orbits the Earth, which can potentially also reduce required delta-v, depending on how the captured object moves relative to it. Since we're talking of NEO asteroids here that would already be at orbital velocities close to those of the Earth-Moon system, that difference wouldn't be small. What Mark answers in numbers, while I don't have any doubt is correct, also feels right. – TildalWave Jan 28 '14 at 15:19

Safety of our blue planet. Eventually, gravity anomalies would cause even a perfectly orbited object (a moonlet?) to preces and hit the body it orbits around. Since orbiting an asteroid means reducing large fraction of its momentum to bring it closer to celestials it naturally orbits (NASA's plans involve capturing a near-Earth asteroid, or NEO, as part of its Asteroid Initiative program) it wouldn't impact with such velocity as the Chelyabinsk meteor did (required LEO velocity is only about 8 km/s, which would have to be reduced to deorbit, while the Chelyabinsk meteor entered Earth's atmosphere at an estimated velocity of 18.6 km/s), which caused it to disintegrate to smaller fragments in the atmosphere due to friction and aerodynamic heating. So it would still pose a significant threat to Earth, even if it was much smaller than the Chelyabinsk superbolide was. I'm myself much more in favor of this happening on the Moon than on the Earth.

It also wouldn't make much difference for the safety of the astronauts studying this asteroid (while it's much bigger than about 1 m in diameter, it can't be considered a meteoroid) or much more difficult to reach, since it couldn't be orbited in Low Earth Orbit (LEO) due to still sufficient atmospheric drag decaying its orbit, and where astronauts would still be somewhat protected from solar wind and high-energy proton particles of cosmic radiation by the Van Allen radiation belt, or they could be reached easier or sent supplies by currently available orbital launch vehicles in case of something going sour.

It is also easier (smaller required delta-v) to later haul the rock back into a safe orbit in cis-lunar space once you're done studying it, if it's orbited around a smaller mass body that's already a significant portion away from the Earth's Hill sphere.

Weapon of Mass Destruction

Any large body in orbit is a potential weapon of mass destruction. Drop a 20-ton rock from orbit, and it may not survive, but if you do it right, it creates a crater some 100m across.

While NASA is not planning on using deadfall ortillery, the possibility is a political issue.

Ownership and access

Legally, all natural bodies in space are common property of all Earth's peoples.

Putting it in LEO would make it accessible to a variety of states that are seen as politically untrustworthy. The key example is North Korea - they have demonstrated the ability to launch SRBM's, which are precursors to a space program as well as to ICBMs. (The initial US space launchers were in fact developed as part of ICBM programs.) The concern that these nations may reach and deorbit a body as a form of terrorism is a real and present (but low probability) threat.

A Lunar Orbit location has far less access to everyone, however, it's a near total bar to most nations. The US, Russia, China, India, and the EU have the capability to get there, as evidenced by the ability to land lunar rovers and install lunar orbiters. It's still close enough for most real-time experimental controls.

More margin for error

The lack of atmosphere ➀ allows for a miss to not result in slowing and impact, nor to result in surface melting of the body.

Lower ∆V

Less overall energy is needed to put items into a stable Lunar orbit than a stable Earth orbit. The speed change needed is less, and thus the duration of acceleration and total thrust can be less.

Less to impact

Fundamentally, inserting anything into Earth orbit involves hitting the mark at the right time and speed to not hit anything else in Earth orbit. There are huge numbers (thousands) of tracked items, and hundreds of useful items, including a manned facility, that might be affected.

Lunar orbit is far less crowded, and no lunar orbiting equipment is essential infrastructure.

Much less impact on satellite orbits.

If I recall correctly, the intent is to move an asteroid of sub-kilometer diameter - but preferably over 100m diameter. These are objects large enough to have a noticeable effect upon other orbiting bodies. The cumulative effect could be catastrophic. The impact is tiny, but would be persistent and affect all other orbiting structures notably.

Less tidal stress

Tidal stress being the difference in pull between the ends of an object in orbit around a more massive object. Lunar orbit results in far less tidal stress than Earth orbit, due to the much lower lunar mass. (Note that Earth tidal stress will still exist in lunar orbit, but will be considerably less than even geosynch.

Proving the tech doesn't require a Earth orbit.

The technology in question is the ability to put asteroids into desired orbits. The principles are exactly the same whether that's Earth, Lunar, or Martian orbit. Success will result in proving the technology.

A base for future science

A lunar orbiting asteroid is a good base for further lunar and deep space oriented science. Being closer to the moon, it provides relay access for far-side rovers, as well as a location outside the van allen belts to test various practical radiation shielding approaches without having to land on the moon itself.

➀ Lunar atmosphere is present, but is at so low a fraction of a pascal at surface as to be safely ignored for orbital purposes.

• Thanks for your answers. Low deltaV is the only valid reason, I think, and it might be enough. But the terrorist paranoia is an illness and planet safety is a joke. The max size will be 8 meter with a mass of maybe 3 tons. A solar ion thruster could never do anything with a real asteroid for planetary defence or of economic importance. The meteoroid retrieval mission has obviously been designed only to use the ion thruster, SLS and Orion. Not for science, security or resource extraction. I read online that many are of the opinion that this crazy mission will never happen. – LocalFluff Feb 16 '14 at 11:00
• The initial proposal I read about was for a 100m asteroid - and using a mass driver, not an ion drive. The ion drive is more readily fundable. And NASA has done crazier things. The Kittinger jump comes to mind. – aramis Feb 17 '14 at 4:23
• For small-asteroid science, they could go to Chelyabinsk instead. They have half a ton pieces of a 20 meter asteroid highly accesible right there on the ground. That'd save a number of billions of dollars. But what to use the SLS+Orion and ion thruster for now? Cancellatiooon... – LocalFluff Feb 18 '14 at 11:14
• @LocalFluff you're utterly missing the point of moving an asteroid into orbit - not the least of which is determining if 0.01G is significantly different physiologically and psychologically, from freefall. A 20m asteroid is worthless for those purposes - a 100m asteroid, as the early proposals were, was sufficient to start looking at the effects separate from freefall. – aramis Feb 19 '14 at 7:22
• A 100 m asteroid would be different, yes. For planetary defence, for futuristic mining visions, and perhaps scientifically if they generally form quite differently than 7 m asteroids do. But now, 7 m it is! (Wouldn't it be much cheaper to put the largest recovered part of the Chelyabinsk meteorite in Lunar orbit instead?) :-p – LocalFluff Feb 19 '14 at 11:42


text:
# Central limit theorem and normality assumption of asset return distribution

Can central theorem justify normality assumption of assets return distribution? And if it can why the empirical evidence show this assumption, which many finance models are based on, is a far cry from reality?

