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# On the conjectured nonexistence of even almost perfect numbers (other than powers of two) and odd perfect numbers [closed]

(Note: This question has been cross-posted to MSE.)

Let $\sigma(a) = \sigma_{1}(a)$ be the sum of the divisors of the positive integer $a$.

A number $M$ is called almost perfect if $\sigma(M) = 2M - 1$. $N$ is called perfect if $\sigma(N) = 2N$.

$M_s = 2^s$ where $s \geq 0$ are almost perfect, with $M_0 = 1$ being the only odd almost perfect number that is currently known. If $M \neq 2^t$ is an even almost perfect number, then Antalan and Tagle showed that $M$ must have the form $$M = {b^2}{2^r}$$ where $r \geq 1$ and $b$ is an odd composite.

On the other hand, only $49$ even perfect numbers have been discovered, and they are of the form $$N_e = 2^{p-1}\left(2^p - 1\right)$$ where $2^p - 1$ and $p$ are primes. It is currently unknown whether there are any odd perfect numbers, but Euler showed that they are of the form $$N_o = n^2 q^c$$ where $q$ is prime with $q \equiv c \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$.

Notice the following:

For numbers that we know exist

Even Perfect Numbers

Assuming $N_e \neq 6$ (because it is squarefree), $$\dfrac{\sigma(2^{(p-1)/2})}{2^p - 1} = \dfrac{2^{(p+1)/2} - 1}{2^p - 1}< 1 < 4 \le 2^{(p+1)/2} = \dfrac{\sigma(2^p - 1)}{2^{(p-1)/2}}.$$

Note that $$\dfrac{\sigma(2^p - 1)}{2^p - 1} \leq \dfrac{8}{7} < \sqrt{\dfrac{7}{4}} < \dfrac{\sigma(2^{(p-1)/2})}{2^{(p-1)/2}}.$$

Almost Perfect Numbers (Powers of Two)

Since $s \geq 0$, $$\dfrac{\sigma(\sqrt{1})}{2^s} \leq 1 \leq 2^{s+1} - 1 = \dfrac{\sigma(2^s)}{\sqrt{1}}.$$

Note that $$\dfrac{\sigma(\sqrt{1})}{\sqrt{1}} = 1 \leq \dfrac{\sigma(2^s)}{2^s}.$$

Spoof Odd Perfect Numbers

In Descartes' example $D = km$, we have $$m = 22021 = {{19}^2}\cdot{61}$$ and $$\sqrt{k} = {3}\cdot{7}\cdot{11}\cdot{13}.$$

Note that $$\dfrac{\sigma(\sqrt{k})}{m} = \dfrac{5376}{22021} < 1 < \dfrac{22022}{3003} < \dfrac{m+1}{\sqrt{k}}$$ and $$\dfrac{m+1}{m} = \dfrac{22022}{22021} < \dfrac{5376}{3003} = \dfrac{\sigma(\sqrt{k})}{\sqrt{k}}.$$

For numbers that are conjectured not to exist

Even Almost Perfect Numbers (Other Than Powers of Two)

$$\dfrac{\sigma(2^r)}{b} < 1 < 2 < \dfrac{\sigma(b)}{2^r}$$

Note that $$\dfrac{\sigma(b)}{b} < \dfrac{4}{3} < \dfrac{3}{2} \leq \dfrac{\sigma(2^r)}{2^r}.$$ "Some New Results On Even Almost Perfect Numbers Which Are Not Powers Of Two" - Theorem 2.2, page 5

Odd Perfect Numbers

The following inequalities are conjectured in New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II:

$$\dfrac{\sigma(q^c)}{n} < 1 < \sqrt{\dfrac{8}{5}} < \dfrac{\sigma(n)}{q^c}.$$

Note that $$\dfrac{\sigma(q^c)}{q^c} < \dfrac{5}{4} < \sqrt{\dfrac{8}{5}} < \dfrac{\sigma(n)}{n}.$$

(Added June 27 2016 - In fact, in a recent preprint, Brown claims a partial proof for $q^c < n$, which would be consistent with the conjecture here.)

Here are my questions:

(1) If $K = {x^2}{y^z}$ is a (hypothetical) number satisfying $\sigma(K) = 2K + \alpha$ (with $y$ prime, $\gcd(x,y)=1$, and where $\alpha$ could be zero or negative), might there be a specific reason why the inequalities $$\dfrac{\sigma(x)}{y^z} \leq 1 \leq \dfrac{\sigma(y^z)}{x}$$ seem to guarantee existence of such numbers $K$?

(2) If $L = {u^2}{v^w}$ is a (hypothetical) number satisfying $\sigma(L) = 2L + \beta$ (with $v$ prime, $\gcd(u,v)=1$ and where $\beta$ could be zero or negative), might there be a specific reason why the inequalities $$\dfrac{\sigma(v^w)}{u} < 1 < \dfrac{\sigma(u)}{v^w}$$ seem to predict nonexistence of such numbers $L$?

Question (1) is illustrated (as detailed above) in the case of even perfect numbers, almost perfect numbers which are powers of two, and spoof odd perfect numbers (otherwise known in the literature as Descartes numbers).

Question (2) is illustrated (as detailed above) in the case of even almost perfect numbers which are not powers of two, and odd perfect numbers.

## closed as unclear what you're asking by Steven Landsburg, Stefan Kohl, user1688, Jan-Christoph Schlage-Puchta, Ben LinowitzJun 27 '16 at 17:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What is the difference between a "reason" and a "special reason"? – Steven Landsburg Jun 27 '16 at 13:00
• @StevenLandsburg - A non-special reason may be superficial, but for me a special reason is necessarily unique. Should I change the word special to specific? I think that is where you are driving at. – Jose Arnaldo Bebita-Dris Jun 27 '16 at 13:08


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# Changing order of integration (double) in polar coord

## Homework Statement

Change the order of the limits of integration of the following double integral and evaluate.

## Homework Equations

$\int_{0}^\frac{\pi}{2} \int_{0}^{cos(\theta)} cos(\theta)\,dr\,d\theta$

## The Attempt at a Solution

Evaluating as it is, I arrive an answer of $\frac{\pi}{4}$.

I know the region to be integrated is the semicircle bounded by the polar axis, with corner points at r = 0, and r = 1, with a height of 1/2. I know that normally, in the cartesian case, to change the order of integration requires the limits to be written from y(x) to x(y), with the x or y intervals adjust accordingly.

Thus, for this problem, the original region is bounded by:

$0<r<cos(\theta)$ and $0<\theta<\frac{\pi}{2}$.

Changing the form, I would write,

$0<\theta<cos^{-1}(r)$ and $0<r<1$

Trying to evaluate in this manner, I end up at

$\int_{0}^1 sin(cos^{-1}(r))\,dr$,

after which I cannot go further. I have very little experience of changing limit orders in the polar case. Any hints would be appreciated!

Last edited:

## Answers and Replies

CompuChip
Homework Helper
Hint: note that sin x = sqrt ( 1 - cos^2 x ).

Cool, thanks; Using the hint: I get

## \int_{0}^1 \sqrt{1-cos(r)}\,dr ##

Multiplying by a factor of sin(r)/sin(r), I get,

## \int_{0}^1 \frac{sin(r)}{\sqrt{1+cos(r)}}\,dr ##.

I use a substitution of

## u = 1 + cos(r) ## with
## \,du = -sin(r)\,dr ##.

I substitute the limits by using

## u_0 = 1+cos(0) = 2 ##
## u_1 = 1+cos(1) = ~1.5403... ##

Evaluating the new integral, through

## -[2u^{\frac{1}{2}}] ##

between u1=1.5403... and u0 = 2, I get an answer of ~0.173144 which is not quite ##\frac{\pi}{4} ##

Are my limits wrong, or did I blunder somewhere else?

