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(Redirected from Fluence)

In radiometry, radiant exposure or fluence is the radiant energy received by a surface per unit area, or equivalently the irradiance of a surface, integrated over time of irradiation, and spectral exposure is the radiant exposure per unit frequency or wavelength, depending on whether the spectrum is taken as a function of frequency or of wavelength. The SI unit of radiant exposure is the joule per square metre (J/m2), while that of spectral exposure in frequency is the joule per square metre per hertz (J⋅m−2⋅Hz−1) and that of spectral exposure in wavelength is the joule per square metre per metre (J/m3)—commonly the joule per square metre per nanometre (J⋅m−2⋅nm−1).

## Mathematical definitions

Radiant exposure of a surface, denoted He ("e" for "energetic", to avoid confusion with photometric quantities), is defined as[1]

${\displaystyle H_{\mathrm {e} }={\frac {\partial Q_{\mathrm {e} }}{\partial A}}=\int _{0}^{T}E_{\mathrm {e} }(t)\,\mathrm {d} t,}$

where

• ∂ is the partial derivative symbol;
• Qe is the radiant energy;
• A is the area;
• T is the duration of irradiation;

### Spectral exposure

Spectral exposure in frequency of a surface, denoted He,ν, is defined as[1]

${\displaystyle H_{\mathrm {e} ,\nu }={\frac {\partial H_{\mathrm {e} }}{\partial \nu }},}$

where ν is the frequency.

Spectral exposure in wavelength of a surface, denoted He,λ, is defined as[1]

${\displaystyle H_{\mathrm {e} ,\lambda }={\frac {\partial H_{\mathrm {e} }}{\partial \lambda }},}$

where λ is the wavelength.

Quantity Unit Dimension Notes
Name Symbol[nb 1] Name Symbol Symbol
Radiant energy density we joule per cubic metre J/m3 ML−1T−2 Radiant energy per unit volume.
Radiant flux Φe[nb 2] watt W = J/s ML2T−3 Radiant energy emitted, reflected, transmitted or received, per unit time. This is sometimes also called "radiant power".
Spectral flux Φe,ν[nb 3] watt per hertz W/Hz ML2T−2 Radiant flux per unit frequency or wavelength. The latter is commonly measured in W⋅nm−1.
Φe,λ[nb 4] watt per metre W/m MLT−3
Radiant intensity Ie,Ω[nb 5] watt per steradian W/sr ML2T−3 Radiant flux emitted, reflected, transmitted or received, per unit solid angle. This is a directional quantity.
Spectral intensity Ie,Ω,ν[nb 3] watt per steradian per hertz W⋅sr−1⋅Hz−1 ML2T−2 Radiant intensity per unit frequency or wavelength. The latter is commonly measured in W⋅sr−1⋅nm−1. This is a directional quantity.
Ie,Ω,λ[nb 4] watt per steradian per metre W⋅sr−1⋅m−1 MLT−3
Radiance Le,Ω[nb 5] watt per steradian per square metre W⋅sr−1⋅m−2 MT−3 Radiant flux emitted, reflected, transmitted or received by a surface, per unit solid angle per unit projected area. This is a directional quantity. This is sometimes also confusingly called "intensity".
Spectral radiance Le,Ω,ν[nb 3] watt per steradian per square metre per hertz W⋅sr−1⋅m−2⋅Hz−1 MT−2 Radiance of a surface per unit frequency or wavelength. The latter is commonly measured in W⋅sr−1⋅m−2⋅nm−1. This is a directional quantity. This is sometimes also confusingly called "spectral intensity".
Le,Ω,λ[nb 4] watt per steradian per square metre, per metre W⋅sr−1⋅m−3 ML−1T−3
Flux density
Ee[nb 2] watt per square metre W/m2 MT−3 Radiant flux received by a surface per unit area. This is sometimes also confusingly called "intensity".
Spectral flux density
Ee,ν[nb 3] watt per square metre per hertz W⋅m−2⋅Hz−1 MT−2 Irradiance of a surface per unit frequency or wavelength. This is sometimes also confusingly called "spectral intensity". Non-SI units of spectral flux density include jansky (1 Jy = 10−26 W⋅m−2⋅Hz−1) and solar flux unit (1 sfu = 10−22 W⋅m−2⋅Hz−1 = 104 Jy).
Ee,λ[nb 4] watt per square metre, per metre W/m3 ML−1T−3
Radiosity Je[nb 2] watt per square metre W/m2 MT−3 Radiant flux leaving (emitted, reflected and transmitted by) a surface per unit area. This is sometimes also confusingly called "intensity".
Spectral radiosity Je,ν[nb 3] watt per square metre per hertz W⋅m−2⋅Hz−1 MT−2 Radiosity of a surface per unit frequency or wavelength. The latter is commonly measured in W⋅m−2⋅nm−1. This is sometimes also confusingly called "spectral intensity".
Je,λ[nb 4] watt per square metre, per metre W/m3 ML−1T−3
Radiant exitance Me[nb 2] watt per square metre W/m2 MT−3 Radiant flux emitted by a surface per unit area. This is the emitted component of radiosity. "Radiant emittance" is an old term for this quantity. This is sometimes also confusingly called "intensity".
Spectral exitance Me,ν[nb 3] watt per square metre per hertz W⋅m−2⋅Hz−1 MT−2 Radiant exitance of a surface per unit frequency or wavelength. The latter is commonly measured in W⋅m−2⋅nm−1. "Spectral emittance" is an old term for this quantity. This is sometimes also confusingly called "spectral intensity".
Me,λ[nb 4] watt per square metre, per metre W/m3 ML−1T−3
Radiant exposure He joule per square metre J/m2 MT−2 Radiant energy received by a surface per unit area, or equivalently irradiance of a surface integrated over time of irradiation. This is sometimes also called "radiant fluence".
Spectral exposure He,ν[nb 3] joule per square metre per hertz J⋅m−2⋅Hz−1 MT−1 Radiant exposure of a surface per unit frequency or wavelength. The latter is commonly measured in J⋅m−2⋅nm−1. This is sometimes also called "spectral fluence".
He,λ[nb 4] joule per square metre, per metre J/m3 ML−1T−2
Hemispherical emissivity ε N/A 1 Radiant exitance of a surface, divided by that of a black body at the same temperature as that surface.
Spectral hemispherical emissivity εν
or
ελ
N/A 1 Spectral exitance of a surface, divided by that of a black body at the same temperature as that surface.
Directional emissivity εΩ N/A 1 Radiance emitted by a surface, divided by that emitted by a black body at the same temperature as that surface.
Spectral directional emissivity εΩ,ν
or
εΩ,λ
N/A 1 Spectral radiance emitted by a surface, divided by that of a black body at the same temperature as that surface.
Hemispherical absorptance A N/A 1 Radiant flux absorbed by a surface, divided by that received by that surface. This should not be confused with "absorbance".
Spectral hemispherical absorptance Aν
or
Aλ
N/A 1 Spectral flux absorbed by a surface, divided by that received by that surface. This should not be confused with "spectral absorbance".
Directional absorptance AΩ N/A 1 Radiance absorbed by a surface, divided by the radiance incident onto that surface. This should not be confused with "absorbance".
Spectral directional absorptance AΩ,ν
or
AΩ,λ
N/A 1 Spectral radiance absorbed by a surface, divided by the spectral radiance incident onto that surface. This should not be confused with "spectral absorbance".
Hemispherical reflectance R N/A 1 Radiant flux reflected by a surface, divided by that received by that surface.
Spectral hemispherical reflectance Rν
or
Rλ
N/A 1 Spectral flux reflected by a surface, divided by that received by that surface.
Directional reflectance RΩ N/A 1 Radiance reflected by a surface, divided by that received by that surface.
Spectral directional reflectance RΩ,ν
or
RΩ,λ
N/A 1 Spectral radiance reflected by a surface, divided by that received by that surface.
Hemispherical transmittance T N/A 1 Radiant flux transmitted by a surface, divided by that received by that surface.
Spectral hemispherical transmittance Tν
or
Tλ
N/A 1 Spectral flux transmitted by a surface, divided by that received by that surface.
Directional transmittance TΩ N/A 1 Radiance transmitted by a surface, divided by that received by that surface.
Spectral directional transmittance TΩ,ν
or
TΩ,λ
N/A 1 Spectral radiance transmitted by a surface, divided by that received by that surface.
Hemispherical attenuation coefficient μ reciprocal metre m−1 L−1 Radiant flux absorbed and scattered by a volume per unit length, divided by that received by that volume.
Spectral hemispherical attenuation coefficient μν
or
μλ
reciprocal metre m−1 L−1 Spectral radiant flux absorbed and scattered by a volume per unit length, divided by that received by that volume.
Directional attenuation coefficient μΩ reciprocal metre m−1 L−1 Radiance absorbed and scattered by a volume per unit length, divided by that received by that volume.
Spectral directional attenuation coefficient μΩ,ν
or
μΩ,λ
reciprocal metre m−1 L−1 Spectral radiance absorbed and scattered by a volume per unit length, divided by that received by that volume.


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## Which plane convex arcs have the smallest maximum curvature?

Let p,q be positive real numbers with p less than q. Suppose that H(p,q) is the class of all convex arcs c in the Cartesian x-y plane which satisfy the following conditions: (1)The y-axis is an axis of symmetry of c. (2)The points (-p,0) and (p,0) are the end-points of c and the point (0,q) is also a point of c. (3)The curvature of c is defined and continuous at each point of c and never changes sign (which can always be taken to be non-negative).

QUESTION: Given p and q is there a formula for the greatest lower bound of the maximum curvature that an arc c belonging to the class H(p,q) can have? Is there a particular arc in this class which actually attains this greatest lower bound?