No. I just published a paper on this. If return is defined as $$r_t=\frac{p_{t+1}q_{t+1}}{p_tq_t},$$ and since returns are not data while prices and volumes are, then it follows that the distribution of returns depends entirely upon the distribution of prices and the distribution of quantities. For example, in bankruptcy, $q_{t+1}=0$.

As it is a very long paper, there is no single distribution or even family of distributions involved in returns. Under the standard stated assumptions in Markowitzian style models, the distribution of returns would integrate out to be $$\frac{1}{\pi}\frac{\sigma}{\sigma^2+(r_t-\mu)^2}.$$ This assumption follows from the fact that there are many buyers and sellers and that stocks are sold in a double auction so the winner's curse does not obtain. If the winner's curse were present, as in antique auctions, then the distribution would be the ratio of two Gumbel distributions.

Regardless, with some interesting special exceptions, returns on equity securities cannot have a mean. Since there cannot be a mean return, then it follows that $\beta$ as defined in models such as the CAPM cannot exist. An assumption in all of these finance models has been that the parameters are known with probability one, but if you drop that assumption, you will find that no estimator exists that could converge to the population parameter.

Ratio distributions are well known and taught as standard graduate work in statistics. The most common method to find a ratio distribution if the underlying are continuous densities is that if $Z=\frac{Y}{X}$, then the density of $Z$ is $$g(z)=\int_{-\infty}^\infty|x|f(x,zx)\mathrm{d}x.$$

With this you can derive just about any form of return except single period discount bonds and cash-for-stock mergers as well as accounting ratios and growth rates.

As a rule of thumb, the central limit theorem is strongly violated for any financial return data, as well as quite a bit of macroeconomic data. Further, again as a rule of thumb, no non-Bayesian estimator exists for financial data. I will be presenting that along with a replacement for Black-Scholes at a conference in Albuquerque in a few weeks. The short form of this would go as follows:

For an equation such as $$x_{t+1}=\beta{x}_t+\epsilon_{t+1},\beta>1$$ only non-parametric regression methods such as Theil's regression or quantile regression could work. Bayesian methods are never less risky than Frequentist methods, though the converse is not true. If we assume that the Frequentist tool chosen has the same sampling density as the posterior density if the prior density were flat and nuisance parameters were marginalized out, then the Frequentist density would still be inadmissible.

The issue is that the Frequentist sampling distribution is symmetric and would include regions where $\beta\le{1}$. If that region has a density of $K$ and the region greater than unity has area $1-K$ then it follows that the Bayesian posterior density with a prior where $\Pr(\beta\le{1})=0$ will have zero mass in the region less than or equal to one, but whose density would be $$\frac{1-K}{K}$$ times greater than the equivalent non-Bayesian component.

Because admissibility can be defined in term of stochastic dominance, it follows that the Bayesian estimator always stochastically dominates a Frequentist estimator. As such, no admissible non-Bayesian method is available for most finance problems and most macroeconomic problems where growth is involved.

• Would you please give a link to the paper - Thanks – vonjd Feb 20 '18 at 21:51
• Is it this one? file.scirp.org/pdf/JMF_2017083015172459.pdf – vonjd Feb 20 '18 at 21:54
• Yes. You will be able to pick up the black-scholes paper in the proceedings of the SWFA Conference. – Dave Harris Feb 21 '18 at 4:07

Normality is just an assumption which can be used to inform a model. No model is ever right, but some might be useful. Also, the assumption of normality doesn't really depend on the actual process being normal. Rather, it just requires that an investor's decisions are driven by mean and variance. While the underlying process of asset price returns is almost never normal - per se - we usually assume normality because:

1. Returns are approximately normal;
2. When an outcome is produced by many small effects acting additively and independently, its distribution will be close to normal; and,
3. It is much harder to do economics without normality

An investor who assumes joint normality need not be interested in anything else because the normal distribution is maximum entropy distribution which is completely described by two parameters: mean $\mu$ and variance $\sigma^2$. Therefore, the assumption of normality imposes the minimal prior structural constraint beyond these moments. An investor who only cares about mean and variance who doesn't use normality is taking into account additional assumptions which may or may not be correct or robust.

There are theories which define investor preferences for moments beyond the first two, but not much evidence of which I am aware that supports such a position. See: The possible preferences of investors for higher than first 2 moments of return distribution?

Approximate normality of logarithmic returns is classically demonstrated through the Lindeberg–Lévy central limit limit theorem (CLT). Normality also arises due to a variation of the CLT in which the sample of means of any distribution will tend toward normal. So, when you begin to look at a large number of assets and returns, the distributions will just tend towards normality.

If however you decide not to assume normality, you should probably have a compelling reason since it is much more difficult to do economics without this assumption.

For example, variance minimization techniques, such as those defined under Modern Portfolio Theory (MPT), are implicitly predicated on the assumption of joint normality. Even though there will be a set of portfolio weights which minimizes variance regardless of the underlying distributions, correlation (which also does not assume normality) is only a complete measure of association if the joint multivariate distribution is normal; i.e., covariance is only an exhaustive measure of co-movement if the joint distributions are themselves normal. We can see this is true because the joint distribution of X and Y is defined by joint normality:

${\frac {1}{2\pi \sigma _{X}\sigma _{Y}{\sqrt {1-\rho ^{2}}}}}\iint _{X\,Y}\exp \left[-{\frac {1}{2(1-\rho ^{2})}}\left({\frac {X^{2}}{\sigma _{X}^{2}}}+{\frac {Y^{2}}{\sigma _{Y}^{2}}}-{\frac {2\rho XY}{\sigma _{X}\sigma _{Y}}}\right)\right]\,\mathrm {d} X\,\mathrm {d} Y$

Which through a proof can be show to produce:

$\sigma _{X+Y}={\sqrt {\sigma _{X}^{2}+\sigma _{Y}^{2}+2\rho \sigma _{X}\sigma _{Y}}},$

If now, we define $\omega_i \sigma^2_i=\sigma_X$, and $\omega_j \sigma^2_j=\sigma_Y$, then we get back the equation which is used as the basis of mean variance optimization of a two asset portfolio:

$\mathbb{E}[\sigma _{p}^{2}]=\omega_{i}^{2}\sigma _{i}^{2}+\omega_{j}^{2}\sigma _{j}^{2}+2\omega_{i}\omega_{j}\sigma _{i}\sigma _{j}\rho _{ij}$

So while the portfolio covariance matrix can always be computed, to the extent that underlying assets have returns which are not normal the optimization is likely to be spurious.


text:
# Building up to understand integrals/area under curve.