CompuChip

format: plain


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# Determining a point's coordinates on a circle

So I have a circle (I know its center's coordinates and radius) and a point on the circle (I know its coordinates) and I have to determine the coordinates of another point on the circle which is exactly at the distance L from the first point.

-
What did you try? –  Harald Hanche-Olsen Feb 24 '13 at 8:45

If I have undresttod corectly, assuming the equation of the circle as $$(x-x_0)^2+(y-y_0)^2=r^2$$ the fixed point $(x_1,y_1)$ and the desired point $(X,Y)$ then you need to solve the following equations simultaneously: $$(X-x_0)^2+(Y-y_0)^2=r^2$$ and $$(X-x_1)^2+(Y-y_1)^2=L^2$$

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Further suggestion: Expand the squares in both equations and subtract them from another to get a linear equation in $X$ and $Y$. Use it to eliminate $Y$ from one of the quadratic equations. –  Harald Hanche-Olsen Feb 24 '13 at 8:50
+1 I think you "nailed it", and with the suggestion from @Harald - the OP is on his/her way! –  amWhy Feb 24 '13 at 13:36

If the equation of the circle be $(x-h)^2+(y-k)^2=r^2$ any point on the circle can be parametrized as $(r\cos\theta+h,r\sin\theta+k)$ and if the known point on the circle be $S(p,q)$

So, we need to know the two values of $\theta$ from $(r\cos\theta+h-p)^2+(r\sin\theta+k-q)=L^2$

$\implies 2r(h-p)\cos\theta+2r(k-q)\sin\theta=L^2-r^2-(h-p)^2-(k-q)^2$

Now, we know $-\sqrt{c^2+d^2}\le c\cos A+d\sin A\le \sqrt{c^2+d^2}$(Proof below)

There will be two or zero solutions according as $c\cos A+d\sin A$ lies within or outside the range.

[

Proof: Let $c=R\cos B,d=R\sin B$ where real $R\ge0$

So, $c^2+d^2=R^2\implies R=\sqrt{c^2+d^2}$ and $\tan B=\frac dc$

and $c\cos A+d\sin A=\sqrt{c^2+d^2}\cos(A-B)$

As $-1\le\cos(A-B)\le1, -\sqrt{c^2+d^2}\le\sqrt{c^2+d^2}\cos(A-B)\le \sqrt{c^2+d^2}$

$\implies -\sqrt{c^2+d^2}\le \cos A+d\sin A\le \sqrt{c^2+d^2}$

]

-


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# A disc of moment of inertia  is rotating with angular velocity  about an axis perpendicular to its plane and passing through its centre. If another disc of moment of inertia  about the same axis is gently placed over it, then the new angular velocity of the combined disc will be Option 1) Option 2) Option 3) Option 4)

As we discussed in

Law of conservation of angular moment -

- wherein

If net torque is zero

i.e.

angular momentum is conserved only when external torque is zero .

Li = L

Option 1)

incorrect

Option 2)

correct

Option 3)

incorrect

Option 4)

incorrect

### Preparation Products

##### Knockout BITSAT 2021

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# Hausdorff Dimension

We work out the area of a square by multiplying the side length by itself. We think of the area as a measure of the size of the square. This value remains meaningful however we think of the square. Whether as a region of R^2 (the plane), or as a subset of the complex numbers, viewed as the Argand diagram, or as an object sitting in the 3-dimensional real world.

But this then becomes a problem if we want to keep using the word ‘measure’. Because the standard Lebesgue measure on R^2 coincides with this definition of area, but if we are in R^3, then the measure of a square is zero. Of course, we are really articulating the idea that dimension is key. The area of a square may be large or small, but its volume will always be zero. Similarly, if we were to try and measure the ‘length’ of a square, in some sense, the only sensible answer could be that the length is infinite. Why? Well, any measure must satisfy the constraint of being increasing under containment. In this sense, any square contains an arbitrarily large number of disjoint lines of length k/2, where k is the side-length of the square, and so the only possibility for the length measure is infinite.

So we realise that dimension is key to resolving this problem. And we are clear in our minds that a square is two-dimensional, since it can be embedded in R^2 but not in R. But this is all rather vague. We would ideally like to have a definition of dimension that is independent of choice of embedding. Not least, the property used in the case of the square has an algebraic flavour that is clear here, but which will certainly not generalise. In vector space terminology, a square is not a subspace, but it certainly generates a subspace of dimension 2. But we want a definition that exploits topological properties. Not least because, for example, the surface of a sphere sits in R^3 and indeed generates R^3 as a vector space, but is very much two-dimensional under any sensible definition.

Ideally we want a way to find the dimension of a space, given only its metric properties. Heuristically, we feel that a measure is the metric to the power of the dimension, so this would resolve all the problems we have been considering.

The rest of this post will talk about one possible solution, the Hausdorff dimension, which specifies the dimension of a metric space in an embedding-independent manner. The motivation is as follows. We can cover a square with lots of smaller squares. Indeed, a square of side-length 1 can be decomposed into $(1+o(1))r^{-2}$ squares of side-length r, as $r \rightarrow 0$. Although they do not tessellate as nicely, the same result holds for small balls. Such a square can be covered by $\Theta(r^{-2})$ balls of radius r. This is a useful observation, because a ball is specified only by metric properties of the space.

From this idea, we obtain a way of constructing a family of measures on the Borel sets of a metric space, which specify the size of subsets of any dimension. We define the $\alpha$-Hausdorff measure of A by:

$\mathcal{H}^\alpha(A):= \liminf_{\delta\downarrow 0} \left\{\sum_{i=1}^\infty \text{diam}(E_i)^\alpha:(E_i)\text{ a covering of }A, \text{ diam}(E_i)<\delta.\right\}$

In other words, we take a covering of A by small sets of diameter at most $\delta$. These might as well be balls at this point. Then we calculate the infimum of the $\alpha$-dimensional measure of these balls, across all such coverings. Letting $\delta\downarrow 0$ gives a measure of the $\alpha$-dimensional volume required to cover a subset with small $\alpha$-dimensional balls. We expect this to be zero when the set has dimension less than $\alpha$, and infinite when the set has dimension greater than $\alpha$. Naturally then, we define the Hausdorff dimension by

$\text{dim}(E)=\inf\{s\geq 0: \mathcal{H}^s(E)=0\}.$

An inquisitive child might well ask the following question: we are happy with the notion of 2-dimensional space, and 3-dimensional space, but what about other non-integer values? Can we have a 2.5-dimensional space, and what does it mean?

As an example of why the integers might not be enough to specify dimension, it is useful to consider fractals. A fractal is some object that has properties of self-similarity, which might mean that, for example, it is impossible to tell how zoomed in a view is. These are typically constructed as the limit of an iterative procedure where each unit present is broken down into several smaller units in a self-similar way. Examples include the Koch snowflake, where the middle third of a line is replaced the other two sides of the equilateral triangle with base equal to that middle third. The same operation is then successively applied to each smaller line segment and so on. The Sierpinski gasket is constructed by applying a similar operation to the areas of triangles.