The (upper half of the) ellipse whose equation is "(x^2)/(p^2)+(y^2)/(q^2)=1" has a maximum curvature of q/(p^2) and is an arc belonging to the class H(p,q), but I cannot prove-and do not think- that q/(p^2) is actually a greatest lower bound for the whole class (except in the case p=q which is specifically excluded).

-
The end of your post is missing; if I were you, I would look at a stadium curve (union of half circles and lines). – Benoît Kloeckner Apr 26 2010 at 14:57
If the circle arc through the three points in question is an admissible curve, it is the answer. – fedja Apr 26 2010 at 15:01
What is a "convex arc"? If it is anything with curvature of constant sign, then the lower bound for the max curvature is zero. If it is an arc that lies on the boundary of its own convex hull, then the answer is a circle arc. If it is a graph of a convex function, the answer is either circle arc or a stadium curve depending on which of p and q is larger. – Sergei Ivanov Apr 26 2010 at 15:19
Thanks for your answer, Sergei. The convex arcs I am considering are always graphs of convex functions and I should have made this clear. I am especially interested in cases where p is "very small" and q is "much greater" than p. Your answer then indicates that the greatest lower bound would be attained by a stadium curve-call it "s(p,q)". But I would still like to know whether, given p and q, there is a formula expressing the maximum curvature of s(p,q) as a function of p and q. Does the curvature of stadium curves have discontinuities? If so, they could not be in the class H(p,q). – Garabed Gulbenkian Apr 29 2010 at 20:47
If $q>p$, the answer is $1/p$. This is the greatest lower bound which is not attained. The stadium curve where it would be attained is not admissible (it is not a graph of a function, and its curvature is discontinuous). But is can be approximated by admissible curves with slightly larger maximum curvature. – Sergei Ivanov May 4 2010 at 14:15

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## The Harmonic Archipelago

Another wild space that has been receiving a lot of recent attention is the so-called Harmonic Archipelago $\mathbb{HA}$ which is the following subspace of $\mathbb{R}^3$.

Harmonic Archipelago

You can describe the construction like this: Start by drawing the usual Hawaiian earring $\mathbb{H}$ onto a solid disk $D$ in the xy-plane. Now between the 1st and 2nd hoops, draw a small disk and push it up so that it becomes a smooth hill with unit height. Do the same thing between the 2nd and 3rd hoops of $\mathbb{H}$  and then the 3rd and 4th hoops and so on. Notice that the Hawaiian earring $\mathbb{H}$ is naturally a subspace of $\mathbb{HA}$ and each hill is hollow underneath. Also, the diameters of the hills tend to $0$.

If you can believe it, the fundamental group of this space is even crazier than the fundamental group of the Hawaiian earring $\mathbb{H}$!

First, let’s get some notation down:

• Let $C_n=\left\{(x,y,0)|\left(x-\frac{1}{n}\right)^2+y^2=\frac{1}{n^2}\right\}$ be the n-th circle so that $\mathbb{H}=\bigcup_{n\geq 1}C_n$ with basepoint $x_0=(0,0,0)$.
• Let $\mathbb{H}_{\geq m}=\bigcup_{n\geq m}C_n$ be the smaller copies of the Hawaiian earring.
• Let $B_n\subset \mathbb{HA}$ be the open  disk between $C_n$ and $C_{n+1}$ which contains the n-th hill in the archipelago.

Now some observations:

1) $\mathbb{HA}$ is path-connected and locally path connected.

2) $\mathbb{HA}$  is non-compact since it does not include limit points on the z-axis. For instance, the sequence given by the top of the hills converges to $(0,0,1)$ which is not included. This means that if $X$ is compact and $f:X\to \mathbb{HA}$ is continuous, then the image $f(X)$ can only hit finitely many of the hills.

3) If $\ell_n:S^1\to C_n$ is the loop that traverses $C_n$ once counterclockwise in the xy-plane, then $\ell_{n}$ can be deformed over finitely many hills (but not infinitely many!). So the homotopy classes $[\ell_{n}]=[\ell_{n+1}]$ in $\pi_1(\mathbb{HA},x_0)$ are the same for all $n\geq 1$ but yet are non-trivial since deforming $\ell_1$ over every hill should violate continuity.

Lemma 1: Every based loop $\alpha:S^1\to \mathbb{HA}$ is homotopic to a loop $\beta_{m}\colon S^1 \to\mathbb{H}_{\geq m}$ for each $m\geq 1$.

Proof: As mentioned above, the image of any based loop $\alpha:S^1\to \mathbb{HA}$ can intersect at most finitely many hills $B_n$. Thus $\alpha$ must have image in one of the spaces that looks like this:

Notice the holes get smaller and smaller so this subspace is a deformation retract of one of the form

$B_1\cup B_2\cup\cdots\cup B_{m-1}\cup\mathbb{H}_{\geq m+1}.$

The retraction is given by expanding each little circle in the xy-plane to the entire hole usually present in the Hawaiian earring. At this point, we can choose $m$ to be as large as we want (by adding some hills back in). The resulting space looks like the one-point union of a smaller copy of the Hawaiian earring and a bumpy region that is homotopy equivalent to a circle.

Where there are no hills, we see a copy of the Hawaiian earring. So the region between the 3rd/4th, 4th/5th,… circles is empty.

Now deform the bumpy region $B_1\cup B_2\cup\cdots\cup B_{m-1}$ onto the smallest circle $C_m$. The composition of these deformation retracts provides a homotopy of $\alpha$ to a loop in $\mathbb{H}_{\geq m}$. Since we could choose $m$ to be arbitrarily large, the lemma is proven.$\square$

Lemma 1 basically says that every based loop is homotopic to arbitrarily small loops.

Corollary 2: The homomorphism on fundamental groups $\phi :\pi_1(\mathbb{H},x_0)\to\pi_1(\mathbb{HA},x_0)$ induced by inclusion is surjective and $\phi([\ell_m])=\phi([\ell_n])$ for all $n,m\geq 1$.

More generally, if $g_n=[\ell_n]\in\pi_1(\mathbb{H},x_0)$, then $g_{k_1}^{\epsilon_1}g_{k_2}^{\epsilon_2} \cdots g_{k_p}^{\epsilon_p}\in\ker\phi$ whenever $\sum_{j}\epsilon_{j}=0$. However, the fundamental group $\pi_1(\mathbb{H},x_0)$ is way bigger than the free subgroup $F_{\infty}=$ so we should not expect that these are the only elements of $\ker\phi$.

Let’s make sure no other surprising homotopies of loops can show up.

Let $s_{n}:\mathbb{H}_{\geq n}\to \mathbb{H}_{\geq n+1}$ be the natural retraction which collapses $C_n$ homeomorphically onto $C_{n+1}$. These maps induce retractions $c_n:\pi_1(\mathbb{H}_{\geq n},x_0)\to\pi_1(\mathbb{H}_{\geq n},x_0)$ which together form a directed system:

Notice that if $\phi_n:\pi_1(\mathbb{H}_{\geq n},x_0)\to\pi_1(\mathbb{HA},x_0)$ is the homomorphism induced by inclusion, then we have, by Corollary 2, that $\phi_{n+1}\circ c_n=\phi_n$ for each $n\geq 1$. Consequently, we get a canonical homomorphism $\Phi$ from the direct limit:

Theorem 3: $\Phi:\varinjlim_{n}\pi_1(\mathbb{H}_{\geq n},x_0)\to\pi_1(\mathbb{HA},x_0)$ is an isomorphism of groups.

Proof: Since $\phi$ is surjective (Corollary 2), so is $\Phi$. Since each $c_n$ is a retraction it suffices to show that if $[\alpha]\in\pi_1(\mathbb{H},x_0)=\pi_1(\mathbb{H}_{\geq 1},x_0)$ and $\phi([\alpha])=1$, then $c_{n-1}\circ c_{n-2}\circ\dots\circ c_1([\alpha])=1$ for some $n>1$. Since $\phi([\alpha])=1$, there is a homotopy $H:[0,1]\times[0,1]\to\mathbb{HA}$ contracting $\alpha$ to the constant loop at ${x_0}$. By compactness, the image of $H$ can intersect only finitely many hills. Apply the composition of deformation retracts from the proof of Lemma 1 to obtain an $n$ and a homotopy $G:[0,1]\times[0,1]\to\mathbb{H}_{\geq n}$ which contracts $s_{n-1}\circ s_{n-2}\circ\dots\circ s_1\circ\alpha$ to the constant loop ${x_0}$. Thus $c_{n-1}\circ c_{n-2}\circ\dots\circ c_1([\alpha])=1$ in $\pi_1(\mathbb{H}_{\geq n},x_0)$. $\square$

Identifying  $\pi_1(\mathbb{HA},x_0)$ as a direct limit illustrates a kind of “universal property.”

Corollary 4: Suppose $Y$ is a space which is first countable at it’s basepoint $y_0$. For every shrinking sequence of based loops $\beta_n\to y_0$ such that $\beta_n\simeq\beta_{n+1}$ for all $n\geq 1$, there is a unique induced homomorphism $f:\pi_1(\mathbb{HA},x_0)\to \pi_1(Y,y_0)$ such that $f(g_n)=[\beta_n]$.

Here is one last interpretation of Theorem 4: Recall that we can represent a homotopy class $[\alpha]\in \pi_1(\mathbb{H},x_0)$ as a sequence $(w_1,w_2,...)\in\varprojlim_{n}F_n$, i.e. where $w_n$ is the word in the free group $F_n$ on letters $g_1,...,g_n$ obtained by removing all appearances of the letter $g_{n+1}$ from $w_{n+1}$. Also, the number of times a given letter $g_k$ can appear in $w_n$ stabilizes as $n\to \infty$ (in other words, $(w_1,w_2,...)$ is locally eventually constant).

