text:
1. rationalize the denominator

If you rationalize the denominator of
then you will get

2. Re: rationalize the denominator

Multiply the fraction with $\frac{5x\sqrt{5}+3y\sqrt{3}}{5x\sqrt{5}+3y\sqrt{3} }$


text:
# Emacs Etags Shortcut Functions

A few functions to let me manage TAGS files more easily. Typically, my projects contain at least one sub-folder. I got sick of manually updating, so I wrote this to help me update a single TAGS file per project (always in the projects' root directory).

Please share your thoughts (aside from "Why are you using Emacs?", you're not likely to change my mind).

One thing I've noticed that could trip me up is that it only tags files of one language (and if I had a project that had both Haskell and JS code, for example, I could tag one or the other). I would particularly appreciate advice on better ways to write find-parent-tags than just building each path from available sub-strings.

(defun create-tag-table ()
"This will recursively tag all files of a given type starting with the current buffers' directory.
It overwrites the old TAGS file, if one exists.
Haskell files assume you've installed hasktags."
(interactive)
(let ((file-type (get-taggable-extension))
(cur-dir default-directory)
(tags-file (find-parent-tags default-directory)))
(if (equalp file-type "hs")
(shell-command "hasktags --ignore-close-implementation --etags find . -type f -name \"*.*hs\"")
(shell-command (concat "find -name \"*." file-type "\" -print | etags -")))))

(defun find-parent-tags (dir)
"Traverses the directory tree up to /home/[user]/ or / whichever comes first.
Returns either nil or the directory containing the first TAGS file it finds."
(interactive (list default-directory))
(find-parent-tags-rec (build-tag-paths dir)))

(defun find-parent-tags-rec (list-of-filepath)
(cond ((null list-of-filepath) nil)
((file-exists-p (car list-of-filepath)) (car list-of-filepath))
(t (find-parent-tags-rec (cdr list-of-filepath)))))

(defun build-tag-paths (dir-string)
(build-tag-paths-rec (remove-if #'empty-string? (split-string dir-string "/")) (list "/")))

(defun build-tag-paths-rec (steps acc)
(if (null steps)
(mapcar (lambda (p) (concat p "TAGS")) acc)
(build-tag-paths-rec (cdr steps)
(cons (concat (car acc) (car steps) "/") acc))))

(defun get-taggable-extension ()
"Either returns the current file's extension (if it's appropriate) or asks the user to pick one with completion"
(let ((b-name (buffer-file-name (current-buffer)))
(valid-exts (list "lisp" "py" "c" "hs" "rb" "ss" "scm" "js" "erl" "el")))
(string-match "\\.\$$.*?\$$\$" b-name)
(let ((current-filetype (match-string 1 b-name)))
(if (member current-filetype valid-exts)
current-filetype
(completing-read "File type: " valid-exts nil 'confirm)))))

(defun empty-string? (s) (equalp s ""))

(provide 'tagariffic)
-

### General notes

• Always prefix your functions with the name of your package or project, e.g. tagariffic-create-tag-table and so on. Emacs has no package scoping.
• Whenever possible (i.e. in the absence of mutual recursion), define your functions before using them. It's a lot clearer for the reader. It's also necessary if you ever define macros.
• You're using some functions from the cl package, so you should start with (require 'cl).
• The usual style for predicates in Lisp is name-p, not name? (that's Scheme).
• Emacs has functions to manipulate file and directory names. Use them rather than going around concatenating slashes; this way your code will work on exotic platforms such as Windows.

### Finding the tags file

Emacs 23 has a function locate-dominating-file (in files.el).

Look up the directory hierarchy from FILE for a file named NAME.
Stop at the first parent directory containing a file NAME, and return the directory. Return nil if not found.

Your (find-parent-tags default-directory) is almost equivalent to (locate-dominating-file (buffer-file-name) "TAGS"). If you want to use the current directory rather than the directory of the current buffer's file (they're almost always the same), use (locate-dominating-file (concat (default-directory) ".") "TAGS"). locate-dominating-file stops at your home directory, too.

The code you posted does nothing with the tags-file variable. Did you mean to run the tag generation command in that directory, or failing that in the current directory? If so, this should do it:

(let ((default-directory (if tags-file (file-name-directory tags-file) default-directory)))
(shell-command …))

### From file type to shell commands

Rather than build in an exception for Haskell, the usual style would be to define an alist from extensions to commands.

(defvar tagariffic-command-alist
(t . "… etags …")))

However, there's a better approach than separate commands for each type; see below.

### The shell commands

Your hasktags command will choke on file or directory names that contain shell special characters: \[?* and whitespace. This isn't very common in source trees, but it could happen. Generally speaking, never use command substitution to generate a list of file names. Instead, let find call the command.

find . -type f -name '*.hs' -exec hasktags --ignore-close-implementation --etags {} +

A second problem is that if there are too many files, the command you wrote will die of a command line too long error; the one I just showed will run hasktags several times with command lines of a permitted size, causing the tag file to be overwritten. As a workaround, start by creating an empty tags file and tell hasktags to append to the tags file.

: >TAGS
find . -type f -name '*.hs' -exec hasktags --ignore-close-implementation -e -a {} +

An improvement to the find commands would be to ignore certain directories belonging to version control systems. Especially with svn, this could speed things up.

find . -type d $$-name .bzr -o -name .git -o -name .hg -o -name .svn -o -name CVS -o -name _darcs$$ -prune -o \
-type f …

Above I mentioned a different approach to choosing the tag generation command. Your approach assumes that a given project will be in a single language. It's easy enough to cater for mixed-language projects: build a single tags file containing the output generated by etags and hasktags.