In the case of the Koch snowflake, in each operation, the total length of the object increases by a factor of 4/3, and so the length of the limiting object is infinite, even though it is constructed from lines, and the endpoints are still distance 1 apart. So it is not clear what the dimension of this object should be. It doesn’t fill space to quite the same extent as a plane area, but by self-similarity, it fills space a bit more than a line locally to any given point.

It’s perhaps easiest to consider as an example for calculation the Cantor Set. Recall this is constructed by starting with the unit interval [0,1], then removing the middle third to leave [0,1/3] u [2/3,1] and then successively removing the middle third of every remaining interval. What remains are precisely those reals which have no ‘1’ in their ternary expansion. Unsurprisingly, this is nowhere dense, and has zero Lebesgue measure, since the measure after the nth iteration is $(\frac23)^n\rightarrow 0$.

However, we can calculate its Hausdorff dimension, by taking advantage of its self-similarity. We suppose that this dimension is s, and we consider the s-Hausdorff measure. It is clear from the definition that the measure of a disjoint union of sets is equal to the sum of the individual measures. So we apply this to the decomposition of the Cantor set C as:

$C=(C\cap [0,\frac13] )\cup (C\cap [\frac23,1]).$

Note that $C\cap[0,\frac13]$ and $C\cap[\frac23,1]$ are isometric to $\frac{C}{3}$. So a covering of C with diameter less than d induces a covering of C/3 with diameter less than d/3, and the s-dimensional volume of the covering sets will be scaled by a factor of $3^{-s}$. We conclude that

$H^s(C)=2H^s(\frac{C}{3})=\frac{2H^s(C)}{3^s}$

$\Rightarrow H^s(C)=\frac{\log 2}{\log 3},$

giving us an example of a non-integral Hausdorff dimension. It turns out that a space with any real Hausdorff dimension can be constructed in this way by adjusting the constants of self-similarity in the iterative construction.

Unsurprisingly, once we are comfortable with objects which are self-similar, it makes sense to look at the Hausdorff dimension of random objects which are self-similar in distribution. The primary example is Brownian motion. Again, we might assume a priori that since Brownian motion is a function of a one-dimensional variable, typically time, it should have dimension one. In fact, recalling that the variation of BM is infinite, it is clear that we have the same situation as in the Koch snowflake example discussed above. Some of the tools involved to deal with the dimension of BM in the two cases d=1 and d=>2 are interesting and non-trivial, so I will postpone discussion of this until a possible further post.


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# ImageTools

Author: Tuomo Valkonen tuomov@iki.fi

This repository contains some tools for doing mathematical imaging in Julia:

• Effective visualisation based directly on GR instead of Plots. A sample visualisation main loop that calls AlgTools-style step routines.
• Sub-pixelresolution image translation and sub-image extraction
• Calculation of discrete image gradients (2D, 3D 3D vector fields)
• A very simple image filtering code that seems more efficient than the ImageFiltering package.

The code is used, for example, by http://dx.doi.org/10.5281/zenodo.3659180.

## Installation

To install this package, first clone the repository and AlgTools with Mercurial:

$hg clone https://tuomov.iki.fi/repos/AlgTools/$ hg clone https://tuomov.iki.fi/repos/ImageTools/

(Canonical public repository URLs indicated here.) AlgTools has to be installed under the some location as ImageTools. Then add the repository to Julia with Pkg.develop:

julia>] develop LOCATION_OF/ImageTools/

Here replace LOCATION_OF/ImageTools with path where you cloned ImageTools. (If you did not change directory after cloning, simply use ImageTools.) Afterwards, you may use the package with:

julia> using ImageTools

This package is not and is not planned to be available via Pkg.add as it is based on the worst and must unusable version control system ever invented: Git.

## Testing

To run a simple denoising test using FISTA, use

\$ julia --project=DenoiseTest
julia> using DenoiseTest
julia> test_denoise()

hg clone https://tuomov.iki.fi/repos/ImageTools/

If you would like to contribute to this project, simply clone the repository and email me a hg bundle of your changes.


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<meta http-equiv="refresh" content="1; url=/nojavascript/">

# Vertex Form of a Quadratic Equation

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Practice Vertex Form of a Quadratic Equation
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Vertex Form of a Quadratic Equation

What if you had a quadratic function like $y - 2 = x^2 + 4x$ ? How could you rewrite it in vertex form to find its vertex and intercepts? After completing this Concept, you'll be able to rewrite and graph quadratic equations like this one in vertex form.

### Guidance

Probably one of the best applications of the method of completing the square is using it to rewrite a quadratic function in vertex form. The vertex form of a quadratic function is

$y - k = a(x - h)^2$

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point $(h, k)$ .

It is also simple to find the $x-$ intercepts from the vertex form: just set $y = 0$ and take the square root of both sides of the resulting equation.

To find the $y-$ intercept, set $x = 0$ and simplify.

#### Example A

Find the vertex, the $x-$ intercepts and the $y-$ intercept of the following parabolas:

a) $y - 2 = (x - 1)^2$

b) $y + 8 = 2(x - 3)^2$

Solution

a) $y - 2 = (x - 1)^2$

Vertex: (1, 2)

To find the $x-$ intercepts,

$\text{Set} \ y = 0: & & -2 & = (x - 1)^2 \\\text{Take the square root of both sides}: & & \sqrt{-2} & = x - 1 && \text{and} && -\sqrt{-2} = x - 1$

The solutions are not real so there are no $x-$ intercepts.

To find the $y-$ intercept,

$\text{Set} \ x = 0: & & y - 2 & = (-1)^2\\\text{Simplify}: & & y - 2 & = 1 \Rightarrow \underline{y = 3}$

b) $y + 8 = 2(x - 3)^2$

$& \text{Rewrite}: & & y - (-8) = 2(x - 3)^2\\& \text{Vertex}: & & \underline{(3, -8)}$

To find the $x-$ intercepts,

$\text{Set} \ y = 0: & & 8 & = 2 (x - 3)^2\\\text{Divide both sides by} \ 2: & & 4 & = (x - 3)^2 \\\text{Take the square root of both sides}: & & 2 & = x - 3 && \text{and} && -2 = x - 3\\\text{Simplify}: & & & \underline{\underline{x = 5}} && \text{and} && \underline{\underline{x = 1}}$

To find the $y-$ intercept,

$\text{Set} \ x = 0: & & y + 8 & = 2(-3)^2\\\text{Simplify}: & & y + 8 & = 18 \Rightarrow \underline{\underline{y = 10}}$

To graph a parabola, we only need to know the following information:

• the vertex
• the $x-$ intercepts
• the $y-$ intercept
• whether the parabola turns up or down (remember that it turns up if $a > 0$ and down if $a < 0$ )

#### Example B

Graph the parabola given by the function $y + 1 = (x +3)^2$ .

Solution

$& \text{Rewrite}: & & y - (-1) = (x - (-3))^2\\& \text{Vertex}: & & \underline{(-3, -1)} && \text{vertex}:(-3, -1)$

To find the $x-$ intercepts,

$&\text{Set} \ y = 0: & & 1 = ( x + 3)^2\\&\text{Take the square root of both sides}: & & 1 = x + 3 \qquad \text{and} \qquad -1 = x + 3\\&\text{Simplify}: && \underline{\underline{x = -2}} \qquad \quad \text{and} \qquad \quad \ \underline{\underline{x = -4}}\\&&& x-\text{intercepts}: \ (-2, 0) \ \text{and} \ (-4, 0)$

To find the $y-$ intercept,

$& \text{Set} \ x = 0: & & y + 1 = (3)^2\\& \text{Simplify:} & & \underline{\underline{y = 8}} && y-\text{intercept}: (0, 8)$

And since $a > 0$ , the parabola turns up.