If $k, let $\sigma_n(w_k)=1$  and if $k\geq n$, let $\sigma_n(w_k)$ be the reduced word in $F_n$ obtained after each letter $g_1,...,g_k$ is replaced by $g_n$. Now let

$\sigma_n(w_1,w_2,...)=(\sigma_n(w_1),\sigma_n(w_2),...)=(1,1,...,1,\sigma_n(w_n),\sigma_n(w_{n+1}),...)$

where the first possible non-trivial word appears in the n-th position. It is pretty straightforward to check that $\sigma_n(w_1,w_2,...)$ is still a locally eventually constant element of $\varprojlim_{n}F_n$.

Corollary 5: If $[\alpha]\in\pi_1(\mathbb{H},x_0)$ corresponds to the sequence $(w_1,w_2,...)\in\varprojlim_{n}F_n$, then $[\alpha]\in \ker\phi$ if and only if there is an $n\geq 1$ such that $\sigma_n(w_1,w_2,...)=(1,1,...,1,\sigma_n(w_n),\sigma_n(w_{n+1}),...)$ is the trivial element of $\varprojlim_{n}F_n$.

Corollary 6: $\ker\phi$ is the conjugate closure of the free subgroup of $\pi_1(\mathbb{H})$ generated by the elements $g_{n}g_{n+1}^{-1}$, $n\geq 1$.

References.

Apparently the first appearance of the harmonic archipelago (where it was also named) was in the following unpublished note:

[1] W.A. Bogley, A.J. Sieradski, Universal Path Spaces, Unpublished preprint. http://people.oregonstate.edu/~bogleyw/research/ups.pdf

Some unpublished notes on understanding the fundamental group of the harmonic archipelago:

[2] P. Fabel, The fundamental group of the harmonic archipelago, preprint. http://arxiv.org/abs/math/0501426.


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# 8.1 Energy

 Page 1 / 3

## External and internal forces

In Grade 10, you saw that mechanical energy was conserved in the absence of external forces. It is important to know whether a force is an internal force or an external force in the system, because this is related to whether the force can change an object's total mechanical energy when it does work on an object.

When an external force (for example friction, air resistance, applied force) does work on an object, the total mechanical energy (KE + PE) of that object changes. If positive work is done, then the object will gain energy. If negative work is done, then the object will lose energy. The gain or loss in energy can be in the form of potential energy, kinetic energy, or both. However, the work which is done is equal to the change in mechanical energy of the object.

## Investigations : external forces

We can investigate the effect of external forces on an object's total mechanical energy by rolling a ball along the floor from point A to point B.

Find a nice smooth surface (e.g. a highly polished floor), mark off two positions, A and B, and roll the ball between them.

The total mechanical energy of the ball, at each point, is the sum of its kinetic energy (KE) and gravitational potential energy (PE):

$\begin{array}{ccc}\hfill {E}_{\mathrm{total},\mathrm{A}}& =& {\mathrm{KE}}_{\mathrm{A}}+{\mathrm{PE}}_{\mathrm{A}}\hfill \\ & =& \frac{1}{2}m{v}_{A}^{2}+mg{h}_{A}\hfill \\ & =& \frac{1}{2}m{v}_{A}^{2}+mg\left(0\right)\hfill \\ & =& \frac{1}{2}m{v}_{A}^{2}\hfill \end{array}$
$\begin{array}{ccc}\hfill {E}_{\mathrm{total},\mathrm{B}}& =& {\mathrm{KE}}_{\mathrm{B}}+{\mathrm{PE}}_{\mathrm{B}}\hfill \\ & =& \frac{1}{2}m{v}_{B}^{2}+mg{h}_{B}\hfill \\ & =& \frac{1}{2}m{v}_{B}^{2}+mg\left(0\right)\hfill \\ & =& \frac{1}{2}m{v}_{B}^{2}\hfill \end{array}$

In the absence of friction and other external forces, the ball should slide along the floor and its speed should be the same at positions A and B. Since there are no external forces acting on the ball, its total mechanical energy at points A and B are equal.

$\begin{array}{ccc}\hfill {v}_{A}& =& {v}_{B}\hfill \\ \hfill \frac{1}{2}m{v}_{A}^{2}& =& \frac{1}{2}m{v}_{B}^{2}\hfill \\ \hfill {E}_{\mathrm{total},\mathrm{A}}& =& {E}_{\mathrm{total},\mathrm{B}}\hfill \end{array}$

Now, let's investigate what happens when there is friction (an external force ) acting on the ball.

Roll the ball along a rough surface or a carpeted floor. What happens to the speed of the ball at point A compared to point B?

If the surface you are rolling the ball along is very rough and provides a large external frictional force, then the ball should be moving much slower at point B than at point A.

Let's now compare the total mechanical energy of the ball at points A and B:

$\begin{array}{ccc}\hfill {E}_{\mathrm{total},\mathrm{A}}& =& {\mathrm{KE}}_{\mathrm{A}}+{\mathrm{PE}}_{\mathrm{A}}\hfill \\ & =& \frac{1}{2}m{v}_{A}^{2}+mg{h}_{A}\hfill \\ & =& \frac{1}{2}m{v}_{A}^{2}+mg\left(0\right)\hfill \\ & =& \frac{1}{2}m{v}_{A}^{2}\hfill \end{array}$
$\begin{array}{ccc}\hfill {E}_{\mathrm{total},\mathrm{B}}& =& {\mathrm{KE}}_{\mathrm{B}}+{\mathrm{PE}}_{\mathrm{B}}\hfill \\ & =& \frac{1}{2}m{v}_{B}^{2}+mg{h}_{B}\hfill \\ & =& \frac{1}{2}m{v}_{B}^{2}+mg\left(0\right)\hfill \\ & =& \frac{1}{2}m{v}_{B}^{2}\hfill \end{array}$

However, in this case, ${v}_{A}\ne {v}_{B}$ and therefore ${E}_{\mathrm{total},\mathrm{A}}\ne {E}_{\mathrm{total},\mathrm{B}}$ . Since

$\begin{array}{ccc}\hfill {v}_{A}& >& {v}_{B}\hfill \\ \hfill {E}_{\mathrm{total},\mathrm{A}}& >& {E}_{\mathrm{total},\mathrm{B}}\hfill \end{array}$

Therefore, the ball has lost mechanical energy as it moves across the carpet. However, although the ball has lost mechanical energy, energy in the larger system has still been conserved. In this case, the missingenergy is the work done by the carpet through applying a frictional force on the ball. In this case the carpet is doing negative work on the ball.

When an internal force does work on an object by an (for example, gravitational and spring forces), the total mechanical energy (KE + PE) of that object remains constant but the object's energy can change form. For example, as an object falls in a gravitational field from a high elevation to a lower elevation, some of the object's potential energy is changed into kinetic energy. However, the sum of the kinetic and potential energies remain constant. When the only forces doing work are internal forces, energy changes forms - from kinetic to potential (or vice versa); yet the total amount of mechanical energy is conserved.

Which equation do u wanna balance
Rifumo
i tried to write it but my phone just can't write it Cr2 O72-(aq) + H2S (aq) Cr3 (aq) + S(s)
Brian
.
Brian
hello. what is the difference between a primary, secondary and tertiary alcohol
primary the C=bonded to 1 Carbon atom.... secondary=bonded to 2 carbon atoms tertiary=bonded to 3 carbon atoms
Christina
Christina
thanks Christina
Baningi
My pleasure
Christina
Thank you Christina. This is very helpful 😀👍👍💯
Thokozani
💯💯
Thokozani
Anytime Thokozani
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thank you guys i didn't know about the primary, secondary and tertiary alcohols
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thank you Christina
Brian
can somebody help me with functional isomers
function isomer has the same molecular formula but different functional group
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what's momentum
the product of an objects mass times velocity. it is mainly prevalent in collisions.
tyrique
momentum
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Methane contains C and H. This compound is
ketones's functional group
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carboxylic acids
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thanks
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difference between A.c and D.c
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AC has slip rings DC has split rings commutator
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to avoid rust and to improve strength and conductivity
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the variable whose value doesn't depend on other is known as independent variable ex time
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no
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what's this?
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if there is extensive exposure, it can lead to skin cancer
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yes depend upon their intensity
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i hv a problem with this chapter because i also have problem with Newton's laws of motion
I have a problem with calculation in Newton's law of motion
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me also VuyokAzi
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what's the question
Christina
Which area in particular?
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just get the hang if free body diagrams,know the directions.
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I know how to draw free diagram I have a problem with calculation in second law
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guys what is momentum
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@Luvelo... Momentum is the product of an object's mass and its velocity.
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the model based on wave nature of electron
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State Coulomb's law in words
the magnitude of the electrostatic force exerted point on each other is directly proportional to the product on magnitude of charges and inversely proportional to the square distance between them
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the change in frequency of the sound detected by a listener because sound source and listener have different velocity
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Newton's first law of motion
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yes that's true
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Botho
Is the normal force always 0
no. newtons 3rd law states that if something exerts a force into something else it will experience that same magnitude of force but in the opposite direction. so the net force is equal to the force the object applies to the surface but in the opposite direction
Courtney
define the term rate of reaction in word
I think it a chemical process in which substance act mutually on each other.
Sphesihle
how many bones are in the human body?
Asali
it's the rate at which the reactants are able to change to products
Courtney
206 bones
Elizabeth


text:
# How to define this application g to meet $Im \space g \subseteq T$?