## Homework Statement

A particle starts at rest, then accelerates at a constant rate of 1 meter per second squared.

Perhaps a(t) = t

## The Attempt at a Solution

I have a series of calculus-related questions based on this statement, but first, I just want to know if the equation above, a(t)=t, and the graph below correctly represents the statement, "A particle starts at rest, then accelerates at a constant rate of 1 meter per second squared."

EDIT:

Or, is the correct equation for the statement: a(t) = 1

and the graph for the statement:

Last edited:

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
Yes your second graph/equation is correct. Your first graph is the correct graph for the velocity of the particle. I don't know how much calculus you know already, but acceleration is the derivative of velocity, so you can get your second graph by differentiating the first.

What you are asking is to solve the second-order differential equation $\displaystyle \frac{d^2 y}{dt^2}=1$. Luckily for you, solving this differential equation requires near to no work compared to other second-order equations.

What you do is to take the integral of the right hand side twice. You will end up with a function involving two constants because you are integrating twice. You can then find those constants giving the starting velocity and the starting position some values.

And about your graph, the second one represents the acceleration. What does the first one represent?


text:
# Example 01: Calculation of free energy#

In this example, a simple calculation of the Helmholtz free energy $$F(NVT)$$ is illustrated. We use a Fe BCC structure at 100K.

## Solid free energy#

The input file for calculation of solid BCC Fe at 100 K is provided in the folder. A detailed description of the input file is available here. The calculation can be started from the terminal using:

calphy -i input.yaml

This should give a message:

Total number of 1 calculations found

calphy is running in the background executing the calculation. A new folder called fe-BCC-100-0, and files called fe-BCC-100-0.sub.err and fe-BCC-100-0.sub.out will also be created.

After the calculation is over, the results are available in the report.yaml file. The file is shown below:

average:
spring_constant: '3.32'
vol/atom: 11.994539429692749
input:
concentration: '1'
element: Fe
lattice: bcc
pressure: 0.0
temperature: 100
results:
error: 0.0
free_energy: -4.263568357143783
pv: 0.0
reference_system: 0.015029873513789175
work: -4.278598230657573

The calculated free energy is $$-4.2636$$ eV/atom, the value reported in the publication is $$-4.2631147(1)$$ eV/atom. The calculated value can be further improved by increasing the system size, and by increasing the switching time.


text:
# Physical and Chemical Properties of Matter

We are all surrounded by matter on a daily basis. Anything that we use, touch, eat, etc. is an example of matter. Matter can be defined or described as anything that takes up space, and it is composed of miniscule particles called atoms. It must display the two properties of mass and inertia.

### Introduction

The different types of matter can be distinguished through two components: composition and properties. The composition of matter refers to the different components of matter along with their relative proportions. The properties of matter refer to the qualities/attributes that distinguish one sample of matter from another. These properties are generally grouped into two categories: physical or chemical.

Figure 1: Visual With Examples. Content from S.M.

### Physical (Properties and Changes)

Physical Property: A physical property is one that is displayed without any change in composition. (Intensive or Extensive)

Intensive properties: A physical property that will be the same regardless of the amount of matter.

• density: m/v
• color: The pigment or shade
• conductivity: electricity to flow through the substance
• malleability: if a substance can be flattened
• luster: how shiny the substance looks

Extensive Properties: A physical property that will change if the amount of matter changes.

• mass: how much matter in the sample
• volume: How much space the sample takes up
• length: How long the sample is

Physical Change: Change in which the matter's physical appearance is altered, but composition remains unchanged. (Change in state of matter)

• Three main states of matter are: Solid, Liquid, Gas
• Solid is distinguished by a fixed structure. Its shape and volume do not change. In a solid, atoms are tightly packed together in a fixed arrangement.
• Liquid is distinguished by its malleable shape (is able to form into the shape of its container), but constant volume. In a liquid, atoms are close together but not in a fixed arrangement.
• Gas is made up of atoms that are separate. However, unlike solid & liquid, a gas has no fixed shape and volume.

Example 1: Physical Change

When liquid water ($$H_2O$$) freezes into a solid state (ice), it appears changed; However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen  and 88.81% oxygen by mass.

Figure 2: Ice Melting is a physical change

### Chemical (Properties and Changes)

• Chemical Property: Any characteristic that gives a sample of matter the ability/inability to undergo a change that alters its composition. Examples: Alkali metals react with water; Paper's ability to burn.
• Chemical Change: Change in which one or more kinds of matter are transformed to new kinds of matter with altered compositions (or Chemical Reaction).
Example 2: Chemical Change

Magnesium + Oxygen → Magnesium Oxide

$2 Mg + O_2 \rightarrow 2 MgO$

Iron + Oxygen → Iron Oxide/ Rust

$4 Fe + 3O_2 \rightarrow 2 Fe_2O_3$

Using the components of composition and properties, we have the ability to distinguish one sample of matter from the others.

### Problems

1. Milk turns sour. This is a ________________
• Chemical Change

• Physical Change

• Chemical Property

• Physical Property

• None of the above

2. HCl being a strong acid is a __________, Wood sawed in two is ___________
• Chemical Change, Physical Change

• Physical Change, Chemical Change

• Chemical Property, Physical Change

• Physical Property, Chemical Change

• None of the above

3. CuSO4 is dissolved in water
• Chemical Change

• Physical Change

• Chemical Property

• Physical Property

• None of the above

4. Aluminum Phosphate has a density of 2.566 g/cm3
• Chemical Change

• Physical Change

• Chemical Property

• Physical Property

• None of the above

5. Which of the following are examples of matter?

• A Dog

• Carbon Dioxide

• Ice Cubes

• copper (II) nitrate

• A Moving Car

6. The formation of gas bubbles is a sign of what type of change?

7. True or False: Bread rising is a physical property.

8. True or False: Dicing potatoes is a physical change.

9. Is sunlight matter?

10. The mass of lead is a _____________property.

### Solutions

1)chemical change 2) chemical property, physical change 3) physical change 4) physical property 5) All of the above 6) chemical 7) False 8) True 9) No 10) physical property