: >TAGS
find . -type d $$-name .bzr -o …$$ -prune -o \
-type f -name '*.hs' -exec hasktags -e -a {} + -o \
-type f $$-name '*.[CScls]' -o -name '*.el' -o …$$ -exec etags {} +
-
FWIW - a detail: I disagree with define your functions before you use them. Sometimes that is good practice, sometimes not. Sometimes things are more readable top-down, with the detailed utility functions defined after the high-level ones. IOW, master-detail order. And this is more the case in a declarative (e.g. functional, logic) language. [It is of course the case that sometimes definition order is imposed, as for macros, as you pointed out.] – Drew Sep 16 '14 at 22:29

format: plain


text:
# Sodium is always kept in kerosene. or In the laboratory, sodium is immersed in kerosene. - Science and Technology 1

Explain
Short Note

Write scientific reason.

Sodium is always kept in kerosene.

In the laboratory, sodium is immersed in kerosene.

#### Solution 1

Sodium is a very reactive metal. It is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide. This is a strongly exothermic reaction and lot of heat is generated.

#### Solution 2

1. Sodium reacts with moisture present in air, even at room temperature, to form sodium hydroxide and hydrogen gas.
$\ce{\underset{\text{Sodium}}{2Na_{(s)}} + \underset{\text{Water}}{2H2O}_{(l)} -> \underset{\text{hydroxide}}{\underset{\text{Sodium}}{2NaOH}_{(aq)}} + \underset{\text{Hydrogen}}{2H_{2(g)}}}$
2. The reaction is exothermic and therefore, the released hydrogen gas readily catches fire on exposure to air.
Hence, to prevent accidents, sodium is stored in kerosene.
Concept: Type of Elements: Metals
Is there an error in this question or solution?


text:
# Authorea Help

This post showcases some complex tables created using LaTeX. In Authorea Beta, Insert -¿ LaTeX and use the source code added after each table.

$$m$$ $$\Re\{\underline{\mathfrak{X}}(m)\}$$ $$-\Im\{\underline{\mathfrak{X}}(m)\}$$ $$\mathfrak{X}(m)$$ $$\frac{\mathfrak{X}(m)}{23}$$ $$A_{m}$$ $$\varphi(m)\ /\ ^{\circ}$$ $$\varphi_{m}\ /\ ^{\circ}$$


text:
# covariant derivative vs. exterior derivative

I have the following question. Let $M$ be a Riemannian manifold with metric $g$ and $\nabla$ the Levi-Civita connection. Let furthermore $\alpha \in \Omega^{k}(M)$ be a $k$-form such that $\nabla \alpha = 0$. Why is then $d \alpha = 0$?

thanks, jan


text:
## Solving Ramanujan’s Puzzling Problem

Consider a sequence of functions as follows:- $f_1 (x) = \sqrt {1+\sqrt {x} }$ $f_2 (x) = \sqrt{1+ \sqrt {1+2 \sqrt {x} } }$ \$ f_3 (x) = \sqrt {1+ \sqrt…

## Its a Mystery! – Simulacrum in Eagle Nebula

This is one of the top mysteries of universe.

## CV Raman, Raman Effect on Raman Spectroscopy and Raman Scattering

Chandrasekhar Venkat Raman Chandrasekhar Venkat Rama, also known as Sir CV Raman was a Physicist, Mathematician and a Nobel Laureate. Venkat (his first name) was a Tamil Brahmin and was the second of the eight…

## Scientific Temper in Indian Students

Read these two facts: There are 49 billionaires in India. There are 73 Nobel Laureates in MIT. I have heard people talking about the quality of education in India. Everyone assumes that it is in…

## Environment Initiatives : Few Good Steps for a Happier Future

Our earth is at a very crucial stage. From this position, only we can make sure that where would our future go? Few Authorities and Initiatives are trying their most to save our earth &…

## Cloud Computing

Intro The much talked about Cloud Computing promises to change the way Information Technology People look at delivering IT and the way business people perceive it. This one is now a general and important part…

## Clean Technology – Why it is the time to be Green?

What is the 'Need' of the, "Clean Technology"? Why Not? There is a desperate need to meet the challenges caused by climate change and environmental degradation. In the past, it was the entrepreneurial spirit that…

## Touch Screen Systems

A typical touchscreen input system is basically an input device like a mouse or trackpad. If counting as a definition, a touch screen is a computer display screen that is sensitive to human touch, allowing…

## Elementary Particle Physics

Particle physics is the study of Particles, from what everything is made of. In this section of physics we study the fundamental particles that make up all of matter, and their mutual interaction. Everything around…


text:
Main Content

# Solve Partial Differential Equation of Nonlinear Heat Transfer

This example shows how to solve a partial differential equation (PDE) of nonlinear heat transfer in a thin plate.

The plate is square, and its temperature is fixed along the bottom edge. No heat is transferred from the other three edges since the edges are insulated. Heat is transferred from both the top and bottom faces of the plate by convection and radiation. Because radiation is included, the problem is nonlinear.

The purpose of this example is to show how to represent the nonlinear PDE symbolically using Symbolic Math Toolbox™ and solve the PDE problem using finite element analysis in Partial Differential Equation Toolbox™. In this example, perform transient analysis and solve the temperature in the plate as a function of time. The transient analysis shows the time duration until the plate reaches its equilibrium temperature at steady state.

### Heat Transfer Equations for the Plate

The plate has planar dimensions 1 m by 1 m and is 1 cm thick. Because the plate is relatively thin compared with the planar dimensions, the temperature can be assumed to be constant in the thickness direction, and the resulting problem is 2-D.

Assume that convection and radiation heat transfers take place between the two faces of the plate and the environment with a specified ambient temperature.

The amount of heat transferred from each plate face per unit area due to convection is defined as

${Q}_{c}={h}_{c}\left(T-{T}_{a}\right)$,

where ${T}_{a}$ is the ambient temperature, $T$ is the temperature at a particular $x$ and $y$ location on the plate surface, and ${h}_{c}$ is a specified convection coefficient.