Let $$S$$ be the subspace of matrix 2x2 in $$\mathbb R$$ formed by symetrical matrix and $$T$$ the one formed by the matrix of trace zero.

Let $$A=\left(\begin{array}{cc} 1 & 0\\ 2 & 0 \end{array}\right)$$ and $$f:S\rightarrow T$$ linear given by $$f(M)=AM-MA.$$

So the first question is to get $$\mathcal B_1$$ and $$\mathcal B_2$$ basis os $$S$$ and $$T$$ so the matrix of $$f$$ is $$A=\left(\begin{array}{cc} I_r & 0\\ 0 & 0 \end{array}\right)$$.

So I ended up with:

$$\mathcal B_1$$ :$$\left(\begin{array}{cc} 0& 0\\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)$$.

$$\mathcal B_2$$ : $$\left(\begin{array}{cc} 0& 0\\ -1 & 0 \end{array}\right)\left(\begin{array}{cc} -2 & 1\\ 0 & 2 \end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 1 & 0 \end{array}\right)$$

then the matrix of $$f$$ is: $$\left(\begin{array}{cc} 1 & 0 & 0\\ 0 & 1 &0\\0&0&0\end{array}\right)$$.

But now they ask me to find the applications $$g:S\rightarrow S$$ with $$f\circ g=f$$ and express their matrix respect $$\mathcal B_1$$(g is linear).

So here I got that those g are the ones with matrix: $$\left(\begin{array}{cc} 1 & 0 & 0\\ 0 & 1 &0\\a&b&c\end{array}\right)$$ .

With $$a,b,c\in \mathbb R$$. But I don't know how to start with the next question. Of all those g, find the ONLY ONE that meets $$Im\space g \subseteq T$$.

How to start? Any hint?

• How are the members of your $\mathcal B_i$ elements of $S$ or $T$? – Berci Dec 31 '18 at 0:39

• note that $$S\cap T = \langle\left(\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}\right) \rangle$$
• $$f\left(\begin{matrix}m_{11} & m_{12} \\ m_{21}& m_{22} \end{matrix}\right)=\left(\begin{matrix} 0 & -m_{12} \\ -m_{21}& 0 \end{matrix}\right)$$


text:
# Interpolation and model theoretic forcing – some new perspectives (Campinas, Brazil, 12/18)

Interpolation and model theoretic forcing – some new perspectives.

(A lecture in Cantor Meets Robinson – Set theory, model theory and their philosophy. University of Campinas, Brazil, December 2018.)

Model Theoretic Forcing has been interweaved with interpolation theorems in infinitary logic since the early work of Mostowski, Vaught, Harnik and others. I will present some of these historical connections and their effect on Shelah’s much more recent logic $L^1_\kappa$. In particular I will focus on some connections between model theoretic forcing and the model theory of abstract elementary classes.

# Infinitary logic, large cardinals and AECs: some reflections (Montseny, Catalunya, 11/18)

Reflections on Set Theoretic Reflection – Montseny, Catalunya, nov. 2018.

Abstract:

The interaction between infinitary logic and the model theory of abstract elementary classes has had a serious imprint of large cardinals since the inception of AECs. Although later developments in AECs have emphasized a more purely model theoretic treatment, capturing independence-like relations, there are various fundamental questions on the relation between various logics and AECs — and, in some of these, large cardinals are central.
I will discuss some work by Boney on these connections, as well as some recent joint work by Väänänen and myself.

# Between Infinitary Logic and Abstract Elementary Classes (Barranquilla, 6/18)

(slides)

Encuentro de Sociedad Matemática Mexicana y Sociedad Colombiana de Matemáticas, Universidad del Norte, Barranquilla, mayo y junio de 2018.

A lecture on the role of the problem of capturing abstract elementary classes and Shelah’s logic $L^1_\kappa$.

# Logics underlying Abstract Elementary Classes (Warsaw, 4/18)

Abstract: I will first describe Abstract Elementary Classes as a global generalization of Infinitary Logic. I will emphasize constructions such as Galois types, the Representation Theorem and various open problems. In the second half, I will focus on some recent research on logics underlying AECs – with special emphasis on Shelah’s $L^1_\kappa$ logic (satisfying Interpolation and weak remnants of compactness) and the role it plays in controlling Abstract Elementary Classes. This second part contains recent results of research and several open questions.
Logic Seminar, University of Warsaw.

# Around the Small Index Property (on quasiminimal classes). Beersheva and Jerusalem, 11/16

Two lectures on the same topic (Reconstruction Problem, SIP, quasiminimal classes), with different emphasis, given at Ben Gurion University in Beersheva and at the Hebrew University of Jerusalem, in November 2016.

Slides and notes for blackboard lecture.

# Reflection Principles & Abstract Elementary Classes. Bellaterra, Catalonia, 9/16

A lecture on Reflection Principles & Abstract Elementary Classes given at the Centre de Recerca Matemàtica (Bellaterra, Catalunya), in the context of the Workshop on Set-Theoretical Aspects of the Model Theory of Strong Logics, in September of 2016.

# Language, Logic and Nonelementary Classes: External/Internal Interactions. Helsinki, 5/16.

A minicourse in Helsinki on Language, Logic and Nonelementary Classes: External/Internal Interactions, given at the University of Helsinki’s Mathematics Department in May of 2016.

The topics were

• Around The Presentation Theorem
• The Small Index Property for Homogeneous Classes
• Categories, AECs, Interpolation.

Here is the course material (presented on the blackboard).


text:
# Please, for the incorrect ones, provide calculations/steps to get the result if possible..... On ...

Please, for the incorrect ones, provide calculations/steps to get the result if possible.....

On January 14, Peerless Rocks Inc., a marble contractor, issued for cash 50,000 shares of $15 par common stock at$51, and on March 17, it issued for cash 90,000 shares of $8 par preferred stock at$11.

a. Journalize the entries for January 14 and March 17.

For a compound transaction, if an amount box does not require an entry, leave it blank.

Jan. 14 Cash 2550000 750000 1800000 Paid-In Capital in Excess of Pa Mar. 17 Cash 990000 720000 Paid-In Capital in Excess of Pa 270000 b. hat is the total amount invested (total paid-in capital) by all stockholders as of March 17? 3540000


text:
# Question Bank on Basic Computer for all Competitive Exam

This Question Bank on Basic Computer may vary from the Actual paper. Use the Question Bank on Basic Computer as a reference for the exam preparation. Check the Question Bank on Basic Computer from this page.

## Basic Computer Knowledge Question Bank

1. Which of the following is not a multitasking operating system?
(a) DOS
(b) Windows
(c) Unix
(d) Linux

2. What type of software would you need to create an invitation to your birthday party?
(b) Database
(c) Word processing
(d) Desktop publishing

3. In MS Word, the key F12 opens a
(a) Save As dialog box
(b) Open dialog box
(c) Save dialog box
(d) Close dialog box

4. Portrait and Landscape are
(a) Page Orientation
(b) Paper Size
(c) Page Layout
(d) All of above

5. Which of the following is not a font style?
(a) Bold
(b) Italics
(c) Regular
(d) Superscript

6. Which of the following is not essential component to perform a mail merge operation in MS Word?
(a) Main document
(b) Data source
(c) Merge fields
(d) Word fields

7. Small squares, called __________, on the selection rectangle that surrounds a graphic can be used to change the dimensions of the graphic.
(a) scroll boxes
(b) sizing handles
(c) status indicators
(d) move handles

8. The minimum number of rows and columns in MS Word document is
(a) 1 and 1
(b) 2 and 1
(c) 1 and 2
(d) 2 and 2

9. MS-Word automatically moves the text to the next line when it reaches the right edge of the screen, this is called
(a) Carriage Return
(b) Enter
(c) Word Wrap
(d) None of these

10. Thesaurus tool in MS Word is used for
(a) Deletion
(b) Synonyms and Antonyms words
(c) Grammar options
(d) All of above

11. To autofit the width of column in MS Word
(a) Double click the left border of column
(b) Double click the right border of column
(c) Double click the column header
(d) All of these

12. AutoCorrect was originally designed to replace __________ words as you type.
(a) Grammatically incorrect
(b) Sentence incorrect
(c) Misspelled
(d) None of these

13. The Word Count command on the Tools menu displays the number of words as well as the number of __________ in the current document.
(a) Lines
(b) Characters
(c) Paragraphs
(d) All of these

14. What is a quick way to extend these numbers to a longer sequence, for instance 1 through 20?
(a) Select both cells, and then drag the fill handle over the range you want, for instance 18 more rows
(b) Select the range you want, include both cells, point to fill on the Edit menu, and then click down.
(c) Copy the second cell, click in the cell below it, on the standard toolbar click the down arrow on the Paste button, and then click Paste Special
(d) All of the above

(a) %
(b) =
(c) +
(d) –

16. The cell reference for cell range of K2 to N12 is
(a) K2.N12
(b) K2;N12
(c) K2:N12
(d) K2-N12

17. What is the AutoComplete feature of Excel?
(a) It automatically completes abbreviated words
(b) It completes text entries that match an existing entry in the same column
(c) It completes text and numeric entries that match an existing entry in the same column
(d) It completes text entries that match an existing entry in the same worksheet

18. Which of the following is correct syntax in Excel?
(a) IF(LogicalTest, TrueResult, FalseResult)
(b) IF(LogicalTest, (TrueResult, FalseResult))
(c) IF(LogicalTest, TrueResult) (LogicalTest, FalseResult)
(d) IF(LogicalTest, TrueResult), IF(LogicalTest, FalseResult)

19. If the cell B1 contains the formula =$A$1, which of the following statements is true?
(a) There is a relative reference to cell A1
(b) There is an absolute reference to cell A1
(c) Further changes in value of A1 will not affect the value of B1
(d) Further changes in value of B1 will affect the value of A1