Graph all the points and connect them with a smooth curve:

#### Example C

Graph the parabola given by the function $y = - \frac{1}{2} (x - 2)^2$ .

Solution:

$& \text{Rewrite} & & y - (0) = - \frac{1} {2} (x - 2)^2\\& \text{Vertex:} & & \underline{(2, 0)} && \text{vertex:} (2, 0)$

To find the $x-$ intercepts,

$\text{Set} \ y = 0: & & 0 & = - \frac{1} {2} (x - 2)^2 \\\text{Multiply both sides by} \ -2: & & 0 & = (x - 2)^2 \\\text{Take the square root of both sides}: & & 0 & = x - 2\\\text{Simplify}: & & & \underline{\underline{x = 2}} && x-\text{intercept:} (2, 0)$

Note: there is only one $x-$ intercept, indicating that the vertex is located at this point, (2, 0).

To find the $y-$ intercept,

$\text{Set} \ x = 0: & & y & = -\frac{1} {2}(-2)^2 \\\text{Simplify:} & & y & = - \frac{1} {2} (4) \Rightarrow \underline{\underline{y = -2}} && y- \text{intercept:}(0, -2)$

Since $a < 0$ , the parabola turns down.

Graph all the points and connect them with a smooth curve:

Watch this video for help with the Examples above.

### Vocabulary

• The vertex form of a quadratic function is

$y - k = a(x - h)^2$

This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point $(h, k)$ .

• To find the $x-$ intercepts from the vertex form: just set $y = 0$ and take the square root of both sides of the resulting equation.
• To find the $y-$ intercept, set $x = 0$ and simplify.

### Guided Practice

Graph the parabola given by the function $y = 4(x +2)^2-1$ .

Solution:

$& \text{Rewrite} & & y - (-1) = 4(x +2)^2\\& \text{Simplify} & & y +1 = 4(x +2)^2\\& \text{Vertex:} & & \underline{(-2, -1)} && \text{vertex:} (-2, -1)$

To find the $x-$ intercepts,

$\text{Set.} \ y = 0: & & 0 & = 4(x +2)^2-1 \\\text{Subtract 1 from each side}: & & 1 & = 4(x +2)^2 \\\text{Divide both sides by 4}: & & \frac{1}{4} & = (x +2)^2 \\\text{Take the square root of both sides}: & & \frac{1}{2} & = \pm (x + 2)\\\text{Separate}: & & & \frac{1}{2}=-(x+2) && \frac{1}{2}=x+2)\\\text{Simplify}: & & & \underline{\underline{x = -2.5}} && \underline{\underline{x = -1.5}}$

The $x-$ intercepts are $(-2.5, 0)$ and $(-1.5, 0)$ .

To find the $y-$ intercept,

$\text{Set} \ x = 0: & & y & = 4(0 +2)^2-1 \\\text{Simplify:} & & y & = 15 \Rightarrow \underline{\underline{y = 15}} && y- \text{intercept:}(0, 15)$

Since $a < 0$ , the parabola turns up.

Graph all the points and connect them with a smooth curve:

### Explore More

Rewrite each quadratic function in vertex form.

1. $y= x^2 - 6x$
2. $y + 1 = -2x^2 -x$
3. $y = 9x^2 + 3x - 10$
4. $y = -32x^2 + 60x + 10$

For each parabola, find the vertex; the $x-$ and $y-$ intercepts; and if it turns up or down. Then graph the parabola.

1. $y - 4 = x^2 + 8x$
2. $y = -4x^2 + 20x - 24$
3. $y = 3x^2 + 15x$
4. $y + 6 = -x^2 + x$
5. $x^2-10x+25=9$
6. $x^2+18x+81=1$
7. $4x^2-12x+9=16$
8. $x^2+14x+49=3$
9. $4x^2-20x+25=9$
10. $x^2+8x+16=25$


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# Centripetal force of a roller coaster car

I got all the other centripetal force questions but this one has me stumped...

As a roller coaster car crosses the top of a 30m-diameter loop-the-loop, its apparent weight is the same as its true weight. What is the car's speed at the top?

The formula I used for all of the other questions was F=mv^2/r. The problem is the only number I have is the diameter (therefore radius). I'm guessing because the apparent weight is the same as true weight the net force is zero? There's also no way to get mass. Maybe there's another formula? Any assistance with this problem would be much appreciated!

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
The true weight of something is $$F=mg$$, so then you don't need to figure out the mass.

Last edited:
rcgldr
Homework Helper
If the apparent weight is the same as the true weight, then the passengers at the top of the loop are experiencing the equivalent of 1 g of force from their seats.


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# Joint distribution of density forecasts

I have a panel data set and I have created a model and finally I have obtained some density forecasts.

That is, I run my model for the $y_{it}$ and i obtain predictions for $\hat{y_{i,t+1}}$ , $\hat{y_{i,t+2}}$ ,..., $\hat{y_{i,t+h}}$ etc.

The $\hat{y_{i,t+1}}$ , $\hat{y_{i,t+2}}$ etc, all of them follow a normal distribution, but with different mean and variance. so $\hat{y_{i,t+1}} \sim N(\mu_{1},\sigma_{1}^{2})$ , $\hat{y_{i,t+2}} \sim N(\mu_{2},\sigma_{2}^{2})$, ..., $\hat{y_{i,t+h}} \sim N(\mu_{h},\sigma_{h}^{2})$.

My question is, what is their joint distribution ? And how can I obtain it ?

Edit: The $y_{it}$'s are the daily consumption of a product of individual $i$ at time $t$, so yes, they are highly correlated.

Thank you for your help :)


text:
# Relation between metric tensor and second fundamental form

I'm confused with these definitions. The metric of certain space and the second fundamental form seem to be the same object.

I don't know what else to say, this is a pretty straight forward question.

-


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Certain properties of the Dziok--Srivastava operator. (English) Zbl 1081.30021
Making use of Hadamard product (or convolution), a certain linear operator $H_{p,q,s}(\alpha _1)$ was recently introduced and studied in a series of papers by Dziok and Srivastava. In this paper, some other properties of the operator $H_{p,q,s}(\alpha _1)$ are derived with the help of differential subordination.