### References

1. Petrucci, Bissonnette, Herring, Madura. General Chemistry: Principles and Modern Applications. Tenth ed. Upper Saddle River, NJ 07458: Pearson Education Inc., 2011. www.prenhall.com/petrucci
2. Cracolice, Peters. Basics of introductory Chemistry An active Learning Approach. Second ed. Belmont, CA 94001:Brooks/Cole, 2007. www.cengage.com/permissions

### Contributors

• Samantha Ma (UC Davis)

08:42, 15 Sep 2014

## Tags

### Textbook Maps

An NSF funded Project


text:
# Leibniz-Hopf algebra

Let $\mathbf{Z}\langle Z \rangle$ be the free associative algebra on $Z = \{Z_1,Z_2,\ldots\}$ over the integers. Give $\mathbf{Z}\langle Z \rangle$ a Hopf algebra structure by means of the following co-multiplication, augmentation, and antipode: $$\mu(Z_n) = \sum_{i+j=n\,;\,\,i,j \ge 0} Z_i \otimes Z_j$$ where $$Z_0 = 1$$ $$\epsilon(Z_n) = 0,\ \ n=1,2,\ldots$$ $$\iota(Z_n) = \sum_{i_1+\cdots+i_k = n} (-1)^k Z_{i_1} \cdots Z_{i_k} \ ,$$ where the sum is over all strings $i_1,\ldots,i_k$, $i_j \ge 1$, such that $i_1+\cdots+i_k = n$. This makes $\mathbf{Z}\langle Z \rangle$ a Hopf algebra, called the Leibniz–Hopf algebra. This Hopf algebra is important, e.g., in the theory of curves of non-commutative formal groups (see Formal group) [a1], [a2], [a5]. Its commutative quotient $\mathbf{Z}[Z]$, with the same co-multiplication, is the underlying Hopf algebra of the (big) Witt vector functor $R \mapsto W(R)$ (see Witt vector) and it plays an important role in the classification theory of unipotent commutative algebraic groups and in the theory of commutative formal groups (amongst other things) [a3].

The Leibniz–Hopf algebra $\mathbf{Z}\langle Z \rangle$ is free as a $\mathbf{Z}$-module and graded. Its graded dual is also a Hopf algebra, whose underlying algebra is the overlapping shuffle algebra $\mathrm{OSh}(\mathbf{N})$. As a $\mathbf Z$-module, $\mathrm{OSh}(\mathbf{N})$ is free with basis $\mathbf{N}^*$, the free monoid (see Free semi-group) of all words in the alphabet $\mathbf{N}$ with the duality pairing $\mathbf{Z}\langle Z \rangle \times \mathrm{OSh}(\mathbf{N}) \rightarrow \mathbf{Z}$ given by $$\left({ w , Z_{i_1}\cdots Z_{i_r} }\right) = \begin{cases}1,& w = i_1\cdots i_r ,\\ 0 &\text{otherwise}.\end{cases}$$

The overlapping shuffle product of two such words $u=(a_1,\ldots,a_s)$, $v = (b_1,\ldots,b_t)$ is equal to $$a \times_{\mathrm{OSh}} b = \sum_{f,g} (c_1,\ldots,c_r)$$ where the sum is over all $r \in \mathbf{N}$ and pairs of order-preserving injective mappings $f : \{1,\ldots,s\} \rightarrow \{1,\ldots,r\}$, $g : \{1,\ldots,t\} \rightarrow \{1,\ldots,r\}$ such that $\mathrm{im}(f) \cup \mathrm{im}(g) = \{1,\ldots,r\}$, and where $$c_i = a_{f^{-1}(i)} + b_{g^{-1}(i)}\ ,\ \ \ i = 1,\ldots,r$$ with $a_{f^{-1}(i)} = 0$ if $f^{-1}(i) = \emptyset$ , and similarly for $b_{g^{-1}(i)}$.

For example, $$(a) \times_{\mathrm{OSh}} (b_1,b_2) = (a,b_1,b_2) + (b_1,a,b_2) + (b_1,b_2,a) + (a+b_1,b_2) + (b_1,a+b_2) \ .$$ The terms of maximal length of the overlapping shuffle product form the shuffle product, see Shuffle algebra.

A word $w \in \mathbf{N}^*$, $w = (a_1,\ldots,a_s)$, is elementary if the greatest common divisor of $a_1,\ldots,a_s$ is $1$. With this terminology, the Ditters–Scholtens theorem [a4], [a5] says that, as an algebra over $\mathbf{Z}$, the overlapping shuffle algebra $\mathrm{Osh}(\mathbf{Z})$ is the free commutative polynomial algebra with as generators the elementary concatenation powers of elementary Lyndon words (see Lyndon word). (E.g., the third concatenation power of $(a_1,a_2)$ is $(a_1,a_2,a_1,a_2,a_1,a_2)$.) In contrast with the case of the shuffle algebra, this theorem already holds over $\mathbf{Z}$ (not just over $\mathbf{Q}$).

#### References

 [a1] E.J. Ditters, "Curves and formal (co)groups" Invent. Math. , 17 (1972) pp. 1–20 [a2] E.J. Ditters, "Groupes formels" , Lecture Notes , Univ. Paris XI: Orsay (1974) [a3] M. Hazewinkel, "Formal groups and applications" , Acad. Press (1978) MR0506881 MR0463184 Zbl 0454.14020 [a4] M. Hazewinkel, "The Leibniz Hopf algebra and Lyndon words" Preprint AM CWI , 9612 (1996) [a5] A.C.J. Scholtens, "$S$-Typical curves in non-commutative Hopf algebras" , Free Univ. Amsterdam (1996) (Thesis)

#### References

 [b1] Michiel Hazewinkel, "The Algebra of Quasi-Symmetric Functions is Free over the Integers", Advances in Mathematics 164 (2001) 283–300 DOI 10.1006/aima.2001.2017
How to Cite This Entry:
Leibniz-Hopf algebra. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Leibniz-Hopf_algebra&oldid=36844
This article was adapted from an original article by M. Hazewinkel (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article


text:
Simon Plouffe has found a formula which produces 50 prime numbers in a row:

\begin{align} a_0 &= 10^{500}+961 \\ a_{n+1} &= a_n^{\frac{101}{100}} \end{align}

$\lfloor a_n \rfloor$ is a prime for the first 50 $a_n$!

He's trying to find better formulas than those discovered by Wright and Mills, which produce infinitely many primes but grow really quickly.

Paper on the arXiv: arxiv.org/abs/1901.01849

(thanks @erou)

A Mastodon instance for maths people. The kind of people who make $\pi z^2 \times a$ jokes.