20. When all the numbers between 0 and 100 in a range should be displayed in Red Color,
(a) Use =if() function to format the required numbers red
(b) Apply Conditional Formatting command on Format menu
(c) Select the cells that contain number between 0 and 100 then click Red color on Text Color tool
(d) All of above

21. Which of the following can you paste selectively using Paste Special command?
(a) Validation
(b) Formats
(c) Formulas
(d) All of these

22. Which of the following series type is not valid for Fill Series dialog box?
(a) Linear
(b) Growth
(c) Autofill
(d) Time

23. A numeric value can be treated as label value if __________ precedes it.
(a) Apostrophe (‘)
(b) Exclamation (!)
(c) Hash (#)
(d) Tilde (~)

24. Which Chart can be created in Excel?
(a) Area
(b) Line
(c) Pie
(d) All of these

25. Getting data from a cell located in a different sheet is called
(a) Accessing
(b) Referencing
(c) Updating
(d) Functioning

## More Question Set on Basic Computer Knowledge

26. What is a slide-title master pair?
(a) The title area and text area of a specific slide
(b) A slide master and title master merged into a single slide
(c) A slide master and title master for a specific design template
(d) All of the above

27. In the context of animations, what is a trigger?
(a) An action button that advances to the next slide
(b) An item on the slide that performs an action when clicked
(c) The name of a motion path
(d) All of above

28. Special effects used to introduce slides in a presentation are called
(a) Effects
(b) Custom animations
(c) Transitions
(d) Present animations

29. How can you create a uniform appearance by adding a background image to all slides?
(a) Create a template
(b) Edit the slide master
(c) Use the autocorrect wizard
(d) All of these

30. What is defined by the handout master?
(a) Slide formatting
(b) Layout of audience handout notes
(c) Handout content formatting for Microsoft Word export
(d) All of the above

31. Which command brings you to the first slide in your presentation?
(a) Next slide button
(b) Page up
(c) Ctrl + Home
(d) Ctrl + End

32. To maintain the perspective (height and width ratio) of an object when resizing, you need to
(a) Press and hold the shift key while dragging a corner sizing handle
(b) Press and hold the alt + ctrl keys while dragging a middle handle
(c) Drag a corner sizing handle
(d) (a) and (c)

33. Which statement is valid about computer program?
(a) High level languages must be converted into machine language to execute
(b) High level language programs are more efficient and faster to execute
(c) It is more difficult to identify errors in high level language program than in low level programs
(d) All of these

34. Software intended to satisfy a user’s specific processing needs are called
(a) System software
(b) Process software
(c) Documentation
(d) Application software

35. In magnetic disks data is organized on the platter in a concentric sets or rings called
(a) Sector
(b) Track
(d) block

36. The performance of cache memory is frequently measured in terms of a quantity called
(a) miss ratio
(b) hit ratio
(c) latency ratio

37. A CPU contains
(a) a card reader and a printing device
(b) an analytical engine and a control unit
(c) a control unit and an arithmetic logic unit
(d) an arithmetic logic unit and a card reader

38. The first computers were programmed using
(a) assembly language
(b) machine language
(c) source code
(d) object code

39. Register is a group of _________
(a) OR gates
(b) OR & AND gate
(c) Flip-flops
(d) None of these

40. The address bus flow in _________
(a) Bidirection
(b) Unidirection
(c) Mulidirection
(d) Circular

41. Status register is also called as _________
(a) Accumulator
(b) Stack
(c) Counter
(d) Flags

42. Which of the following statements is true?
(a) Minicomputer works faster than Microcomputer
(b) Microcomputer works faster than Minicomputer
(c) Speed of both the computers is the same
(d) The speeds of both these computers cannot be compared

43. A computer cannot ‘boot’ if it does not have ______
(a) Compiler
(c) Operating System
(d) Assembler

44. What characteristic of read-only memory (ROM) makes it useful?
(a) ROM information can be easily updated.
(b) Data in ROM is non-volatile, that is, it remains there even without electrical power.
(c) ROM provides very large amounts of inexpensive data storage.
(d) ROM chips are easily swapped between different brands of computers.

45. The output quality of a printer is measured by
(a) Dot per inch
(b) Dot per sq. inch
(c) Dots printed per unit time
(d) All of the above

46. Which of the printers used in conjunction with computers uses dry ink powder?
(a) Daisy wheel printer
(b) Line printer
(c) Laser printer
(d) Thermal printer

47. Main component of first generation computer was
(a) Transistors
(b) Vacuum Tubes and Valves
(c) Integrated Circuits
(d) None of the above

48. Result of logical operation is _________
(a) Boolean
(b) Integer
(c) Character
(d) String

49. Where is data stored when it is brought into processor?
(a) Auxiliary
(b) Main memory
(c) Register
(d) Magnetic tape

50. What is the register that contains the address of the next instruction?
(a) Instruction register
(b) Program counter
(c) Accumulator
(d) Container


text:
# Sample Covariance for 2 vectors

I'm trying to calculate sample covariance between these two vectors. I defined a function with two input variables. I don't know if it is correct? Also my formula for sample covariance won't run. Can anyone help me write it out in R?

  xv = c(1., 5.5, 7.8, 4.2, -2.7, -5.4, 8.9)
yv = c(0.1, 1.5, 0.8, -4.2, 2.7, -9.4, -1.9)
sampleCov= function(x,y){
cov(xv,yv) = frac{sum_{i=1}^{n}(x_i-\mu_x)(y_i-\mu_y)}{n-1}].
return (Cov(xv,yv)
}

-
Your function doesn't work because it has syntax errors. Fix those first. Have you tried the cov function? – Joshua Ulrich Oct 17 '12 at 19:35
To be honest i'm very new to programming and dont know how to fix them. I've been stuck on this for a while. How can I improve my function? – Bill Oct 17 '12 at 19:40
You can improve your function by reading the introductory manual, especially Section 10, Writing your own functions. – Joshua Ulrich Oct 17 '12 at 19:44
For starters, don't use LaTeX markup as R code; it just won't parse. It is hard to know where to begin here because you have got totally the wrong end of the stick if you thought R could even begin to understand that code. – Gavin Simpson Oct 17 '12 at 19:44

There is a base function in R called cov that does exactly what you want, but if you want to write a function (no need to do that) you can try the following:

COV<- function(x,y) {
if(length(x)!=length(y)) {stop('x must have the same length as y ')}
x.bar <- mean(x)
y.bar <- mean(y)
N <- length(x)

Cov <- (sum((x-x.bar)*(y-y.bar))) / (N-1)
return(Cov)
}

COV(xv, yv)
[1] 8.697381

cov(xv, yv)
[1] 8.697381

As you can see COV gives the same result as cov so you don't have to write a function for that.

Besides, the body of your function doesn't have R syntax, instead you wrote LaTex syntax which are not the same.

-
Ok thanks a bunch. This was very helpful. I appreciate it – Bill Oct 17 '12 at 19:51

Just use the internal cov() function:

xv <- c(1., 5.5, 7.8, 4.2, -2.7, -5.4, 8.9)
yv <- c(0.1, 1.5, 0.8, -4.2, 2.7, -9.4, -1.9)
cov(xv, yv)

R> cov(xv, yv)
[1] 8.697381

If you really want to reinvent the wheel:

sampleCov <- function(x,y){
stopifnot(identical(length(x), length(y)))
sum((x - mean(x)) * (y - mean(y))) / (length(x) - 1)
}

R> sampleCov(xv, yv)
[1] 8.697381

-
Thanks, this worked and was very helpful. – Bill Oct 17 '12 at 19:50


text:
Easy-to-use, Comprehensive and Affordable Systems Management

Damon Warren氏、IT担当役員
Burnet Institute

# How to suppress the prompt when calling a VBS script using the K1000 appliance

When calling a Visual Basic script using a batch file, there is an option to suppress the command prompt from popping up when executing the Visual Basic script.

The cscript option is //NOLOGO

Example:

C:\Windows\System32\cscript.exe //NOLOGO ScriptName.vbs

Reference for Visual Basic different Options from the net.

Categories for this entry


text:
# Maximum likelihood estimator of SNR for a known signal superimposed in AWGN

Assuming the following model (Fir instant, received signal of a communication / RADAR system):

$$x[n] = \alpha s[n] + w[n]$$

Where:

• $s[n]$ is a known signal of length $L$ and known energy $p = \sum_{n=0}^{L-1} {s}^{2}[n]$.
• $\alpha$ is unknown attenuation factor. Though it is known $\alpha \leq 1$.
• $w[n]$ is additive white gaussian noise (AWGN) where $E[w[n]] = 0$ and $E[w[m]w[n] = \delta[n-m] {\sigma}^{2}$. The noise variance ${\sigma}^{2}$ is unknown parameter.
• There are $N$ samples, namely, $n \in [0, \ldots, N-1]$ where $N > L$.

The question is, what is the Maximum Likelihood Estimator of the SNR of the received signal - $\frac{{\alpha}^{2}p}{{\sigma}^{2}}$?

Using Matched Filter / Correlation one could calculate the ML estimator of $\alpha$ and ${\sigma}^{2}$. Yet since the SNR isn't one to one function of the parameters, it cannot be calculated using them.

Hence the question is: what's the MLE of the SNR of the received signal?