##### MSC:
 30C45 Special classes of univalent and multivalent functions 30C50 Coefficient problems for univalent and multivalent functions
Full Text:
##### References:
 [1] Dziok, J.; Srivastava, H. M.: Classes of analytic functions associated with the generalized hypergeometric function. Appl. math. Comput 103, 1-13 (1999) · Zbl 0937.30010 [2] Dziok, J.; Srivastava, H. M.: Some subclasses of analytic functions with fixed argument of coefficients associated with the generalized hypergeometric function. Adv. stud. Contemp. math 5, 115-125 (2002) · Zbl 1038.30009 [3] Dziok, J.; Srivastava, H. M.: Certain subclasses of analytic functions associated with the generalized hypergeometric function. Integral transform. Spec. funct 14, 7-18 (2003) · Zbl 1040.30003 [4] Gangadharan, A.; Shanmugam, T. N.; Srivastava, H. M.: Generalized hypergeometric functions associated with k-uniformly convex functions. Comput. math. Appl 44, 1515-1526 (2002) · Zbl 1036.33003 [5] Kim, Y. C.; Srivastava, H. M.: Fractional integral and other linear operators associated with the Gaussian hypergeometric function. Complex variables theory appl 34, 293-312 (1997) · Zbl 0951.30010 [6] Liu, J. -L.: Strongly starlike functions associated with the dziok--Srivastava operator. Tamkang J. Math 35 (2004) · Zbl 1064.30006 [7] Liu, J. -L.; Srivastava, H. M.: Classes of meromorphically multivalent functions associated with the generalized hypergeometric function. Math. comput. Modelling 38 (2004) · Zbl 1049.30008 [8] Miller, S. S.; Mocanu, P. T.: Differential subordinations and univalent functions. Michigan math. J 28, 157-171 (1981) · Zbl 0439.30015 [9] S.S. Miller, P.T. Mocanu, Differential Subordinations: Theory and Applications, Series on Monographs and Textbooks in Pure and Applied Mathematics, Vol. 225, Marcel Dekker, New York, 2000 [10] Owa, S.; Srivastava, H. M.: Univalent and starlike generalized hypergeometric functions. Canad. J. Math 39, 1057-1077 (1987) · Zbl 0611.33007 [11] Srivastava, H. M.; Owa, S.: Some characterization and distortion theorems involving fractional calculus, generalized hypergeometric functions, Hadamard products, linear operators, and certain subclasses of analytic functions. Nagoya math. J 106, 1-28 (1987) · Zbl 0607.30014 [12] Srivastava, H. M.; Owa, S.: Univalent functions, fractional calculus, and their applications. (1989) · Zbl 0683.00012 [13] Srivastava, H. M.; Owa, S.: Current topics in analytic function theory. (1992) · Zbl 0976.00007


text:
vec_as_names() takes a character vector of names and repairs it according to the repair argument. It is the r-lib and tidyverse equivalent of base::make.names().

vctrs deals with a few levels of name repair:

• minimal names exist. The names attribute is not NULL. The name of an unnamed element is "" and never NA. For instance, vec_as_names() always returns minimal names and data frames created by the tibble package have names that are, at least, minimal.

• unique names are minimal, have no duplicates, and can be used where a variable name is expected. Empty names, ..., and .. followed by a sequence of digits are banned.

• All columns can be accessed by name via df[["name"]] and df$name  and with(df, name). • universal names are unique and syntactic (see Details for more). • Names work everywhere, without quoting: df$name and with(df, name) and lm(name1 ~ name2, data = df) and dplyr::select(df, name) all work.

universal implies unique, unique implies minimal. These levels are nested.

vec_as_names(
names,
...,
repair = c("minimal", "unique", "universal", "check_unique"),
quiet = FALSE
)

## Arguments

names A character vector. These dots are for future extensions and must be empty. Either a string or a function. If a string, it must be one of "check_unique", "minimal", "unique", or "universal". If a function, it is invoked with a vector of minimal names and must return minimal names, otherwise an error is thrown. Minimal names are never NULL or NA. When an element doesn't have a name, its minimal name is an empty string. Unique names are unique. A suffix is appended to duplicate names to make them unique. Universal names are unique and syntactic, meaning that you can safely use the names as variables without causing a syntax error. The "check_unique" option doesn't perform any name repair. Instead, an error is raised if the names don't suit the "unique" criteria. By default, the user is informed of any renaming caused by repairing the names. This only concerns unique and universal repairing. Set quiet to TRUE to silence the messages.

## minimal names

minimal names exist. The names attribute is not NULL. The name of an unnamed element is "" and never NA.

Examples:

Original names of a vector with length 3: NULL
minimal names: "" "" ""

Original names: "x" NA
minimal names: "x" ""

unique names are minimal, have no duplicates, and can be used (possibly with backticks) in contexts where a variable is expected. Empty names, ..., and .. followed by a sequence of digits are banned. If a data frame has unique names, you can index it by name, and also access the columns by name. In particular, df[["name"]] and df$name and also with(df, name) always work. There are many ways to make names unique. We append a suffix of the form ...j to any name that is "" or a duplicate, where j is the position. We also change ..# and ... to ...#. Example: Original names: "" "x" "" "y" "x" "..2" "..." unique names: "...1" "x...2" "...3" "y" "x...5" "...6" "...7"  Pre-existing suffixes of the form ...j are always stripped, prior to making names unique, i.e. reconstructing the suffixes. If this interacts poorly with your names, you should take control of name repair. ## universal names universal names are unique and syntactic, meaning they: • Are never empty (inherited from unique). • Have no duplicates (inherited from unique). • Are not .... Do not have the form ..i, where i is a number (inherited from unique). • Consist of letters, numbers, and the dot . or underscore _ characters. • Start with a letter or start with the dot . not followed by a number. • Are not a reserved word, e.g., if or function or TRUE. If a vector has universal names, variable names can be used "as is" in code. They work well with nonstandard evaluation, e.g., df$name works.

vctrs has a different method of making names syntactic than base::make.names(). In general, vctrs prepends one or more dots . until the name is syntactic.

Examples:

 Original names:     ""     "x"    NA      "x"
universal names: "...1" "x...2" "...3" "x...4"

Original names: "(y)"  "_z"  ".2fa"  "FALSE"
universal names: ".y." "._z" "..2fa" ".FALSE"

rlang::names2() returns the names of an object, after making them minimal.

The Names attribute section in the "tidyverse package development principles".

## Examples

# By default, vec_as_names() returns minimal names:
vec_as_names(c(NA, NA, "foo"))#> [1] ""    ""    "foo"
# You can make them unique:
vec_as_names(c(NA, NA, "foo"), repair = "unique")#> New names:
#> * NA -> ...1
#> * NA -> ...2#> [1] "...1" "...2" "foo"
# Universal repairing fixes any non-syntactic name:
vec_as_names(c("_foo", "+"), repair = "universal")#> New names:
#> * _foo -> ._foo
#> * + -> .#> [1] "._foo" "."


text:
• On the Iwasawa Algebra Associated to a Normal Element of $\mathbb{C}_p$

• # Fulltext

https://www.ias.ac.in/article/fulltext/pmsc/120/01/0045-0055

• # Keywords

Iwasawa algebra; local fields; distributions.

• # Abstract

Given a prime number 𝑝 and the Galois orbit $O(x)$ of a normal element 𝑥 of $\mathbb{C}_p$, the topological completion of the algebraic closure of the field of 𝑝-adic numbers, we study the Iwasawa algebra of $O(x)$ with scalars drawn from $\mathbb{Q}_p$ and relate it with $\mathbb{Q}_p$-distributions and functionals.