The amount of heat transferred from each plate face per unit area due to radiation is defined as

${Q}_{r}=ϵ\sigma \left({T}^{4}-{T}_{a}^{4}\right)$,

where $ϵ$ is the emissivity of the face and $\sigma$ is the Stefan-Boltzmann constant. Because the heat transferred due to radiation is proportional to the fourth power of the surface temperature, the problem is nonlinear.

The PDE describing the temperature in this thin plate is

`$\rho {C}_{p}{t}_{z}\frac{\partial T}{\partial t}-k{t}_{z}{\nabla }^{2}T+2{Q}_{c}+2{Q}_{r}=0$`

where $\rho$ is the material density of the plate, ${C}_{p}$ is its specific heat, ${t}_{z}$ is its plate thickness, $k$ is its thermal conductivity, and the factors of two account for the heat transfer from both of its faces.

### Define PDE Parameters

Set up the PDE problem by defining the values of the parameters. The plate is composed of copper, which has the following properties.

```kThermal = 400; % thermal conductivity of copper, W/(m-K) rhoCopper = 8960; % density of copper, kg/m^3 specificHeat = 386; % specific heat of copper, J/(kg-K) thick = 0.01; % plate thickness in meters stefanBoltz = 5.670373e-8; % Stefan-Boltzmann constant, W/(m^2-K^4) hCoeff = 1; % convection coefficient, W/(m^2-K) tAmbient = 300; % the ambient temperature emiss = 0.5; % emissivity of the plate surface```

### Extract PDE Coefficients

Define the PDE in symbolic form with the plate temperature as a dependent variable $T\left(t,x,y\right)$.

```syms T(t,x,y) syms eps sig tz hc Ta rho Cp k Qc = hc*(T - Ta); Qr = eps*sig*(T^4 - Ta^4); pdeeq = (rho*Cp*tz*diff(T,t) - k*tz*laplacian(T,[x,y]) + 2*Qc + 2*Qr)```
```pdeeq(t, x, y) =  ```

Now, create coefficients to use as inputs in the PDE model as required by Partial Differential Equation Toolbox. To do this, first extract the coefficients of the symbolic PDE as a structure of symbolic expressions using the `pdeCoefficients` function.

`symCoeffs = pdeCoefficients(pdeeq,T,'Symbolic',true)`
```symCoeffs = struct with fields: m: 0 a: 2*hc + 2*eps*sig*T(t, x, y)^3 c: [2x2 sym] f: 2*eps*sig*Ta^4 + 2*hc*Ta d: Cp*rho*tz ```

Next, substitute the symbolic variables that represent the PDE parameters with their numeric values.

```symVars = [eps sig tz hc Ta rho Cp k]; symVals = [emiss stefanBoltz thick hCoeff tAmbient rhoCopper specificHeat kThermal]; symCoeffs = subs(symCoeffs,symVars,symVals);```

Finally, since the fields `symCoeffs` are symbolic objects, use the `pdeCoefficientsToDouble` function to convert the coefficients to the `double` data type, which makes them valid inputs for Partial Differential Equation Toolbox.

`coeffs = pdeCoefficientsToDouble(symCoeffs)`
```coeffs = struct with fields: a: @makeCoefficient/coefficientFunction c: [4x1 double] m: 0 d: 3.4586e+04 f: 1.0593e+03 ```

### Specify PDE Model, Geometry, and Coefficients

Now, using Partial Differential Equation Toolbox, solve the PDE problem using finite element analysis based on these coefficients.

First, create the PDE model with a single dependent variable.

```numberOfPDE = 1; model = createpde(numberOfPDE);```

Specify the geometry for the PDE model—in this case, the dimension of the square.

```width = 1; height = 1;```

Define a geometry description matrix. Create the square geometry using the `decsg` (Partial Differential Equation Toolbox) function. For a rectangular geometry, the first row contains `3`, and the second row contains `4`. The next four rows contain the $x$-coordinates of the starting points of the edges, and the four rows after that contain the $y$-coordinates of the starting points of the edges.

```gdm = [3 4 0 width width 0 0 0 height height]'; g = decsg(gdm,'S1',('S1')');```

Convert the DECSG geometry into a geometry object and include it in the PDE model.

`geometryFromEdges(model,g);`

Plot the geometry and display the edge labels.

```figure; pdegplot(model,'EdgeLabels','on'); axis([-0.1 1.1 -0.1 1.1]); title('Geometry with Edge Labels Displayed');```

Next, create the triangular mesh in the PDE model with a mesh size of approximately 0.1 in each direction.

```hmax = 0.1; % element size msh = generateMesh(model,'Hmax',hmax); figure; pdeplot(model); axis equal title('Plate with Triangular Element Mesh'); xlabel('X-coordinate, meters'); ylabel('Y-coordinate, meters');```

Specify the coefficients in the PDE model.

```specifyCoefficients(model,'m',coeffs.m,'d',coeffs.d, ... 'c',coeffs.c,'a',coeffs.a,'f',coeffs.f);```

### Find Transient Solution

Apply the boundary conditions. Three of the plate edges are insulated. Because a Neumann boundary condition equal to zero is the default in the finite element formulation, you do not need to set the boundary conditions on these edges. The bottom edge of the plate is fixed at 1000 K. Specify this using a Dirichlet condition on all nodes on the bottom edge (edge E1).

`applyBoundaryCondition(model,'dirichlet','edge',1,'u',1000);`

Specify the initial temperature to be 0 K everywhere, except at the bottom edge. Set the initial temperature on the bottom edge E1 to the value of the constant boundary condition, 1000 K.

```setInitialConditions(model,0); setInitialConditions(model,1000,'edge',1);```

Define the time domain to find the transient solution of the PDE problem.