-
"... since the SNR isn't one to one function of the parameters, it cannot be calculated using them" isn't always true. In this case, the MLE of the SNR is exactly what you'd hope, just plug in the MLEs of $\alpha$ and $\sigma^2$.
To see this heuristically, assume the MLE of the SNR, label it $\widehat{SNR}$, is something else. Then the likelihood calculated using $\widehat{SNR}$ and the MLE of, say, $\alpha$ and using the relationship $SNR = \alpha^2p / \sigma^2$ will be lower than it would be if the MLEs of both $\alpha$ and $\sigma^2$ were used, because the estimate of $\sigma^2$ will be different from the MLE of $\sigma^2$. Therefore $\widehat{SNR}$ isn't maximizing the likelihood. Therefore it isn't the MLE of the SNR.


text:
# Matrix Exponential [closed]

i'm working on a paper about quantum entropy and i'm stuck on an analytical step about computing a matrix exponential.

I have the following Hamiltonian $$H=\frac{1}{2}\sum_ip_i^2+\frac{1}{2}\sum_{i,j}x_iK_{ij}x_j$$

so that i get the following wave function $$\psi_0(x_1,x_2,...,x_N)=\frac{1}{2}(\det\Omega)^{1/4}\exp(-x\Omega x/2)$$ where $$\Omega$$ is the square root of K: If $$K=U^TK_DU$$ where $$K_D$$ is diagonal and U is orthogonal, then $$\Omega=U^T\sqrt{K_D}U$$.

To find the density matrix for the N-n oscillators, we trace over the first N ones, so we get : $$\rho(x_{n+1},...,x_N,x'_{n+1},...,x'_N)=\int\prod_i^ndx_i\psi_0(x_1,...,x_n,x_{n+1},...,x_N)\psi_0^*(x_1,...,x_n,x'_{n+1},...,x'_N)$$

The paper solves this integral writing $$\Omega=\left(\begin{matrix} A&B\\ B^T&C\\ \end{matrix}\right)$$ where $$A=n\times n$$ and $$C=(N-n) \times (N-n)$$.

Then on the paper the $$\rho(x_{n+1},...,x_N,x'_{n+1},...,x'_N)$$ is written like $$\rho(x,x')\sim \exp\left(-\frac{x\gamma x+x'\gamma x'}{2} +x\beta x'\right)$$ where $$\gamma=C-\beta$$, $$\beta=\frac{1}{2}B^TA^{-1}B$$ and both x and x' have N-n components.

$$\textbf{How do I get the last result?}$$

I thought that the right way to the solution is to diagonalize $$\Omega$$ but I don't know how because A and C have different ranks.

• You need to proofread what you wrote. – Cosmas Zachos Sep 3 at 19:48

Your formula suggests the use of a block-matrix Gaussian decomposition such as $$\left[\matrix{ A & B\cr C & D}\right]=\left[\matrix{I &BD^{-1} \cr 0 &I }\right] \left[\matrix{A-BD^{-1} C&0\cr 0& D}\right]\left[\matrix{ I &0 \cr D^{-1} C &I}\right],$$ or the one with a lower triangular matrix on the leftmost factor so that $$A^{-1}$$ replaces the $$D^{-1}$$. In these products the $$B$$ and $$C$$ matrices do not have to be square because all the matrix sizes match up.


text:
### Re: OpenCart 1.5.6 Released!

MarketInSG wrote:just noticed the commit. broke quite a number of my scripts which searches for footer.
Yeah it was strange, it didn't have an issue on the original code. Only when you downloaded the ZIP or did a fresh clone did it happen! Probably something on the line breaks going awol..

An awesome Web Development company based in Leeds. E-Commerce Specialists by day, Geeks and Gamers by night

Tel: 0845 634 1838
Web: http://www.welfordmedia.co.uk

Active Member

Posts

Joined
Thu Jul 21, 2011 5:59 pm
Location - Leeds, UK

### Re: OpenCart 1.5.6 Released!

My biggest complaint, aside from this being a bit of a sellout build, is the coding style differs from the original code.
Little things like spacing before the open/closing braces. It doesn't bother me so much in your own openbay files...but when you change the styling of the engine/controller.php file for the hasAction() function.. that is just annoying. OpenCart's coding consistency is what makes it so great.

Also all the unnecessary clutter of folders. We don't need an "amazon", "amazonus", "ebay", "play", etc folder inside of controller directories. That could have easily been done as a single "openbay" folder with subfolders to keep it all self contained and more importantly removed for those of us that have absolutely no use for any of it. When a single mod takes more folders than the core code, it is time to make a subdirectory.

Posts

Joined
Tue Jul 22, 2008 3:02 am

### Re: OpenCart 1.5.6 Released!

Also: in the new added translation files it is missing the \\headingtitle, \\text, \\column etc. markings and all the lines are mixed up.
While in other language files the lines are grouped but still the \\ parts are missing.

To me it looks like someone who did the coding that is not instructed how it should be done or how OpenCart files are set-up.

More "strange" way of coding etc. I've seen.
All the new files are missing the >? closing tags?

I agree with Qphoria, why these additional folders which serve NO purpose at all?
What are they different then the other modules, payment and shipping ones?

Norman in 't Veldt
Moderator OpenCart Forums

_________________ READ and Search BEFORE POSTING _________________

[How to] BTW + Verzend + betaal setup.

Global Moderator

Posts

Joined
Mon Nov 09, 2009 7:00 pm
Location - Winkel - The Netherlands

### Re: OpenCart 1.5.6 Released!

i will go over the code when i have a chance and make it conform to OC style.

need to release something though. the last release was over 7 months ago.

OpenCart®
Project Owner & Developer.

Posts

Joined
Fri Nov 03, 2006 6:57 pm

### Re: OpenCart 1.5.6 Released!

Maybe a better search, checkout and options should have released

Active Member

Posts

Joined
Sat Apr 13, 2013 6:05 pm

### Re: OpenCart 1.5.6 Released!

Agree with Q and i2paq.
Looks like new parts have been grafted on core, with language strings redundant.
Was not it better to let Openbay, Amazon,... in free extensions as majority will not use ? Paypal update is good.
And does corrected bugs in 2.0 have been ported back in 1.5.6 ?

All my extensions

Active Member

Posts

Joined
Wed Feb 10, 2010 8:43 am
Location - France

### Re: OpenCart 1.5.6 Released!

I have to say, saying you need to release something isn't a good enough reason to not look through the code. I agree with others that this seems more about adding more mods for OC to generate affiliate cash more than anything, or money received from the various companies to add in their mods. This code could have been pushed to OC's 1.5.6 branch and left for peer review by others before going live. Instead, we've had basically a release with little more than commercial plugins added - even if it was only a couple of days. No disrespect to Welford media, but they should have thoroughly gone through their code to make sure it was written as the rest of OC. This was never happening when Qphoria was part of the dev team, and as far as I can see nobody else has been allowed to push changes without the code conforming to the rest of the cart

Another issue is that we're forced with all of the OpenBay stuff regardless of wanting it or not. I mean every store now is going to be loading 5 extra classes into memory for all of the OpenBay stuff without any need - for those wondering, it's this code in the index.php files

Code: Select all

//OpenBay Pro
$registry->set('openbay', new Openbay($registry));
$registry->set('play', new Play($registry));
$registry->set('ebay', new Ebay($registry));
$registry->set('amazon', new Amazon($registry));
$registry->set('amazonus', new Amazonus($registry));

Sure they can be commented out, but they shouldn't be in every store as standard without the user enabling them. This really isn't necessary for the vast majority of stores, not to mention they could quite easily be Model files loaded as required, or even added to the registry by the various controllers. No other modifications load like this - Payment gateways, shipping options and so on all load on demand as they should. While I'm sure time has been spent planning this, it's really not the way it should be coded

For now, I would encourage people to stick with the 1.5.5.X release until this code has been properly reviewed, formatted to standard and streamlined to be as fast as it should be

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### Re: OpenCart 1.5.6 Released!

JAY6390 wrote:Another issue is that we're forced with all of the OpenBay stuff regardless of wanting it or not.
Agree!

We, in The Netherlands, will never use it.

I'll hope it will be removed from 2.0 and offered as a separate add-on.

If not, we will end up like PrestaShop, bloated with 3rd party stuff just to please the sponsers.........

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### Re: OpenCart 1.5.6 Released!

Daniel wrote:need to release something though. the last release was over 7 months ago.

Why not let the 1.5.5.1 CE do the job in terms of bug-fixes and release Openbay as a 3rd party add-on and focus on 2.0?

Now we have to support upgrades, release additional langagepacks en many have to update their Extensions.

And most of all: more bugs, code issues etc. ect.

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### Re: OpenCart 1.5.6 Released!

i2Paq wrote:
Daniel wrote:need to release something though. the last release was over 7 months ago.

Why not let the 1.5.5.1 CE do the job in terms of bug-fixes and release Openbay as a 3rd party add-on and focus on 2.0?

Now we have to support upgrades, release additional langagepacks en many have to update their Extensions.

And most of all: more bugs, code issues etc. ect.
I am with you on this. And in between I've also noticed a 1.5.5.2 branch on GitHub. WTH!

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### Re: OpenCart 1.5.6 Released!

OpenBay is a free add on which can be found in the extension store. I'm sure plenty of us can download it if we want. Out of all the stores I had worked on, I only see 1 to 2% of them using OpenBay. I don't think we all need integration with amazon etc.

OpenCart is simple enough at v1.5.5.1, let's keep it simple for the rest of the releases rather than complicating everything and slowly ending up like other systems. A clean up is really much needed now..

The closing tag was missing because most code editor omits it. We all agree with qphoria that the coding styles need to be standardized.

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### R: OpenCart 1.5.6 Released!

... So, as usual... We will need to wait for 1.5.6.1 release!

M

Sent from my DesireHD.