Use $ and $ for inline LaTeX, and $ and $ for display mode.


text:
# Eigenvalue analysis of preconditioned partial differential operator

today, I encountered a confused problem by accident, but I have no ideas to deal with it fully. The question can be described as follows: for example, when we need to use FDM/FEM to discrete the partial differential operator with variable coefficients, $\mathcal{L} = d_1(x,y)\frac{\partial^2 u(x,y)}{\partial x^2} + d_2(x,y)\frac{\partial^2 u(x,y)}{\partial y^2} + v(x,y)u(x,y)$, where the coefficients $d_1(x,y),d_2(x,y),v(x,y)\geq 0$. For simplicity, we use the following notations to explain the discretization by FDM: $$d_1(x,y),\ d_2(x,y),\ v(x,y) \rightharpoonup D_1, D_2, V\ \ (\mathrm{three\ diagonal\ matrices,\ whose\ entries\ are\ all\ nonnegative})\\ \frac{\partial^2 u(x,y)}{\partial x^2}\rightharpoonup L_1 (\mathrm{discretized\ matrix})\\ \frac{\partial^2 u(x,y)}{\partial y^2}\rightharpoonup L_2 (\mathrm{discretized\ matrix})\\$$ so we can obtain the discretized matrix of $\mathcal{L}$ as follows, $A:= D_1L_1 + D_2L_2 + VI$ ($I$ is the identity matrix). Maybe, we want to construct a preconditioner as $P:=d_1L_1 + d_2L_2+\hat{v}I$, where $d_1,d_2,\hat{v}$ are all the mean value of diagonal entries (for example, if $D_1 = diag(s_1,s_2,\ldots,s_n)$, then $d_1 = \frac{s_1 + s_2+\ldots + s_n}{n}$). in fact, this preconditioner can be regard as the discretized matrix of $\mathcal{L} = d_1\frac{\partial^2 u(x,y)}{\partial x^2} + d_2\frac{\partial^2 u(x,y)}{\partial y^2} + \hat{v}u(x,y)$, so I want to know that:

(1) Is this idea reasonable?

(2) if so, how to analyze the spectrum of preconditioned matrix $P^{-1}A$, for the sake of simplicity, take $v(x,y) = 0,$, I can write $$(d_1L_1 + d_2L_2)^{-1}(D_1L_1 + D_2 L_2) = I + (d_1L_1 + d_2L_2)^{-1}[(D_1 - d_1I)L_1 + (D_2-d_2 I) L_2],$$ it means that there are some eigenvalue closed to 1 and this can translate a rapid convergence of Krylov subspace methods? However, this analysis seems to be very rough. So are there some more detailed spectral analysis of preconditioned matrix ? Maybe, some one can provide some references to explain this idea ?

(3) Maybe, some other boundary condition can be also considered.

I have do a small test problem: the spectral distributions of original and preconditioned matrix, please see the following figures:

• Have you tried to do this with a small test problem? – Kirill Nov 2 '14 at 16:19
• @Kirill, thank you for your kind reminder, I have add the spectral distributions of a couple of small matrices. Also I use the GMRES(20) method to solve the resulted linear systems needed 1222 iteration steps, however, if I add the preconditioner $P$, it only needs 22 iteration steps. – Hsien-Ming Ku Nov 2 '14 at 17:37
• This is close to Jacobi preconditioner, so it seems reasonable, but is it better than Jacobi? How are you solving linear systems $P^{-1}x=b$? If you can demonstrate by experiment that it performs well, that alone might be sufficient, because analysing preconditioners mathematically can be difficult. – Kirill Nov 3 '14 at 1:19


text:
# Induction?

The question is quite famous and is generally included under the topic of induction, but I didn't actually catch how to do it. And it goes as follows:

Two sequences $(a_{1},a_{2},...)$ and $(b_{1},b_{2},...)$ are defined as $a_{1}=3, a_{n}=3^{a_{n-1}}$ and $b_{1}=4, b_{n}=4^{b_{n-1}}$. Show that $a_{1000}>b_{999}$.

Thank you to everyone who shall participate in the discussion and help me. Please explain how to do it, as well.

Note by Shourya Pandey
6 years, 4 months ago

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First, you should prove that $a_{n}>b_{n-1}$ using induction. For the base case, let $n=2$, and it is obvious that $a_{n}=a_2=3^{a_1}=3^3=27$ is greater than $b_{n-1}=b_1=4$.

Now we must show that $a_{n+1}>b_n$ if we assume that $a_n>b_{n-1}$. $a_{n+1}=3^{a_n}$ and b_n=4^{b_{n-1}}. \(3^{a_n}>4^{b_{n-1}}

$\log_4{3^{a_n}}>b_{n-1}$

$a_n \cdot \log_4{3} > b_{n-1}$

$\log_4{3}>\frac{b_{n-1}}{a_n}$

This is the best I could do. I'm not sure how to proceed from here.

- 6 years, 4 months ago

I had proceeded in a similar manner, but to no avail. Maybe we need to prove it for certain powers or pattern and complete the proof by reverse induction, because I am sure that reverse induction works. But a reference point is what is required to be found.

- 6 years, 4 months ago

That's is a good start. However, as you realized, you needed to use a statement that is stronger than the induction hypothesis that you currently have. That factor of $\log_4 3$ seems to mess things up.

This suggests that you should make your induction hypothesis stronger, which was what Sebastian was doing.

For example, if the induction hypothesis was that $a_{n+1} > 2 b_n$, then you can see that the previous statement we wanted is true. However, of course, this strengthen the inequality that we get in the induction step, and hence may longer be true. Thus part of this problem is to find the stronger induction statement to prove.

I call this method of induction proof "stronger induction" (this terminology is not standard), because you have to guess at a stronger statement to show than the obvious guess. I am planning a series of posts on induction, and hope to cover this at some point in the future.

Staff - 6 years, 4 months ago

Well so the "stronger induction" here refers to the "stronger inequality" which is being used to prove the statement true.

- 6 years, 4 months ago

The word stronger refers to proving a stronger statement, then that given by the (initial) induction hypothesis. In this particular case, it is a stronger inequality.

Another example of a question approached by "stronger induction" is as follows:

$a, b$ are positive integers such that $\frac {2ab}{a^2+b^2} = \sin \theta$, show that $(a^2+b^2)^n \sin n\theta$ is an integer.

For this example, the induction hypothesis of

Let $P_n$ be the proposition that $(a^2+b^2)^n \sin n\theta$ is an integer.

is not sufficient enough for us to work on the induction step.