```endTime = 10000; tlist = 0:50:endTime;```

Set the tolerance of the solver options.

```model.SolverOptions.RelativeTolerance = 1.0e-3; model.SolverOptions.AbsoluteTolerance = 1.0e-4;```

Solve the problem using `solvepde`. Plot the temperature along the top edge of the plate as a function of time.

```R = solvepde(model,tlist); u = R.NodalSolution; figure; plot(tlist,u(3,:)); grid on title 'Temperature Along the Top Edge of the Plate as a Function of Time' xlabel 'Time (s)' ylabel 'Temperature (K)'```

Based on the plot, the transient solution starts to reach its steady state value after 6000 seconds. The equilibrium temperature of the top edge approaches 450 K after 6000 seconds.

Show the temperature profile of the plate after 10,000 seconds.

```figure; pdeplot(model,'XYData',u(:,end),'Contour','on','ColorMap','jet'); title(sprintf('Temperature in the Plate, Transient Solution (%d seconds)\n', ... tlist(1,end))); xlabel 'X-coordinate, meters' ylabel 'Y-coordinate, meters' axis equal;```

Show the temperature at the top edge at 10,000 seconds.

`u(3,end)`
```ans = 449.8031 ```


text:
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# If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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02 Sep 2018, 02:07
Why the pair (x + y) and (x - y) is not considered? It would yield x² - y² and there is an integer (1) multiplying y.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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02 Sep 2018, 02:24
thiagod wrote:
Why the pair (x + y) and (x - y) is not considered? It would yield x² - y² and there is an integer (1) multiplying y.

Exactly that pair and only that pair gives x^2 -(by)^2. Please re-read the whole discussion.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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02 Sep 2018, 11:29
thiagod wrote:
Why the pair (x + y) and (x - y) is not considered? It would yield x² - y² and there is an integer (1) multiplying y.

That is the pair. That's the only pair that yields the correct product.

And there are 4C2 ways to pick a pair. 4C2 = $$\frac{4!}{2!2!}$$ = 6.

So the there's only 1 pair that satisfies out of the 6 possible pairs that could be chosen. $$\frac{1}{6}$$

Hope this helps!
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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21 Oct 2018, 00:51
x^2-(by)^2
=(x+by)(x-by)
only x+y and x-y can be the pair
All occasions are "selection of 2 out of 4 = 4C2= 6
Probability is 1/6
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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15 Jul 2019, 06:34
total expressions possible 4c2 ; 6
so x^2 -(by)^2 ; 1/6
IMO E

Bunuel wrote:
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 -(by)^2, where b is an integer?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Four expressions can be paired in $$C^2_4=6$$ ways (choosing 2 out of 4);

$$x^2 - (by)^2=(x-by)(x+by)$$ and only one pair is making such expression: $$(x+1*y)(x-1*y)$$, hence probability is 1/6.

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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17 Jul 2019, 14:24
Made the silly mistake of including (x+5y) in the selection range.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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06 Jan 2020, 04:31
Actually it is selecting two( (x+y) and (x-y)) out of four
Therefore the probability is 2/4 = 1/2
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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06 Jan 2020, 15:51
1
charanvemasani wrote:
Actually it is selecting two( (x+y) and (x-y)) out of four
Therefore the probability is 2/4 = 1/2

Hi charanvemasani,

Unfortunately, that is NOT how 'probability math' works. We're given the terms (X+Y), (X+5Y), (X-Y) and (5X-Y). We're asked for the probability that multiplying any randomly chose pair will give a result that is written in the format: X^2 - (BY)^2.

Since there are only 4 terms, and we're MULTIPLYING, there are only 6 possible outcomes.

From the prompt, you should notice that the 'first part' of the result MUST be X^2....and that there should be NO 'middle term'....which limits what the first 'term' can be in each of the parentheses....

By brute-forcing the 6 possibilities, you would have...
(X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
(X+Y)(X-Y) = X^2 - Y^2
(X+Y)(5X-Y) = X^2 + 4XY - Y^2
(X+5Y)(X-Y) = X^2 + 4XY - 5Y^2
(X+5Y)(5X-Y)= 5X^2 +24XY - 5Y^2
(X-Y)(5X-Y) = 5X^2 -6XY + Y^2

Only the second option is in the proper format, so we have one option out of six total options.

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at   [#permalink] 06 Jan 2020, 15:51

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# Math Questions -- Algebra

Discussion in 'Physics & Math' started by Amadeusboy, Apr 4, 2008.

Messages:
22
This is my first post to SciForums and I'm not sure if this is the proper forum for my question. If this is not the correct forum, please direct me to the proper one and accept my apology for the intrusion.

Recently, I have been reviewing calculus and algebra and came across two problems with which I had difficulty. The questions are from page 7 of Calculus by Boyce and DiPrima (Copyright 1988). They are:

Determine the set $S=${$x\mid x^2+x-s>0 \text{ } \forall x\in \mathbb R^1$ },

and

Determine the set $S=$ {$x\mid x^2-3x+s<0 \text { for some }x\in \mathbb R^1$ }.

Is there a difference between the qualifiers "for all" and "for some"? I thought "for some" was equivalent to "for any" and wouldn't this imply "for all"?

If anyone could assist me with these problems, either mathematical or semantical, it would be appreciated.

3. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

Messages:
8,967
You're in the right place...wait for Temur or Quarkhead!

5. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
I have the feeling that one of the x in front of the vertical line | and the x in front of $\in$ should be s. Probably the first x should be s.

Suppose P(y) is some statement that may or may not hold depending on y. As an example, let us assume

P(y) = "y is an odd number."

So P(1) holds, but P(2) does not hold, or equivalently, "not P(2)" holds.

The statement

$P(y) \forall y\in\mathbb{N}$

does not hold, because some natural numbers are not odd. But

$P(y) \textrm{ for some } y\in\mathbb{N}$

holds, because some natural numbers are odd. The latter can be read also as "There exists a natural number y such that P(y)".