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My last mods: Partita IVA e CF | Pro EU VAT Number | Sales Agents | Pricelist Pro
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### Re: OpenCart 1.5.6 Released!

i2Paq wrote:Why not let the 1.5.5.1 CE do the job in terms of bug-fixes and release Openbay as a 3rd party add-on and focus on 2.0?

Now we have to support upgrades, release additional langagepacks en many have to update their Extensions.

And most of all: more bugs, code issues etc. ect.
I totally agree with i2Paq and the guys, I was looking forward to the exciting v2.0 major release too and I am surprised by this 1.5.6 version, which looks more like a patched-up 1.5.5.1.
I think that the community should be invited to participate in minor releases before publication. A 24 hour final bug fixing session with the community on GitHub could be organised through the forum, the day before a release.

Good job still, the official Paypal integration and partnership was needed and that's good for the future.

OpenCart custom solutions @ https://villagedefrance.net

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### Re: OpenCart 1.5.6 Released!

villagedefrance wrote:I think that the community should be invited to participate in minor releases before publication. A 24 hour final bug fixing session with the community on GitHub could be organised through the forum, the day before a release.
Promises made over and over again, and broken even more
Good job still, the official Paypal integration and partnership was needed and that's good for the future.
Totally agree, but a separate release for 1.5.5.1 and older 1.5.x version would have been better.

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### Re: R: OpenCart 1.5.6 Released!

I agree with everything that's been said up to this point. Adding further extensions doesn't bother me so much, but the non-standard loading and coding of them just makes working with OpenCart code messier, and more difficult for regular developers.

It seems like it would be a much better idea to create a plug-in download system for these kind of add-ons that aren't generally used. (A lot of the included payment gateways could be part of this, actually.) I'm sure Daniel could make it happen, especially with the extension installer code for 2.0 as a basis.
madimar wrote:... So, as usual... We will need to wait for 1.5.6.1 release!

M
So true

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### Re: OpenCart 1.5.6 Released!

I agree with the many posters here:
• Many of the new files have missing ?> in the end
• Openbay classes should not be loaded for the registry in the index.php, rather, should be loaded on demand as needed, like other extension modules
• Probably quite a few missing bugfixes

Override Engine * Integrated VQMod * Unused Images Manager * Instant Option Price Calculator * Number Option * Google Rich Snippets * Google Tag Manager * Export/Import Tool * SpamBot Buster * Survey Plus

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### Re: OpenCart 1.5.6 Released!

As far as the coding goes....i would have to agree. Also:

Running PayPal Pro in an iFrame allows merchants to hide the identity of the PayPal payment gateway for their card payments. It also reduces the requirements of a merchant’s PCI compliance, so there is no need for a dedicated server or VPS and penetration testing to become compliant - all payment pages are hosted at PayPal and not on your store.
I just wanted to make a few comments on Pay Flow Link (iFrame). While it's great that Opencart is offering this there needs to be some clarification on just how wonderful it is....or not. I feel qualified to make such a comment because I too work with Pay Pal as a liaison for their merchant tools and understand how complicated their system change has made it for some people whether they be developers or store owners or both.

While all of this sounds enticing, it's actually a bad idea for anyone serious about their eCommerce and wanting to offer the best checkout experience possible. It's a cheap way to have something that act's like DIRECT PAYMENT but never will allow you to make it look like it's a part of your store. Anyone that has used the Pay Flow Link on the new Pay Flow Gateway knows this and it should never be pushed as a great idea but only an alternative for particular stores that aren't in the business of wanting to grow their company OR for those that aren't' really concerned about how their checkout page looks. Yes, there are exceptions. I'm talking about the majority of serious eCommerce store owners.

Furthermore,
1) it looks cheap....and the client will still feel as if they are "not on your website" because you'll never be able to style it the way you want it.
2) some folks won't even be using an SSL because of this type of advice which is horrible for business in general. Whether the pages are hosted on Pay Pal or not, a website should always provide a sense of security....period.

My advice: find a way to get approved for a Pay Pal Pro Account and use the Direct Payment Method. This once was called DoDirectPayment but now uses the Pay Flow Gateway. There is a Pay Flow Pro Direct Payment extension for sale in the store and soon my Recurring Billing extension will be released which is the updated version of Pay Pal Recurring Payments.

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### Re: OpenCart 1.5.6 Released!

don't worry everything can be fixed. i just didn't have time to finish 2.0

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### Re: OpenCart 1.5.6 Released!

Just wondering, do I have to download the entire 1.5.6, or is there an upgrade? I currently have 1.5.5.1.

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### Re: OpenCart 1.5.6 Released!

leatherneck wrote:Just wondering, do I have to download the entire 1.5.6, or is there an upgrade? I currently have 1.5.5.1.
The entire thing


text:
Car range

Calculate the maximum range of car, if you can spend 10 euros, price of diesel is 1.55 Eur/l and car consumption is 3 l/100 km.

Correct result:

x =  215.1 km

Solution:

$V = 10 / 1.55 = 6.45 \ l \ \\ \ \\ x = \dfrac{ 100 \cdot 10 }{ 1.55 \cdot 3} = 215.1 \ \text{km}$

We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!

Tips to related online calculators
Check out our ratio calculator.
Do you want to convert length units?
Tip: Our volume units converter will help you with the conversion of volume units.

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If the Canadian dollar appreciated by C$0.005 relative to the US dollar, what would be the new value of the Canadian dollar per US dollar? Assume the current exchange rate was US$1 = C$0.907. • Hour salary You work for 4 hours on a Saturday and 8 hours on Sunday. You also receive a$50 bonus. You earn $164. How much did you earn per hour? • Parcel 4 To send a parcel by messenger within city limits costs 60 cents for the first pound and 48 cents for each additional pound. What is the cost, in cents, of sending a parcel weighing p=4 pounds? • Trio ratio Hans, Alena and Thomas have a total of 740 USD. Hans and Alena split in the ratio 5: 6 and Alena and Thomas in the ratio 4: 5. How much will everyone get? • Two discounts The MP4 player cost 960 CZK, the price was twice gradually reduced in the ratio 4: 3: 3/2. How much did the player cost after the second discount? • Apples School kitchen bought 36kg apples for 12 kc/kg. How many kilograms of apples 1/4 cheaper can they buy for the same money? • Six workers Six workers earned a total of CZK 12,600 per week on the construction site (5 working days). How much do 7 workers earn in 10 days with the same daily average salary? • Kilo candy One kilogram of chocolate candy cost 10 euros. Kate buys 250 grams and had the 8 pack of candy. How much does cost one bonbon? • Three 43 Three brothers inherited a cash amount of 62,000, and they divided it among themselves in the ratio of 5:4:1. How much more is the largest share than the smallest share? • Babysitting The amount that Susan charges per hour for babysitting are directly proportional to the number of children she is watching. She charges$4.50 for 3 kids. How much would she charge to babysit 5 kids?
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text:
# Internet Problem Solving Contest

## Solution to Problem J – Judicious cuts

The easy subproblem is really really easy. To the point that you’ll probably bang your head against the wall if you didn’t get it. The easiest solution: output r − 1 distinct horizontal lines.

We will now take a look at the hard subproblem.

First let’s look at how to count the number of regions for a given arrangement of lines. Our output format already ensures that there are no lines parallel to the y-axis, but in general, one can slightly rotate the plane to avoid those anyway. Consider any finite amount of non-vertical lines. We can now find the rightmost point of every region. There are two types of regions:

• Those that do not have a rightmost point, as they are unbounded on the right side. Since there are no lines parallel to the y-axis, it is easy to observe that there are exactly ℓ + 1 of these.
• Those that do have a rightmost point. Such rightmost point always lies at the intersection of two or more lines. We call these regions bounded.

In the figure below, the regions filled with stripes do not have a rightmost point – those are the unbounded regions. The other five filled with solid colour are bounded.

When the lines are in a general position (that is, no three lines meet at a point and no two lines are parallel) then every pair of lines has a unique intersection point. Each intersection point serves as a rightmost point to exactly one region. Since there are ${\ell} \choose 2$ such intersections, the total number of regions is
$$\frac{\ell^2 + \ell + 2}{2}.$$
This is also the maximal amount of regions one can make using lines. How do parallel lines and multiple lines meeting at a point affect the result?

The case of parallel lines is simple – if there is a set of k parallel lines, we have $k\choose 2$ fewer intersection points which exactly corresponds to the number of regions we lose.

Let’s look at m lines meeting at a point and ignore all the others for a while. If the lines did not meet at a single point, but they were in general position instead, there would be $m \choose 2$ regions with a rightmost point. Now we have 1 intersection point that borders 2m regions. We already know that m + 1 of them are unbounded, which leaves us with m − 1 bounded regions. This means that the effect of an arrangement of m lines meeting in a single point on the number of regions is a decrease by ${m \choose 2} - (m - 1) = {m-1 \choose 2}$.

The above observations can now be used to count the number of regions for any configuration of lines. (This is sometimes called Roberts’ formula.)

This gives us an idea how to construct the solution for a given r. If there is an such that (ℓ2 + ℓ + 2)/2 = r, output lines in a general position. Otherwise, find the smallest for which the general position has more regions and try to remove some of them by introducing the “flaws” mentioned above. For example, if r = 13, we can take ℓ = 5 lines. If they were in a general position, we would have 16 regions. If we instead place them so that four of them meet at the same point, this will reduce the number of regions by ${4-1 \choose 2} = 3$, leaving us with 13 regions.

There are many different constructions, for instance one can introduce some sets of parallel lines, and ensure that they do not create points where three or more lines intersect. The solution presented below uses the notion of a point-line duality and is able to stay well within the limits on line parameters.