Staff - 6 years, 4 months ago


text:
# Types of Numbers and Algebraic Properties

This section covers:

# Types of Numbers

Before we get too deep into algebra, we need to talk about the types of numbers there are out there. We saw a few of these earlier, and you may not have seen all these types of numbers yet, but you will have to learn them in school. The letters in parentheses indicate how they are abbreviated sometimes. Sorry, it’s not the most exciting stuff to learn….

The types of numbers we’ll talk about include Whole Numbers, Counting Numbers or Natural Numbers, Integers, Rational Numbers, Irrational Numbers, Real Numbers, Imaginary Numbers, and Complex Numbers:

 Type of Number Examples Hints Natural Numbers (Counting Numbers)  ($$\mathbb{N}$$) Numbers you use for counting: $$1, 2, 3,…$$ It’s “natural” to count on your fingers: $$1, 2, 3,…$$ Whole Numbers The natural numbers, plus $$0: 0, 1, 2, 3, …$$ The word “whole” has an “o” in it, so include 0. Integers  ($$\mathbb{Z}$$) Whole numbers, their opposites (negatives), plus $$0: … –2, –1, 0, 1, 2, …$$ Integers can be separated into negative, 0, and positive numbers. Rationals ($$\mathbb{Q}$$) Integers and all fractions, positive and negative, formed from integers. These include repeating fractions, such as $$\displaystyle \frac{1}{3}$$ or $$.33333…$$ or $$.\overline{3}$$. The word “rational” is a derivation of “ratio”, and rational numbers are numbers that can be written as a ratio of two integers. “Q” stands for quotient. Irrationals Numbers that cannot be expressed as a fraction, such as $$\pi ,\,\sqrt{2},\,e$$.  (We’ll learn about these later). If something is “irrational”, it’s not easy to explain or understand. Real Numbers ($$\mathbb{R}$$) Rational numbers and Irrational Numbers. The real number system can be represented on a number line: If a number exists on a number line that you can see, it must be “real”. Note that the “smallest” real number is negative $$(-)$$ infinity $$(-\infty)$$, and the largest real number is infinity $$(\infty)$$. We can never really get to these “numbers” ($$-\infty$$ and $$\infty$$), but we can indicate them as the “end” of the real numbers. Complex Numbers ($$\mathbb{C}$$) Real numbers, plus imaginary numbers (concept only, such as $$\sqrt{{-2}}=\sqrt{2}i$$). “Imaginary” numbers are difficult to imagine, since they are so “complex”.

Here’s a Venn Diagram that shows how the different types of numbers are related. Note that all types of numbers are considered complex. And don’t worry too much about the complex and imaginary numbers; we’ll cover them in the Imaginary (Non-Real) and Complex Numbers section.

# Algebraic Properties

Before we get into Algebra, we also need to talk about some of the properties we’ll use to solve equations.  We’ll need these to get the variable all by itself on one side of the equal sign – which is the basis of algebra.

Let me show you this with just plain numbers — since these work with plain numbers, they also work with variables (letters)!

## Algebraic Properties of Equality

There are what we call the Algebraic Properties of Equality, since they deal with two sides of an equal sign:

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 Example Property of Equality $$\begin{array}{c}5=5\\5+1=5+1\\\,\,\,\,\,\,\,6=6\,\,\,\,\,\surd \end{array}$$ We add 1 to both sides and get $$6=6$$. So addition works! Try it for other numbers. This is called the Additive Property of Equality, since we are adding the same thing to both sides of the equation. $$\begin{array}{c}5=5\\5-3=5-3\\\,\,\,\,\,\,\,\,2=2\,\,\,\,\,\surd \end{array}$$ If we subtract 3 from both sides, it works, too! This is called the Subtraction Property of Equality, since we are subtracting the same thing from both sides of the equation. $$\displaystyle \begin{array}{c}5=5\\5\times 4=5\times 4\\\,\,\,\,\,\,\,\,\,20=20\,\,\,\,\,\surd \end{array}$$ If we multiply both sides by 4, it works, too! This is called the Multiplicative Property of Equality, since we are multiplying the same thing to both sides of the equation. $$\displaystyle \begin{array}{c}6=6\\6\div 2=6\div 2\\\,\,\,\,\,\,\,\,3=3\,\,\,\,\,\surd \end{array}$$ If we divide both sides by 2, it works too! This is called the Division Property of Equality, since we are dividing the same thing on both sides of the equation.

## Commutative and Associative Properties

There are two more properties that will be very useful in solving algebra equations:

 Example Property $$\begin{array}{c}(5+4)+2=5+(4+2)\\9+2=5+6\\\,\,\,\,\,\,\,\,\,11=11\,\,\,\,\,\,\surd \end{array}$$ If we group numbers differently (using parentheses), we are using the Associative Property. This only works for addition and multiplication. Try it yourself with subtraction and division to see that it doesn’t work. I remember this since you associate yourself with different groups. $$\begin{array}{c}\,8\times 4\times 2=4\times 2\times 8\\32\times 2=8\times 8\\\,\,\,\,\,\,\,\,\,\,\,\,\,64=64\,\,\,\,\,\,\surd \end{array}$$ If we change the order of numbers or variables, we are using the Commutative Property. This only works for addition and multiplication. Try it yourself with subtraction and division to see that it doesn’t work. I remember this since the word “commutative” has an “o” for the second letter; this reminds me of “order”.

As an example of why the Associative and Commutative properties are important, we may need to use these to show that “$$5+4+2=2+2+7$$”:

\displaystyle \begin{align}5+4+2&=2+2+7\\5+4+2&=7+2+2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Commutative}\\\left( {5+4} \right)+2&=\left( {7+2} \right)+2\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Associative}\\9+2&=9+2\\11&=11\end{align}

## Distributive Property

There is one other property that is used a lot in algebra; this one is a little different.

Let’s say you are trying to take a collection for your sweet algebra teacher for an end of the year gift. You are collecting $10 from 10 girls and 8 boys. You can see that you will collect$180, but there are two different ways to solve this problem:

 $$10×(10 + 8)=\10×18=\180$$ You add up the girls and boys first to get 18 people, and then multiply by $10. OR $$(\10×10)+(\10×8)=\100+\80=\180$$ You first collect the money from the girls to get$100, and then collect the money from the boys to get $80, and then add the two amounts together to get$180.

This is called the Distributive Property, since we can either leave the 10 on the outside of the parentheses, or distributethrough (or “push it through“) to both the numbers on the inside of the parentheses. We can do this when there is addition or subtraction inside the parentheses.