7. ### D HSome other guyValued Senior Member

Messages:
2,257
There is a huge difference between "for all" and "for some". Suppose a given condition is true for one and only one real value. It certainly is not true for all reals, but it is true "for some" reals.

The forms as written don't make sense. Since the sets are written as S, I think these should have been written as

Determine the set $S= \lbrace s\in {\mathbb R}^1 \, \mid \, x^2+x-s>0 \; \forall x\in {\mathbb R}^1 \rbrace$,

and

Determine the set $S=\lbrace s \in {\mathbb R}^1 \, \mid \, x^2-3x+s<0 \text { for some} x\in {\mathbb R}^1 \rbrace$

Last edited: Apr 4, 2008
8. ### QuarkHeadRemedial Math StudentValued Senior Member

Messages:
1,717
Hi Amadeus! Welcome to the bear-pit (just kidding. We're all really nice people who love our Mums).
While I fully agree with temur and D H about your qualifiers, I don't share their reservations about your notation; it looks like perfectly standard set-builder to me.

Let's look at your first inequality. Note straight away that, if x is real, then $x^2 \gt 0$ for all x in R. Then either $x^2 \gt x$ or that $x^2 \lt x$. Do you see when these assertions are true?

Then this implies that in $x^2 + x - s \gt 0$, either $s \lt x$ or $s \gt -x$. Again, do you see why?

Then, remembering that s was not specified as an element in S, and that x is real, I may have for x in R (again remember that this is part of your condition) two real intervals (sets) $[-x, s)$ and $(s, x]$. But now recall that that this must be true for all x in R, so that, since I require a single set I may form $S = (-\infty,s)\cup(s, \infty)$.

See if you can get the other one

Last edited: Apr 4, 2008
9. ### D HSome other guyValued Senior Member

Messages:
2,257
Well, I disagree. First, $x^2+x-s>0$ does not imply $s<x$ or $s>-x$. In terms of $x$, it implies $x<\frac {-1 - \sqrt{1+4s}} 2$ or $x>\frac{1 + \sqrt{1+4s}} 2$. In terms of $s$, it means $s<x^2+x$, period.

Messages:
22
I'm sorry for the lateness of my response; normally I do not have much free time during the work day.

Temur and D H are correct--I inadvertently substituted a "$x$" for "$s$" in the beginning of both statements. My thanks to both for pointing out the error. The questions should read:

Determine the set $S=\left\{s\mid x^2+x-s>0\text{ }\forall x\in\mathbb{R}^1\right\}$,

and

Determine the set $S=\left\{s\mid x^2-3x+s<0\text{ for some } x\in\mathbb{R}^1\right\}$.

I had never used Tex and was most likely more concerned with getting the Tex command correct at the expense of the content. I regret the error and any time wasted due to this oversight.

After reading D H's second post and remembering that one is looking for only real values of x, it is clear that the radicand must be at least zero for real values of x. Thus,
$1+4s\geq 0 \text { } \Rightarrow \text { } s\geq-\frac{1}{4}$.
However, $s>-\frac{1}{4}$ are the only values that meet the above stipulations implying $S=\left$$-\frac14,\infty\right$$$.
Please correct me if I'm wrong, but I believe this answers the first question, after one takes the correction into account. I'll use this technique to examine the second question again.

Also, my thanks to temur and D H for correcting my misconceptions about "for all" and "for some".

Last edited: Apr 5, 2008
11. ### D HSome other guyValued Senior Member

Messages:
2,257
Almost, but exactly wrong. You want the quadratic $x^2+x-s$ to be positive definite. A quadratic form is positive definite if and only if the coefficient of the $x^2[/itex] term is positive (which is true here) and the radicand is [i]negative[/i]. A negative radicand ensures that the quadratic has no zeros in the reals. In other words, it won't change signs.$

12. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Basically the first set is the set of s for which the minimum of the quadratic polynomial is positive, and the second set is that for which the minimum is negative.

Messages:
22
D H, thanks for helping me with this. I see my error now. A zero or positive radicand implies a real solution(s) for $x$, which was not what was desired.

Messages:
22
I have been reviewing some more and came across another question which has been giving me difficulties. The question comes from page forty of Calculus by Boyce and DiPrima (1988)

"The cost of a taxi is fifty cents for the first fifth of a mile and ten cents for each additional fifth of a mile. Give a formula for the cost $C$ in dollars as a funtion of the miles traveled $x$ for $0\leq x\leq 2$."

The first question I have is interpreting the problem. Are fractional increments of the costs allowed? For example, does riding in the taxi for one tenth of a mile cost a quarter and riding for three tenths of a mile cost fifty-five cents?

If fractional increments of the costs are allowed, the solution would be

$C(x)=\left\{\frac{5x}{2} \tex {, } 0 \leq x \leq \frac{1}{5} \\ \frac{1}{2} + \frac{x-\frac{1}{5}}{2}\tex{, } \frac{1}{5}<x\tex{.}$

In this case $C(\frac{1}{10})=0.25$ and $C(\frac{3}{10})=0.55$.

Or does riding in the taxi for one tenth of a mile cost cost fifty and riding for three tenths of a mile cost sixty cents? (Alternatively does riding in the taxi for three tenths of a mile cost fifty cents and four tenths of a mile cost sixty cents?) In other words, are only whole multiples of the costs are allowed?

For this case, I have come up with a solution that somewhat works:

$C(x)=\left\{\frac{1}{2} \tex {, } 0 \leq x \leq \frac{1}{5} \\ \frac{1}{2} + \frac{[5x]}{10}\tex{, } \frac{1}{5}<x<2\tex{.}$

(Here the brackets [] denote the greatest integer function.) In this case $C(\frac{1}{10})=0.50$ and $C(\frac{3}{10})=0.60$. But $C(\frac{2}{5})=0.70$

The two preceding problems in the text involve the definition and use of the Heaviside unit step function, which they define as follows:

$H(t)=\left\{1\tex{, } t \geq 0\\0\tex{, }t<0\tex{.}$

Since my question immediately follows the questions regarding the Heaviside unit step function, I was given the impression that this function would be part of the answer. The solution I found using this function had many cases. Starting from the beginning, for $0<x\leq\frac{1}{5}$,

$C(x)=\frac{1}{2}H(x)$.