• # Author Affiliations

1. Department of Mathematics, University of Bucharest, Romania
2. Institute of Mathematics of the Romanian Academy, P.O. Box 1-764, Bucharest 014700, Romania
3. Department of Mathematics, University of Illinois at Urbana-Champaign, Altgeld Hall, 1409 W. Green Street, Urbana, IL 61801, USA

• # Proceedings – Mathematical Sciences

Volume 131, 2021
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
# Towards Pattern-Based Reasoning for Friendly Ontology Debugging

@article{Jarrar2008TowardsPR,
title={Towards Pattern-Based Reasoning for Friendly Ontology Debugging},
author={Mustafa Jarrar and Stijn Heymans},
journal={Int. J. Artif. Intell. Tools},
year={2008},
volume={17},
pages={607-634}
}
• Published 18 April 2008
• Computer Science
• Int. J. Artif. Intell. Tools
Reasoning with ontologies is a challenging task specially for non-logic experts. When checking whether an ontology contains rules that contradict each other, current description logic reasoners can only provide a list of the unsatisfiable concepts. Figuring out why these concepts are unsatisfiable, which rules cause conflicts, and how to resolve these conflicts, is all left to the ontology modeler himself. The problem becomes even more challenging in case of large or medium size ontologies…
28 Citations

## Figures from this paper

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The mappings are implemented as an extension to the DogmaModeler tool, which uses Racer as a background reasoning engine, and the expressive, methodological, and graphical capabilities of ORM make it a good candidate for use as a graphical notation for ontology languages.
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• Computer Science, Philosophy
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• 2018
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This research presents a meta-modelling framework that automates the very labor-intensive and therefore time-heavy and therefore expensive and expensive process of manually cataloging and cataloging individual words and phrases in a ontology.
Checking Full Satisfiability of Conceptual Models
• Computer Science
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• 2010
This investigation shows that the full satisfiabil- ity problem is ExpTime-complete in the full scenario, NP-complete if the authors drop isa between relationships, and NLogSpace- complete if they further drop covering over classes.
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• Computer Science
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• 2015
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• Computer Science
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• Computer Science
ER
• 2010
This investigation shows that the full satisfiability problem is ExpTime-complete in the full scenario, NP- complete if the authors drop isa between relationships, and NLogSpace-complete if they further drop covering over classes.

## References

SHOWING 1-10 OF 30 REFERENCES
Debugging OWL-DL Ontologies: A Heuristic Approach
• Computer Science
SEMWEB
• 2005
This paper presents a ‘black boxed' heuristic approach based on identifying common errors and inferences that can be used to create and use OWL ontologies.
Towards Automated Reasoning on ORM Schemes - Mapping ORM into the DLRidf Description Logic
The goal of this article is to formalize Object Role Modeling (ORM) using the DLR description logic. This would enable automated reasoning on the formal properties of ORM diagrams, such as detecting
Mapping ORM into the SHOIN/OWL Description Logic - Towards a Methodological and Expressive Graphical Notation for Ontology Engineering
The mappings are implemented as an extension to the DogmaModeler tool, which uses Racer as a background reasoning engine, and the expressive, methodological, and graphical capabilities of ORM make it a good candidate for use as a graphical notation for ontology languages.
Resolving Inconsistencies in Evolving Ontologies
• Computer Science, Philosophy
ESWC
• 2006
An algorithm to select the axioms from an ontology causing the inconsistency, as well as a set of rules that ontology engineers can use to resolve the detected inconsistency are proposed.
Towards Methodological Principles for Ontology Engineering.
The main goal of this thesis is to present methodological principles for ontology engineering to guide ontology builders towards building ontologies that are both highly reusable and usable, easier to build, and smoother to maintain.
Debugging OWL ontologies
• Computer Science
WWW '05
• 2005
A number of simple debugging cues generated from the description logic reasoner, Pellet, are integrated in the hypertextual ontology development environment, Swoop, to significantly improve the OWL debugging experience, and point the way to more general improvements in the presentation of an ontology to new users.
Context Dependency Management in Ontology Engineering: A Formal Approach
• Computer Science
J. Data Semant.
• 2007
A formal framework for supporting context dependency management processes, based on the DOGMA framework and methodology for scalable ontology engineering, which can be combined to manage complex context dependencies like articulation, application, specialisation, and revision dependencies.
On Using Conceptual Data Modeling for Ontology Engineering
• Computer Science
J. Data Semant.
• 2003
An ontology engineering-framework that enables reusing conceptual modeling approaches in modeling and representing ontologies, and an ontology-engineering tool that implements the ideas presented in the paper.
Explanation Support for OWL Authoring
• Computer Science
ExaCt
• 2005
This paper describes ongoing work towards a systems for explaining subsumption for a significant fraction of the Description Logic (DL) underlying OWL Lite, and uses a tableau-based approach to generate step by step quasi-natural language explanations.
OWL Pizzas: Practical Experience of Teaching OWL-DL: Common Errors & Common Patterns
• Computer Science
EKAW
• 2004
This paper presents the most common difficulties encountered by newcomers to the language, that have been observed during the course of more than a dozen workshops, tutorials and modules about OWL-DL and it’s predecessor languages.


text:
Build neural networks in 1 line

## Project description

nn_builder lets you build neural networks without boilerplate code. You specify the type of network you want and it builds it.

### Install

pip install nn_builder

### Support

Network Type NN CNN RNN
PyTorch :heavy_check_mark: :heavy_check_mark: :heavy_check_mark:
TensorFlow 2.0 :heavy_check_mark: :heavy_check_mark: :heavy_check_mark:

### Examples

On the left is how you can create the PyTorch neural network on the right in only 1 line of code using nn_builder:

Similarly for TensorFlow on the left is how you can create the CNN on the right in only 1 line of code using nn_builder:

### Usage

See this colab notebook for lots of examples of how to use the module. 3 types of PyTorch and TensorFlow network are currently supported: NN, CNN and RNN. Each network takes the following arguments:

Field Description Default
input_dim Dimension of the input into the network. See below for more detail. Not needed for Tensorflow. N/A
layers_info List to indicate the layers of the network you want. Exact requirements depend on network type, see below for more detail N/A
output_activation String to indicate the activation function you want the output to go through. Provide a list of strings if you want multiple output heads No activation
hidden_activations String or list of string to indicate the activations you want used on the output of hidden layers (not including the output layer), default is ReLU and for example "tanh" would have tanh applied on all hidden layer activations ReLU after every hidden layer
dropout Float to indicate what dropout probability you want applied after each hidden layer 0
initialiser String to indicate which initialiser you want used to initialise all the parameters PyTorch & TF Default
batch_norm Boolean to indicate whether you want batch norm applied to the output of every hidden layer False
columns of_data_to_be_embedded List to indicate the column numbers of the data that you want to be put through an embedding layer before being fed through the hidden layers of the network No embeddings
embedding_dimensions If you have categorical variables you want embedded before flowing through the network then you specify the embedding dimensions here with a list of the form: [ [embedding_input_dim_1, embedding_output_dim_1], [embedding_input_dim_2, embedding_output_dim_2] ...] No embeddings
y_range Tuple of float or integers of the form (y_lower, y_upper) indicating the range you want to restrict the output values to in regression tasks No range
random_seed Integer to indicate the random seed you want to use 0
return_final_seq_only Only needed for RNN. Boolean to indicate whether you only want to return the output for the final timestep (True) or if you want to return the output for all timesteps (False) True

Each network type has slightly different requirements for input_dim and layers_info as explained below:

##### 1. NN
• input_dim: # Features in PyTorch, not needed for TensorFlow
• layers_info: List of integers to indicate number of hidden units you want per linear layer.
• For example:
from nn_builder.pytorch.NN import NN
model = NN(input_dim=5, layers_info=[10, 10, 1], output_activation=None, hidden_activations="relu",
dropout=0.0, initialiser="xavier", batch_norm=False)