Here is another example, and one using a variable:

 Example Distributive Property \displaystyle \begin{align}4\times (2+3)&=4\times 5\,\,=4+4+4+4+4\\&=(4+4)+(4+4+4)\\&=4\times 2+4\times 3\,\,\,\,\,\surd \end{align} I just randomly divided up 5 into 2 and 3. When you multiply 4 times 5 ($$2+3$$), it is just like adding 4 five times. Because of the associative property above, you can lump together the first two 4‘s first and the last three 4‘s second.   This “proves” the Distributive Property. $$\displaystyle 2n+3n=n(2+3)=n(5)=5n$$ Distributive Property again; you will see this again a lot in algebra. This is an example of “combining like terms”, or putting together the same variables.

# Summary of Algebraic Properties

Since these are pretty important, here’s another table with these properties (and a couple more) with new examples:

 Name Hints Examples Notes Associative “Grouping” You “associate” with different groups. $$\begin{array}{l}5+\left( {15+4} \right)\\\,\,\,=\left( {5+15} \right)+4\end{array}$$ Works with addition and multiplication, not subtraction or division. Commutative “Ordering” Since Commutative has an “o” in it, think “order”. $$\begin{array}{l}5+4+3\\\,\,=4+3+5\end{array}$$ Works with addition and multiplication, not subtraction or division. Distributive “Distributing or Pushing Through Parentheses” Think of “distributing” something to your friends. $$\displaystyle \begin{array}{*{20}{l}} \begin{array}{l}5\times \left( {3+4} \right)\\\,\,\,=\left( {5\times 3} \right)+\left( {5\times 4} \right)\\\,\,\,=15+20=35\end{array} \end{array}$$   $$\begin{array}{l}5-2\left( {x-3} \right)\\\,\,\,=5-2x+6\\\,\,\,=11-2x\end{array}$$   $$\begin{array}{l}5x+7x\\\,\,\,=\left( {5+7} \right)=12x\end{array}$$ When negatives are on the outside of the parenthesis, make sure you distribute the negatives to second number, too.   Remember that multiplying two negatives results in a positive. Identity “Staying the Same” You always come back to your “identity”. $$\begin{array}{c}9+0=9\\9\times 1=9\end{array}$$ Additive identity is 0.   Multiplicative identity is 1. Inverse “Undoing” When you put your car in “inverse”, you go backwards. \displaystyle \begin{align}9+-9=0\\9\times \frac{1}{9}=1\\\frac{8}{9}\times \frac{9}{8}=1\end{align} Additive inverse is $$-a$$, since $$-a+a=0$$.   Multiplicative inverse is $$\displaystyle \frac{1}{a}$$, since $$\displaystyle \frac{1}{a}\times \frac{a}{1}=1$$; the multiplicative inverse of $$\displaystyle \frac{a}{b}$$ is $$\displaystyle \frac{b}{a}$$, since $$\displaystyle \frac{a}{b}\times \frac{b}{a}=1$$. Multiplicative inverses are reciprocals.

There are actually other properties used in algebra that you’ll be learning, but these are the main ones you’ll be using to solve algebra problems. Remember that your goal in solving algebra problems is to get the variable or unknown to one side all by itself! We’ll see this in the Solving Algebraic Equations section.

# Proper Algebraic Notation

One more boring thing we must talk about before we solve equations is proper algebraic notation, or “grammar”.

Just like English has proper “grammar”, math does to!  The proper way to write the solutions of equations (and inequalities, which we’ll learn shortly) is shown below.

## Sets

A set of numbers (or anything!) is a collection of items that are called elements. A set can be finite, such as the numbers 1, 2, and 3 (written as $$\left\{ {1,2,3} \right\}$$). A set can also be infinite (with an unlimited set of numbers), such as the set of real numbers (including all the fractions) between 0 and 1.

### Union and Intersection

The union of two or more sets includes everything in either of the sets. For example, the union of the sets $$\left\{ {1,2,3} \right\}$$ and $$\left\{ {3,4,5} \right\}$$ would be $$\left\{ {1,2,3,4,5} \right\}$$, since you include everything in both sets, but don’t repeat numbers. You write union as $$\cup$$, so $$\left\{ {1,2,3} \right\}\cup \left\{ {3,4,5} \right\}=\left\{ {1,2,3,4,5} \right\}$$.

The intersection of two or more sets includes only those things that are in both sets.  For example, the intersection of the sets $$\left\{ {1,2,3} \right\}$$ and $$\left\{ {3,4,5} \right\}$$ would be $$\left\{ 3 \right\}$$, since you include only the numbers in both sets, but don’t repeat the numbers. You write intersection as $$\cap$$, so $$\left\{ {1,2,3} \right\}\cap \left\{ {3,4,5} \right\}=\left\{ 3 \right\}$$.

The way these are written (with the brackets) are called roster notation, since you have a “roster” or list of numbers.

### Set Builder, Inequality, and Interval Notation

Other notations are more useful and will be used by your teacher. These include set builder notation, inequality notation, and interval notation, as shown with examples. You can see how there may be many ways to show set builder notation.

 Example Words or Equation Set Builder/Inequality Notation Interval Notation $$x=4$$ $$\left\{ {x|x=4} \right\}$$ or $$\left\{ {x:x=4} \right\}$$ This is read “the set of all $$x$$ such that $$x$$ equals 4”. $$\left[ 4 \right]$$ (Not typically used) $$x\ne 6$$   All real numbers except 6 $$\left\{ {x|x\ne 6} \right\}$$ or $$\left\{ {x\in \mathbb{R}|\,\,x\ne 6} \right\}$$ or $$\left\{ {x|x\in \mathbb{R},\,\,\,x\ne 6} \right\}$$ This is read as “the set of all numbers $$x$$ such that $$x$$ is not 6 and $$x$$ is a real number”. $$\begin{array}{c}\left( {-\infty ,6} \right)\cup \left( {6,\infty } \right)\\\left( {-\infty ,6} \right)\,\,\text{or}\,\,\left( {6,\infty } \right)\end{array}$$   (Either way can be used) All Real Numbers $$\left\{ {x|x\in \mathbb{R}} \right\}$$ $$\left( {-\infty ,\infty } \right)$$ No Solution { } Ø (symbol for null set or “nothing”) $$x=4,5,6$$ $$\{4,5,6\}$$ $$\left[ 4 \right]\cup \left[ 5 \right]\cup \left[ 6 \right]$$ (Not typically used) $$\displaystyle -2\le x<4$$ ($$x$$ is greater than or equal to –2 and less than 4) $$\left\{ {x|-2\le x<4} \right\}$$ or $$\left\{ {x\in \mathbb{R}|-2\le x<4} \right\}$$ Inequality notation: $$-2\le x<4$$ $$[2,4)$$ (The 2 is included, but the 4 is not) $$\displaystyle x\le -2\text{ or}\,\,\text{ }x>2\text{ }$$ ($$x$$ is less than or equal to –2 or greater than 2) $$\left\{ {x\in \mathbb{R}|\,\,x\le -2\,\,\text{or}\,\,x>2} \right\}$$ or $$\left\{ {x|x\le -2\,\,\text{or}\,\,x>2} \right\}$$ Inequality notation: $$x\le -2\,\,\text{or}\,\,x>2$$ $$\left( {-\infty ,-2} \right]\text{ }\cup \text{ }\left( {2,\infty } \right)$$ $$\left( {-\infty ,-2} \right]\text{ or }\left( {2,\infty } \right)$$ Positive even numbers $$\left\{ {x\in \mathbb{N}|\,\,x=2n} \right\}$$ or $$\left\{ {x|\,\,x=2n\,\,\,\text{and}\,\,\,x\in \mathbb{N}} \right\}$$ Not applicable; not an interval