Here$C(\frac{1}{10})=0.50$.

For $\frac{1}{5}<x\leq \frac{2}{5}$,

$C(x)=\frac{1}{2}+\frac{1}{10}H(x-\frac{1}{5})H(\frac{2}{5}-x)$

Here $C(\frac{3}{10})=0.60$.

For $\frac{2}{5}<x\leq \frac{3}{5}$,

$C(x)=\frac{1}{2}+\frac{2}{10}H(x-\frac{2}{5})H(\frac{3}{5}-x)$

Here $C(\frac{1}{2})=0.70$.

Continuing on is this fashion, one could write for $\frac{1}{5}<x$,

$C(x)=\frac{1}{2}+\frac{n}{10}H(x-\frac{n}{5})H(\frac{n+1}{5}-x)$, where $n=[5x]$.

Or alternatively, for $0\leq x\leq2$

$C(x)=\frac{1}{2}H(x)H(2-x)+\frac{n}{10}H(x)H(2-x)$, where $n=[5x]$.

$C(x)=0 \tex{ for }x<0 \tex{ and } x>2$. (It also means that just sitting in the taxi and going nowhere costs you fifty cents.)

Thanks,

15. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
I think your second interpretation is correct, since it is how taxi meters work. For the second part, you should be able to write your function without any conditional statements if you cook up an appropriate combination of Heaviside functions.

Messages:
22
Thanks, Temur
Since the function cuts out at x<0 and x>2, would the last expression given work?

17. ### camilusthe villain with x-ray glassesRegistered Senior Member

Messages:
895
sorry for the off-topic post, but damn, I need to learn set theory! and my college offers no more math classes after differential equations.. I need to change schools i know..

and I could probably teach myself but I've been too busy reading science fiction, H.P. Lovecraft is a god!

18. ### ReikuBannedBanned

Messages:
11,238
Same here! Set theory is going to be a topic i am going to have to learn.

Messages:
22

Firstly, how does one prove that $sin(\theta+2n\pi)=sin\theta$? In other words, how does one prove that $sin 2n\pi=0$?

Since induction is introduced much later in the text, how does one prove this without induction.

Lastly, I had a question about the fundamental period of a function.

First, I will give their definition of the fundamental period and secondly, I will give the problem and my attempt to solve it.

A function $f$ is said to be periodic with period if$p$ if for each $x$ in the domain of $f$ the point $x+p$ is also in the domain of $f$, and $f(x+p)=f(x)$. The smallest positive number $p$ with this property is called the fundamental period.

Show that the functions $f(x)=ax+b$ and $f(x)=ax^2+bx+c \tex{, }a\neq0$ are not periodic with any period. More generally, it can be shown that the only rational function that is periodic is the constant function $f(x)=c$, which can have any period. This problem is from page forty-nine of the aforementioned text.
I proceeded to solve the first part of the question as follows. I need to find $p$such that $f(x+p)=f(x)$. Thus,

$f(x+p)=a(x+p)+b=ax+b=f(x) \tex{ or } ax+ap+b=ax$.

This implies $ap=0$ or, since $a\neq0\tex{, } p=0$. Hence, there is no p which satisfies the given definition.

For the solution to the quadratic, I took a similar approach. Again, I need to find $p$such that $f(x+p)=f(x)$. Thus,

$f(x+p)=a(x+p)^2+b(x+p)+c=ax^2+bx+c=f(x)$

or

$ax^2+2xpa+ap^2+bx+bp+c=ax^2+bx+c$.

This implies $ap^2+p(2xa+b)=0$ or, since $a\neq0\tex{, } p(p+2x+\frac{b}{a})=0$.

Thus $p=\left\{0,-2x-\frac{b}{a}\right\}$.

The first solution for $p$ is identical to the solution in the previous problem. However, the second solution for $p$ is a function of $x$.

Is it because of the fact that $p$ depends on $x$, it is a function and not a number, that the second solution fails to give a period? Or is my interpretation of the definition incorrect? It makes sense that the period cannot be dependent on $x$, but I'm trying to make my solution fit the definition. Thus, I'm looking for the reason why $f(x)=a^2+bx+c$ has no period.

Any assistance is appreciated.

Last edited: Aug 10, 2008
20. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Remember that the relation should hold for each x. This means that if you get a polynomial expression in x then all the coefficients for the powers of x must be 0. In general you will get an expression of form f(p,x)=0 and so the "internal" parameters of f must be such that f(p,x) = F(p) for some F.

ps: I did not check your earlier result very carefully, but from the skills you have demonstrated here I am confident that your results are correct.

21. ### paulfrRegistered Senior Member

Messages:
227

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22
Temur and Paulfr, thank you for your replies.

Paulfr, indeed one can use the identity

$sin(\alpha + \beta)=sin\alpha cos\beta +sin\beta cos\alpha$

and find

$sin(\theta +2n\pi)=sin\theta cos2n\pi+sin2n\pi cos\theta$.

I know that

$sin2n\pi=0$ and $cos2n\pi=1 \tex{; (1)}$

thus,

$sin\theta cos2n\pi+sin2n\pi cos\theta=sin\theta$.

However, I am not sure how to prove statement (1) without using induction. I can construct a geometric proof similar to the one at the site you mentioned for the identity, but I am uncertain how I would construct such a proof for $cos2n\pi \tex{ or }sin2n\pi$.

Does this mean that $p$ is independent of $x$ or is $p\neq p(x)$?