##### 2. CNN
• input_dim: (# Channels, Height, Width) in PyTorch, not needed for TensorFlow
• layers_info: We expect the field layers_info to be a list of lists indicating the size and type of layers that you want. Each layer in a CNN can be one of these 4 forms:
• ["conv", channels, kernel size, stride, padding]
• ["maxpool", kernel size, stride, padding]
• ["avgpool", kernel size, stride, padding]
• ["linear", units]
• For a PyTorch network kernel size, stride, padding and units must be integers. For TensorFlow they must all be integers except for padding which must be one of {“valid”, “same”}
• For example:
from nn_builder.pytorch.CNN import CNN
model = CNN(input_dim=(3, 64, 64),
layers_info=[["conv", 32, 3, 1, 0], ["maxpool", 2, 2, 0],
["conv", 64, 3, 1, 2], ["avgpool", 2, 2, 0],
["linear", 10]],
hidden_activations="relu", output_activation="softmax", dropout=0.0,
initialiser="xavier", batch_norm=True)

##### 3. RNN
• input_dim: # Features in PyTorch, not needed for TensorFlow
• layers_info: We expect the field layers_info to be a list of lists indicating the size and type of layers that you want. Each layer in a CNN can be one of these 4 forms:
• ["lstm", units]
• ["gru", units]
• ["linear", units]
• For example:
from nn_builder.pytorch.CNN import CNN
model = RNN(input_dim=5, layers_info=[["gru", 50], ["lstm", 10], ["linear", 2]],
hidden_activations="relu", output_activation="softmax",
batch_norm=False, dropout=0.0, initialiser="xavier")

## Contributing

Together we can make something that is useful for thousands of people. Anyone is very welcome to contribute via a pull request. Please see the issues page for ideas on the best areas to contribute to and try to:

1. Add tests to the tests folder that cover any code you write
2. Write comments for every function
3. Create a colab notebook demonstrating how any extra functionality you created works

## Project details

Uploaded source
Uploaded py3


text:
Editing matrix in a program: not updating issue
03-17-2015, 09:15 PM (This post was last modified: 03-17-2015 09:17 PM by salvomic.)
Post: #1
 salvomic Senior Member Posts: 1,364 Joined: Jan 2015
Editing matrix in a program: not updating issue
hi,
perhaps I'm using in wrong mode my code, but in my new program for ANOVA two factors I'm searching to editing a matrix (anovamat) trying to get two goals: if the matrix is already set then use it; if not, input a new one.
My programs seems to work but after the first run it use the old data and sure I'm setting variables in a wrong mode. I use something like that with reals and it works, resetting values of variables at every running, but not here...

The whole program is here.

See:
Code:
 export anovamat:=[[0,0],[0,0]]; ... export anova_dfreedom2:={0,0,0,0}; EXPORT ANOVA2() BEGIN local a, ... ... EDITMAT(anovamat); sto(anovamat, anovamat); ...

I'm searching also a mode to set export to get a generic matrix without precise setting for rows and cols in
export anovamat ...

∫aL√0mic (IT9CLU), HP Prime 50g 41CX 71b 42s 12C 15C - DM42 WP34s :: Prime Soft. Lib
03-18-2015, 07:24 AM
Post: #2
 cyrille de brébisson Senior Member Posts: 866 Joined: Dec 2013
RE: Editing matrix in a program: not updating issue
Hello,

Not sure that I understand what you want to do, but

editmat(anovamat), assuming that anovamat is a matrix should edit it. If the user pressed OK at the end, then the value of anovamat should change. If the user presses CANCEL, then it will stay unchanged...

There is no need for the sto underneath...

Since anovamat is a global variable of your program, and assuming that you do not edit the program in between 2 runs (which will reset anovamat), the value of anovamat should stay the same between 2 runs (unless you override it later in your program)...

If you want a 'new' anovamat each run, then you would need to either reset it manually at each run, OR declare it as a local variable.

Cyrille
03-18-2015, 08:41 AM
Post: #3
 salvomic Senior Member Posts: 1,364 Joined: Jan 2015
RE: Editing matrix in a program: not updating issue
(03-18-2015 07:24 AM)cyrille de brébisson Wrote:  Since anovamat is a global variable of your program, and assuming that you do not edit the program in between 2 runs (which will reset anovamat), the value of anovamat should stay the same between 2 runs (unless you override it later in your program)...

If you want a 'new' anovamat each run, then you would need to either reset it manually at each run, OR declare it as a local variable.

Cyrille

Hello,
I don't want another matrix, only that.
What happen if the user press "Enter" instead of OK? I'm trying this...

However after some tries with CAS and Home, in Matrix Editor I saw two matrix: anovamat and anovamat (CAS)...

∫aL√0mic (IT9CLU), HP Prime 50g 41CX 71b 42s 12C 15C - DM42 WP34s :: Prime Soft. Lib
03-19-2015, 06:53 AM
Post: #4
 cyrille de brébisson Senior Member Posts: 866 Joined: Dec 2013
RE: Editing matrix in a program: not updating issue
ENTER and OK, ESC and CANCEL are equivalents.

The Cas namespace is different from the home namespace, even if efforts have been made to merge them, they are for all intend and purposes separated internally.

In your case, you managed to create 2 variables with the same name in the 2 domains. it's actually not that hard to do.
Create the cas version first
then declare the home version as a program exported variable.

Anyhow, try to avoid mixing CAS and non CAS stuff if you can as it tend to lead to troubles.

Good luck

Cyrille
03-19-2015, 10:10 AM
Post: #5
 salvomic Senior Member Posts: 1,364 Joined: Jan 2015
RE: Editing matrix in a program: not updating issue
(03-19-2015 06:53 AM)cyrille de brébisson Wrote:  In your case, you managed to create 2 variables with the same name in the 2 domains. it's actually not that hard to do.
Create the cas version first
then declare the home version as a program exported variable.

hi Cyrille,
I created also the exported variable, to let user edit matrix both inside and outside the program, and I'd like the program could be used in CAS and in Home (and it seems so).
A few times I got the problem with updating of the matrix but now I cannot reproduce the issue (I'm trying other times)...
However the program generally is ok.
I noted also that if I change something in the program and save, the matrix will be reset. But, indeed, one doesn't change things in a program every day

ciao
Salvo

∫aL√0mic (IT9CLU), HP Prime 50g 41CX 71b 42s 12C 15C - DM42 WP34s :: Prime Soft. Lib
03-20-2015, 06:48 AM
Post: #6
 cyrille de brébisson Senior Member Posts: 866 Joined: Dec 2013
RE: Editing matrix in a program: not updating issue
Hello

>I noted also that if I change something in the program and save, the matrix will be reset. But, indeed, one doesn't change things in a program every day

Yes, this is normal and intended.
All the Program global variables will be deleted when the program is changed (after all, it is very possible that editing the program will cause some of these variables to be erased and new ones created).
Then, upon suxesfull compilation of the program, the variables are created and initialized at the specified value.

cyrille
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text:
How would you find the quantum numbers for an atom?

Dec 1, 2015

See explanation.

Explanation:

Quantum numbers are considered as the coordinates that will give an idea about the location of an electron in an atom.

There are four different quantum numbers:

1. Principal Quantum Number (n) has integral values 1, 2, 3, ... This number determines the size and energy of the orbital.
2. Angular Momentum Quantum Number ($l$) associated with the number of different types of subshells in an energy level. $l = 0 , 1 , 2 , 3 , \ldots \left(n - 1\right)$.

3. Magnetic Quantum Number (${m}_{l}$) describes the orientation (direction) in space of each orbital and can have integral values from $- l$ through $+ l$.

4. Electron Spin Quantum Number (${m}_{s}$) describes the direction of spin of the electron(s) in an orbital. There are only two values of ${m}_{s} , + \frac{1}{2} \text{and} - \frac{1}{2}$.