Note that for interval notation, if a “less than” ($$<$$) is used, we use “(“, if a greater than ($$>$$) is used, we use “)”. If a “less than or equal to” ($$\le$$) is used, we use “[“, and if “greater than or equal to” is used ($$\ge$$), we use “]”. Just remember that we used brackets (also called hard brackets) if we are including the endpoint (the actual number at the end), and we use parentheses (also called soft brackets) if we are not including the endpoint.

Also note that since we never actually get to $$\infty$$ or $$-\infty$$, we only use soft brackets (parentheses) with them.

Remember that $$\mathbb{R}$$ means all real numbers (everything on the number line), and $$\mathbb{N}$$ means natural numbers (1, 23, and so on). Also remember that Ø means “no answer” or “no solution”; this happens sometimes in algebra.

We’re going over this now, since we’ll be talking about inequalities soon, and it will get a little more complicated on how to write our answers. Don’t worry too much if you’re overwhelmed with all this (it’s like learning a new language!); you’ll probably give your answers in simple inequality notation or more likely interval notation.

Learn these rules, and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device.  Enjoy!

On to Solving Algebraic Equations – you are ready!

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text:
$f : x \mapsto \frac{1}{2 \pi \cdot \sigma} \exp \left[ -\frac{1}{2} \left( \frac{x - \mu}{\sigma} \right)^2 \right]$
where $$\mu = \mathtt{mean}$$ and $$\sigma = \mathtt{std}$$.
• mean (float, optional, default: 0.5) – Mean of the Gaussian PDF. The default value places the mean in the middle of the particle layer (at $$x = 0.5$$).
• std (float, optional, default: 1/6) – Standard deviation of the Gaussian PDF. The default value is such that the integral of the Gaussian PDF over the $$[\mu - 0.5, \mu + 0.5]$$ interval is about 99.7% (3σ).


text:
1. General Seminars

# CYGNO: an Optical TPC for the search of rare events

## by Davide Pinci (ROMA1)

Europe/Rome
Zoom (Piattaforma Join Zoom Meeting)

### Zoom

#### Piattaforma Join Zoom Meeting

Description

The search for a novel technology able to detect and reconstruct nuclear and electron recoil events in the keV energy range has become more and more important as long as vast regions of high mass WIMP-like Dark Matter candidate have been excluded.
Gaseous Time Projection Chambers (TPC) with optical readout are very promising candidate combining the complete event information provided by TPC technique to the high sensitivity and granularity of last generation scientific light sensors.
CYGNO experiment (a CYGNus module with Optical readout) is developing this approach obtaining good energy and space resolution and a high sensitivity in the keV energy range together with a very good particle identification useful to distinguish nuclear recoils from electronic recoils.
This experiment is part the CYGNUS proto-collaboration which aims at constructing a network of underground observatories for directional Dark Matter search.
A 1 cubic meter demonstrator is expected to be built in 2021/22 aiming to a larger scale apparatus (30m$^3$-100 m$^3$), in a later stage.

Join Zoom Meeting

https://uniroma1.zoom.us/j/84281361690?pwd=VWdncTBHWHNLVHVuNTltcUVzcVMrZz09

Meeting ID: 842 8136 1690
Passcode: 333703

Organized by

L. Cardani, C. Rovelli


text:
TABLE 2

The continuous-time limit predictions of linkage disequilibrium ϒ under a modified Wright–Fisher population model

Timescale
CoalescenceRecombinationLimit process
α > 1β > 1
β = 1
β < 10
α = 1β > 1ϒ3(ϕ, ω)
β = 1ϒ2(η, ϕ, ω)
β < 10
α < 1β > 1ϒ(ω)
β = 1ϒ(ω)
0 < α < β < 1ϒ(ω)
0 < α = β < 1ϒ1(η, ϕ, ω)
0 < β < α < 10
• The different formulas are the limit (9) obtained given different values of α and β. See text for explanation of symbols. Note that .


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# Learning with Techwhiff

#### Trending Questions

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NEED HELP ASAP Valley Company's adjusted trial balance on August 31, its fiscal year-end, follows. It categorizes the following accounts as selling expenses: sales salaries expense, rent expense-selling space, store supplies expense, advertising expense. It categorizes the remaining expenses a...
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Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. P(x) = x(x – 6)(x + 2) -10 1-5 A 5:55 PM 102019 ser...
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A solid sphere of mass m and radius R rolls down a parabolic path PQR from height H without slipping (assume R « H) as shown in the figure below. Path PQ is rough (and so the shell will roll on that path), whereas path QR is smooth, or frictionless (so the shell will only slide, not roll, in t...


text:
Multiplying decimals 3 (standard algorithm)

Practice multiplying two numbers that are written to the tenths, hundredths, or thousandths place.

Problem

58, point, 3, times, 839, equals, question mark


text:
# What is Hess's law?

##### 1 Answer
Apr 16, 2016

A restatement of the principle of conservation of energy.

#### Explanation:

Hess' law of heat summation states that the enthalpy change of a chemical reaction depends upon the initial and final states of the chemical reaction, and is independent of path.

Another way of saying this is:

$\text{TANSTAAFL, There Ain't No Such Thing As A Free Lunch}$