Secondly,
Does this imply that, given the expression

$ap^2+2apx+bp=0\tex{, (2)$

previously posted, the coefficient $2ap=0$; hence, $p=0$, since $a\neq 0$?

Finally,
If I am understanding your statement (please correct me if I am wrong), in this case, the $f(x,p)=0$ you mention is statement (2). Furthermore, combining $p\neq p(x)$ and the implications from statement (2) one finds $F(p)=0$.

23. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
I don't think you can get away without induction altogether. I suspect the idea of the problem was to reduce it to a simpler statement whose rigorous proof requires induction, or prepare you to the next lecture by forcing you to feel the need for something like induction.

$p$ is a constant, so it is independent of x by definition.

Yes.

I have to correct some of my statements here. In some cases we should interpret the requirement that f(x,p) does not depend on x as an equation for p (in general as an equation relating p and the "internal" parameters as I said before, but in some problems the internal parameters are fixed with the problem and we cannot change them). Specifically in this case we have $f(x,p)=ap^2+2apx+bp$, and requiring that $f(x,p)=F(p)$ (i.e., $f$ cannot depend on $x$) we have $p=0$. We get this not even solving the equation for the period (which is F(p)=0), but just from the requirement that the period should be a constant. So in general, the equation f(x,p)=0 will decompose into two equations on p, one requiring that f(x,p) does not depend on x, call it G(p)=0, and another one which is just f(x,p)=0 (here f(x,p)=F(p)).


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Supersymmetry algebra cohomology. IV. Primitive elements in all dimensions from D = 4 to D = 11
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1 Institut für Theoretische Physik, Leibniz Universität Hannover, Appelstraße 2, D-30167 Hannover, Germany
J. Math. Phys. 54, 052302 (2013)
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### References

• By Friedemann Brandt
• Source: J. Math. Phys. 54, 052302 ( 2013 );
1.
1. F. Brandt, “Supersymmetry algebra cohomology. I. Definition and general structure,” J. Math. Phys. 51, 122302 (2010);
http://dx.doi.org/10.1063/1.3515844
1.e-print arXiv:0911.2118v5 [hep-th].
2.
2. A. Van Proeyen, “Tools for supersymmetry,” e-print arXiv:hep-th/9910030v6.
3.
3. F. Brandt, “Supersymmetry algebra cohomology. II. Primitive elements in 2 and 3 dimensions,” J. Math. Phys. 51, 112303 (2010);
http://dx.doi.org/10.1063/1.3515845
3.e-print arXiv:1004.2978v3 [hep-th].
4.
4. F. Brandt, “Supersymmetry algebra cohomology. III. Primitive elements in four and five dimensions,” J. Math. Phys. 52, 052301 (2011);
http://dx.doi.org/10.1063/1.3583554
4.e-print arXiv:1005.2102v2 [hep-th].
5.
5. M. V. Movshev, A. Schwarz and R. Xu, “Homology of Lie algebra of supersymmetries,” e-print arXiv:1011.4731v1 [hep-th].
6.
6. M. V. Movshev, A. Schwarz and R. Xu, “Homology of Lie algebra of supersymmetries and of super Poincare Lie algebra,” Nucl. Phys. B 854, 483 (2012);
http://dx.doi.org/10.1016/j.nuclphysb.2011.08.023
6.e-print arXiv:1106.0335v3 [hep-th].
7.
7. F. Brandt, “Aspects of supersymmetric BRST cohomology,” Strings, Gauge Fields, and the Geometry Behind, edited by A. Rebhan et al. (World Scientific, 2012), p. 87;
7.e-print arXiv:1201.3638v1 [hep-th].
8.
8.In D = 4, 8, …, there is an alternative choice of C for which the matrices ΓaC−1 are antisymmetric1,2 and which therefore requires N ∈ {2, 4, …}, excluding the case N = 1 (more precisely: a case N = 1 with nonvanishing anticommutators in (1.1)).
9.
9.This notation deviates somewhat from the notation used in Ref. 1 where Hg(sgh) denotes H(sgh) in the sector of ghost number g and Hp,*(sgh) denotes H(sgh) in the sector of c-degree p.
10.
10.The first part of Eq. (5.9) and Lemma 5.1 imply for some polynomials pija(ξ) and some . does not depend on ξ2, does not depend on ξ1, and is symmetric under ξ1↔ξ2 whereas all terms in depend on both ξ1 and ξ2 and is antisymmetric under ξ1↔ξ2. As a consequence, the second part of Eq. (5.9) implies that p11 a(ξ), p12 a(ξ), p22 a(ξ) are such that is -exact (or vanishes).
11.
11.In D = 6 one has for (N+, N) = (2, 0) and (N+, N) = (0, 2):where we used that vanishes for (N+, N) = (2, 0) and (N+, N) = (0, 2),12 and that also vanishes because ΓcC−1 is antisymmetric in D = 6, see (3.18).
12.
12.Equations (2.12) and (6.1) imply for (N+, N) = (2, 0), and for (N+, N) = (0, 2). follows from the “completeness relation” of the Γ-matrices in D = 5 which reads and implieswhere we used that (ΓaC−1)(5) and are antisymmetric, whereas (ΓabC−1)(5) is symmetric, see (3.18).
13.
13.For instance:where we used that and vanish because ΓaC−1 is antisymmetric in D = 6, see (3.18).
14.
14.Using the structure of sghc7 (which consists of bilinears and ) and sghc1, …, sghc6 (which only involve bilinears and ), one infers from (2.6) for p = 1 that is at least linear in both ξ1(6) and ξ2(6). This gives with constant 8 × 8-matrices Kr which can be chosen according to Kr ∈ {(C−1)(6),(ΓaC−1)(6),(ΓabC−1)(6),(ΓabcC−1)(6), since these matrices form a basis for 8 × 8-matrices. Equation (2.6) then leads to as in (7.9), up to a coboundary in which can be neglected.
15.
15.For N = 1 all cocycles with p > 0 depend at least linearly on both ξ1(7) and ξ2(7), see first comment at the end of Sec. VIII. In the case p = 1, this implies with constant 8 × 8-matrices Kr which can be chosen according to Kr ∈ {(C−1)(7),(ΓaC−1)(7),(ΓabC−1)(7),(ΓabcC−1)(7)} since these matrices form a basis for 8 × 8-matrices. Equation (2.6) then leads to as in (8.13), up to a coboundary in which can be neglected.
16.
16.The structures of sghc9 (which consists of bilinears and ) and sghc1, …, sghc8 (which only involve bilinears ) imply that with constant 16 × 16-matrices Kr which can be chosen according to Kr ∈ {(C−1)(8),(ΓaC−1)(8), ..., (ΓabcdC−1)(8), , ..., since these matrices form a basis for 16 × 16-matrices. Equation (2.6) then leads to as in (9.11), up to a coboundary in which can be neglected.
17.
17.In the case, p = 1 we use that every cocycle ω1 must be at least linear in the supersymmetry ghosts since no nonvanishing linear combination of the ca with constant coefficients is sgh-closed. This yields as in (10.13). Lemma 9.1 yields as in (10.13). Equation (10.18) follows from the facts that every monomial corresponds to a cocycle in proportional to one of the , and that , analogously to the situation in D = 6.12,13
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# 10