Note that no two electrons may have identical sets of all 4 quantum numbers.

Consider the electron configuration of sodium ""_11Na:

""_11Na:1s^(2)2s^(2)2p^(6)color(red)(3)color(blue)(s)^(color(green)(1))

What would be the set of quantum numbers of the valence electron of sodium:
$\textcolor{red}{n = 3}$: third energy level.
$\textcolor{b l u e}{l = 0}$: sublevel or subshell $s$.
$\textcolor{b l u e}{{m}_{l} = 0}$: one possible orbital.
$\textcolor{g r e e n}{{m}_{s} = + \frac{1}{2}} \mathmr{and} \textcolor{g r e e n}{- \frac{1}{2}}$: the electron spin.


text:
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# M15#18

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M15#18 [#permalink]  19 Nov 2008, 18:24
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A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

(A) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.
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Re: M15#18 [#permalink]  19 Nov 2008, 22:44
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

(C) 2008 GMAT Club - m15#18

* $$\frac{2}{21}$$
* $$\frac{3}{25}$$
* $$\frac{1}{6}$$
* $$\frac{9}{28}$$
* $$\frac{11}{24}$$

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

= (9c1/9c1) (6c1/8c1) (3c1/7c1)
= 1 (6/8) (3/7)
= 9/28
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Re: M15#18 [#permalink]  19 Nov 2008, 23:26
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(3C1 * 3C1 * 3C1) / 9C3
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Re: M15#18 [#permalink]  20 Nov 2008, 10:08
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?
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Re: M15#18 [#permalink]  17 Dec 2008, 07:27
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

This is because we have to find the probability of 3 events combined: 1 blue AND 1 red AND 1 yellow.

Hope this helps.
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Re: M15#18 [#permalink]  02 Mar 2010, 06:17
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

(A) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

p = (3c1*3c1*3c1)/9c3 = 9/28 hence D.
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Re: M15#18 [#permalink]  02 Mar 2010, 06:58
Sorry, how is (3c1*3c1*3c1)/9c3 = 9/28. I keep getting 9/84? 3c1 cubed is 9. Why is 9c3 28? I am sure it has to do with order or something or am I just missing something?

Edit: Crap because 3 cubed is not 9. 27/84 == 9/28. I really hope I don't make stupid mistakes like this on the 17th when I take the test. (It is taking some will power not to edit my question above.)
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Re: M15#18 [#permalink]  02 Mar 2010, 07:25
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Out of 3 blue marbles, 3 red marbles, and 3 yellow marbles, probability of selecting one marble of each color is as follows:

B R Y
(3/9) * (3/8) * (3/7)

B, R, and Y can be arranged in factorial (3) = 3 * 2 * 1 ways.

Thus, resulting probability is:

(3/9) * (3/8) * (3/7) * (3 * 2 * 1) = 9/28

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Re: M15#18 [#permalink]  02 Mar 2010, 08:18
3/9 * 3/8 * 3/7 * 3! = 9/28
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Re: M15#18 [#permalink]  02 Mar 2010, 21:41
I'm still a bit confused. I got the denominator correct (9C3), which enumerates all the possible outcomes. But I only see 6 (3!) desired outcomes...not 27:

BRY
BYR
YRB
YBR
RBY
RYB

What am I missing?

EDIT: I think this is only taking into account only one of three of each color...i.e. B1 Y1 R1, etc.

Let me know if I am right!
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Re: M15#18 [#permalink]  13 Mar 2010, 08:15
(3C1 * 3C1 * 3C1)/9C3

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Re: M15#18 [#permalink]  07 Mar 2011, 20:36
dzyubam wrote:
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

This is because we have to find the probability of 3 events combined: 1 blue AND 1 red AND 1 yellow.

Hope this helps.

Please explain why we have written 9C3 in the denominator.
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Re: M15#18 [#permalink]  08 Mar 2011, 00:30
Denominator contains the number of total possible outcomes. In this case, the number of possible outcomes equals the number of ways 3 marbles can be drawn from the total of 9 marbles available (9C3).
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Re: M15#18 [#permalink]  08 Mar 2011, 22:44
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The faster way to calculate the probability is to follow events manually:
1. The first marble will be of any one color. Probability (P1) is 100% or 1;
2. The second marble should be of other two colors (any of 6 out of 8 remaining marbles). Probability (P2) is 6/8 or 3/4;
3. The third marble should be of the last color (any of 3 out of 7 remaining marbles). Probability (P3) is 3/7;

P=P1*P2*P3=1*3/4*3/7=9/28

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Re: M15#18 [#permalink]  10 Mar 2012, 00:35
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

(A) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

Got D!!!

First Draw - (9 marbles/ 9 marbles) = 1
Second Draw - Since one color is selected & we dont want that same one again... (6 marbles/ 8 remaining marbles) = 3/4
Third Draw - Since two colors are selected & we dont want the same ones again... (3 marbles/ 7 remaining marbles) = 3/7

Multiply all three to get the probability...
1 * (3/4) * (3/7) = 9/28
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Re: M15#18 [#permalink]  11 Mar 2013, 01:41
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For first ball...we can pick any one out of nine..(lets say first is blue)

For second ball of different color...we have choose one ball out of 8 balls
and color of ball should not blue..it may be of red or yellow color...
so probability is 6(3 red and 3 yellow)/8...(lets say we pick a red one here)

For third ball of different color...we have choose one ball out of 7 balls
and color of ball should yellow..
so probability is 3 (yellow)/7

So probability that a marble of each color is among the extracted
$$= \frac{6}{8}*\frac{3}{7}$$
$$= \frac{9}{28}$$
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Re: M15#18 [#permalink]  11 Mar 2013, 01:49
1
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Expert's post
snowy2009 wrote:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

(A) $$\frac{2}{21}$$
(B) $$\frac{3}{25}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{9}{28}$$
(E) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

A. $$\frac{2}{21}$$
B. $$\frac{3}{25}$$
C. $$\frac{1}{6}$$
D. $$\frac{9}{28}$$
E. $$\frac{11}{24}$$

$$P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}$$. Numerator represents ways to select 1 blue marble out of 3, 1 red marble out of 3 and 1 yellow marble out of 3. Denominator represents total ways to select 3 marbles out of 9.

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Re: M15#18 [#permalink]  12 Mar 2013, 22:36
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

So as a basic with Combinations, if we have an event 1 (3C1 - 1 blue ball extracted) followed by event 2 (3C1 - 1 red ball extracted) and event 3 (3C1 - 1 yellow ball extracted), we have to multiply all events.

Instead if this was a case of 'OR', we would have added the events.

Hope that helped.

Can someone tell me what is the difficulty level of this question?
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Re: M15#18 [#permalink]  13 Mar 2013, 01:35
1
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Expert's post
ranjitaarao wrote:
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

So as a basic with Combinations, if we have an event 1 (3C1 - 1 blue ball extracted) followed by event 2 (3C1 - 1 red ball extracted) and event 3 (3C1 - 1 yellow ball extracted), we have to multiply all events.

Instead if this was a case of 'OR', we would have added the events.

Hope that helped.

Can someone tell me what is the difficulty level of this question?

The difficulty level is ~650.
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Re: M15#18 [#permalink]  24 Feb 2014, 10:03
nevermind... i realize i was forgetting the parenthetical part of the selection. in the denominator... its follish lapses like these that affect my scores and keep me in the 600s
Re: M15#18   [#permalink] 24 Feb 2014, 10:03

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