Decision TheoryGame TheoryAI
Frontpage

(This post was originally published on Oct 20th 2017, and is 1 of 4 posts brought forwarded today as part of the AI Alignment Forum launch sequence on fixed points.)

In this post, I give a stronger version of the open question presented here, and give a motivation for this stronger property. This came out of conversations with Marcello, Sam, and Tsvi.

Definition: A continuous function is called ubiquitous if for every continuous function , there exists a point such that .

Open Problem: Does there exist a topological space with a ubiquitous function ?

I will refer to the original problem as the Converse Lawvere Problem, and the new version as the the Ubiquitous Converse Lawvere Problem. I will refer to a space satisfying the conditions of (Ubiquitous) Converse Lawvere Problem, a (Ubiquitous) Converse Lawvere Space, abbreviated (U)CLS. Note that a UCLS is also a CLS, since a ubiquitous is always surjective, since can be any constant function.

Motivation: True FairBot

Let be a Converse Lawvere Space. Note that since such an might not exist, the following claims might be vacuous. Let be a continuous surjection.

We will view as a space of possible agents in an open source prisoner's dilemma game. Given two agents , we will interpret as the probability with which A cooperates when playing against . We will define , and interpret this as the utility of agent when playing in the prisoner's dilemma with .

Since is surjective, every continuous policy is implemented by some agent. In particular, this means gives:

Claim: For any agent , there exists another agent such that . i.e. responds to the way that responds to .

Proof: The function is a continuous function, since is continuous, and evaluation is continuous. Thus, there is a policy in . Since is surjective, this policy must be the image under of some agent , so .

Thus, for any fixed agent , we have some other agent that responds to any the way responds to . However, it would be nice if , to create a FairBot that responds to any opponent the way that that opponent responds to it. Unfortunately, to construct such a FairBot, we need the extra assumption that is ubiquitous.

Claim: If is ubiquitous, then exists a true fair bot in : an agent , such that for all agents .

Proof: Given an agent , there exists an policy such that for all , since is continuous. Further, the function is continuous, since the function and the definition of the exponential topology. Since is ubiquitous, there must be some such that . But then, for all , we have .

Note that we may not need the full power of ubiquitous here, but it is the simplest property I see that gets the result.

Note that this FairBot is fair in a stronger sense than the FairBot of modal combat, in that it always has the same output as its opponent. This may make you suspicious, since the you can also construct an UnfairBot, such that for all . This would have caused a problem in the modal combat framework, since you can put a FairBot and an UnfairBot together to form a paradox. However, we do not have this problem, since we deal with probabilities, and simply have . Note that the exact phenomenon that allows this to possibly work is the fixed point property of the interval which is the only reason that we cannot use diagonalization to show that no CLS exists.

Finally, note that we already have a combat framework that has a true FairBot: the reflective oracle framework. In fact, the reflective oracle framework may have all the benefits we would hope to get out of a UCLS. (other than the benefit of simplicity of not having to deal with computability and hemicontinuity).

This post was originally published on Oct 20th 2017, and has been brought forwarded as part of the AI Alignment Forum launch sequences.

Tomorrow's AIAF sequences post will be 'Iterated Amplification and Distillation' by Ajeya Cotra, in the sequence on iterated amplification.

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text:
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Answer: $\text{ intersection }$ The symbol $\cap$ indicates the intersection of the sets.


text:
# Can someone explain how this unpadded RSA formula works?

$\mathcal{E}(x_1) \cdot \mathcal{E}(x_2) = x_1^e x_2^e$
$\bmod\; m = (x_1x_2)^e \;$
$\bmod\; m = \mathcal{E}(x_1 \cdot x_2)$

What is the superscript $^e$ representing in the first expression, and how do we get from the second line, to the third?

-

## 1 Answer

• $\mathcal{E}$ represents encryption, superscript $e$ represents exponentiation. Textbook RSA encryption is defined as $\mathcal{E}(x) = x^e \mod m$ where $e$ is public exponent and $m$ is the modulus.

• $x_1^e\cdot x_2^e = (x_1\cdot x_2)^e$ is a basic property of exponentiation.

• To get from the third line to the second you use the definition of $\mathcal{E}$ on $x_1\cdot x_2$ yielding $(x_1\cdot x_2)^e$.
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Thanks very much! What do you mean by the definition of $E$, is that the procedure of encryption? –  user7378 Sep 20 '13 at 7:55