Formally a dual point to the line L given by y = mx + b is the point L* = (m, −b). Every line not parallel to y-axis has a dual point, and every point has a corresponding dual line. Duality has many interesting properties that are often useful in similar constructions. We use two of them here without proof:

• Two lines L1 and L2 are parallel iff their duals L1* and L2* have the same x coordinate.
• Three lines L1, L2, L3 meet at a single point iff their dual points L1*, L2*, L3* are collinear. The intersection point is the primal of the line connecting the dual points.

We can describe the construction and also prove its properties easier in the dual representation. In our construction we avoid parallel lines completely by setting mi (the slope of the i-th line) equal to i. We let the y-intercept increase by i between consecutive points. E.g., for five dual points we have {(0, 0),(1, 1),(2, 3),(3, 6),(4, 10)} (depicted in the following figure). What this construction effectively does is increasing the slope of the dual line between x consecutive dual points. It is easy to prove that no three such dual points lie on a dual line: since the slopes of lines connecting different pairs of dual points are different, so is the x-coordinate of intersection of their primals. When the intersection points have different x-coordinate, they are different points. Hence, no three (primal) lines meet at a (primal) point and this approach gives us lines in general position.

How to decrease the number of regions? Let’s start by always outputting a primal line (0, 0). Let x denote the number by which we need to decrease the number of regions to achieve the desired amount. Once we reduce x to zero, we are done. In order to do that, we pick the largest c such that ${c-1 \choose 2} \leq x$. Introducing c collinear dual points reduces x to $x' = x - {c-1 \choose 2}$. The simplest and efficient way to introduce c collinear points is to add c − 1 points collinear with the last. We do this by keeping the same slope for all. For example, the last inserted point is (4, 4) and we demand 3 consecutive points. We use slope 5 for both, that is we output points/lines (5, 9), (6, 14). The point (6, 14) is now the last point, and we can use it for another set of collinear points (e.g., (7, 21), (8, 28)).

The thing left to do is to assess whether this construction is efficient enough. As introducing k collinear dual points removes Θ(k2) regions, and adding an extra line gives us Θ(ℓ) new regions, the intuition is that for large number of lines this construction is sufficient. A bit of case-work reveals the construction does not work for r ∈ {3, 5, 9, 12, 17, 24}, where one would aim to use (2, 3, 4, 5, 6, 7) lines, respectively.

In all cases but r = 5 and r = 17 one can fix the construction by using one set of parallel lines at the end (k parallel lines are sufficient to reduce r by $k \choose 2$, whereas k + 1 intersecting lines are needed; and we can again reuse the last inserted point1), or by hand-crafting the solutions. Case r = 17 can indeed be done using 6 lines, but even the more efficient construction fails to find the solution, we’re left with solution by hand.2

Finally, we’re left with r = 5. This is indeed a special case where we need one more line than the lower bound we established before. We’re left with no choice – the approach used for the easy subproblem is optimal here.

In the scope of the contest, it is advisable to leave the proof out, follow the intuition and then solve by hand the few cases where the construction failed. If we’re lazy, we can drop the negative sign by y-intercept – this corresponds to reflection of the dual along x-axis, which definitely doesn’t affect the solution and the properties of the duality we’re using.

Bonus question: What if parallel lines were not allowed?

1. Notice that when using just parallel lines, one cannot reuse the last point for the next set, as parallelity is transitive. It is also harder to prevent intersecting lines.

2. Hint: one set of 2 parallel lines, one set of 3 parallel lines, and a 3-way intersection.


text:
## Section26Customizations

### Subsection26.1Renaming Document Parts

“Names” for various parts of a document are determined exactly once for each language, ensuring consistency and saving you the bother of always typing them in.

However, you may want to have “Conundrum”s in your document and you have no use for any “Proposition”s. So you can repurpose the proposition tag to render a different name. Or you might have a Lab Manual and want to rename subsection as “Activity”. See the docinfo portion of this sample article to see how this is done, in concert with the example below.

##### Important Notes

If you are renaming many parts of your document, then you may not understand the design philosophy of PreTeXt. In particular, you should not be doing a wholesale shuffle of part, chapter, section, etc. This feature is intended for very limited use and is not considered best practice.


text:
CorrelationTests {coin} R Documentation

## Correlation Tests

### Description

Testing the independence of two numeric variables.

### Usage

## S3 method for class 'formula'
spearman_test(formula, data, subset = NULL, weights = NULL, ...)
## S3 method for class 'IndependenceProblem'
spearman_test(object, distribution = c("asymptotic", "approximate", "none"), ...)

## S3 method for class 'formula'
fisyat_test(formula, data, subset = NULL, weights = NULL, ...)
## S3 method for class 'IndependenceProblem'
fisyat_test(object, distribution = c("asymptotic", "approximate", "none"),
ties.method = c("mid-ranks", "average-scores"), ...)

## S3 method for class 'formula'
quadrant_test(formula, data, subset = NULL, weights = NULL, ...)
## S3 method for class 'IndependenceProblem'
quadrant_test(object, distribution = c("asymptotic", "approximate", "none"),
mid.score = c("0", "0.5", "1"), ...)

## S3 method for class 'formula'
koziol_test(formula, data, subset = NULL, weights = NULL, ...)
## S3 method for class 'IndependenceProblem'
koziol_test(object, distribution = c("asymptotic", "approximate", "none"),
ties.method = c("mid-ranks", "average-scores"), ...)

### Arguments

 formula a formula of the form y ~ x | block where y and x are numeric variables and block is an optional factor for stratification. data an optional data frame containing the variables in the model formula. subset an optional vector specifying a subset of observations to be used. Defaults to NULL. weights an optional formula of the form ~ w defining integer valued case weights for each observation. Defaults to NULL, implying equal weight for all observations. object an object inheriting from class "IndependenceProblem". distribution a character, the conditional null distribution of the test statistic can be approximated by its asymptotic distribution ("asymptotic", default) or via Monte Carlo resampling ("approximate"). Alternatively, the functions asymptotic or approximate can be used. Computation of the null distribution can be suppressed by specifying "none". ties.method a character, the method used to handle ties: the score generating function either uses mid-ranks ("mid-ranks", default) or averages the scores of randomly broken ties ("average-scores"). mid.score a character, the score assigned to observations exactly equal to the median: either 0 ("0", default), 0.5 ("0.5") or 1 ("1"); see median_test. ... further arguments to be passed to independence_test.

### Details

spearman_test, fisyat_test, quadrant_test and koziol_test provide the Spearman correlation test, the Fisher-Yates correlation test using van der Waerden scores, the quadrant test and the Koziol-Nemec test. A general description of these methods is given by Hájek, Šidák and Sen (1999, Sec. 4.6). The Koziol-Nemec test was suggested by Koziol and Nemec (1979). For the adjustment of scores for tied values see Hájek, Šidák and Sen (1999, pp. 133–135).

The null hypothesis of independence, or conditional independence given block, between y and x is tested.

The conditional null distribution of the test statistic is used to obtain p-values and an asymptotic approximation of the exact distribution is used by default (distribution = "asymptotic"). Alternatively, the distribution can be approximated via Monte Carlo resampling by setting distribution to "approximate". See asymptotic and approximate for details.

### Value

An object inheriting from class "IndependenceTest".

### References

Hájek, J., Šidák, Z. and Sen, P. K. (1999). Theory of Rank Tests, Second Edition. San Diego: Academic Press.

Koziol, J. A. and Nemec, A. F. (1979). On a Cramér-von Mises type statistic for testing bivariate independence. The Canadian Journal of Statistics 7(1), 43–52. doi: 10.2307/3315014

### Examples

## Asymptotic Spearman test
spearman_test(CONT ~ INTG, data = USJudgeRatings)

## Asymptotic Fisher-Yates test
fisyat_test(CONT ~ INTG, data = USJudgeRatings)


text:
# On $\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\frac{n}{6^6}\,x\right)\,dx$

Reshetnikov gave the remarkable evaluation,

\begin{align} I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx \\ &=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right)\\ &=0.7857194\dots \end{align} where $\alpha$ is a quartic root. However, it seems this can be simplified a bit and generalized.

I. $p=5$: Given,

$$I(n)=\int_0^1\arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{n}{4^4}\,x\right)\,\mathrm dx$$ Is it true that, in general, $$\frac{12n}{5^5}\, I(n) = -12\Big(1\color{blue}-\frac{n}{5^5}\Big)\arctan\frac1\alpha+6\ln\Big(\frac{2}{1+\alpha^2}\Big)+3\pi+P(\alpha)$$ where $P(\alpha)=(1-\alpha)^3(5+3\alpha)$ and $\alpha$ is the largest positive root of the quintic, $$\alpha^5-\alpha^4+\frac{n}{5^5}=0$$ provided real number $n<4^4\,$? (Note: By sheer coincidence, the choice of $n=4$ in the other post caused the quintic to factor.)

II. $p=7$: Given,

$$J(n)=\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\,\frac{n}{6^6}\,x\right)\,\mathrm dx$$ is it true that, $$\frac{60n}{7^7}\, J(n) = 60\Big(1\color{blue}+\frac{n}{7^7}\Big)\arctan\frac1\alpha-30\ln\Big(\frac{2}{1+\alpha^2}\Big)-15\pi-P(\beta)$$ where $P(\beta) = (1-\beta)^2(27-6\beta-9\beta^2+8\beta^3+10\beta^4)$ and $\beta$ is the largest positive root of, $$\beta^7-\beta^6+\frac{n}{7^7}=0$$ provided real $n<6^6\,$?

Questions:

1. How do we prove the two conjectured equalities?
2. What's the formula for $p=3$? (Mathematica takes too long to evaluate the integral that I couldn't use an integer relations subroutine on it.)
• But Reshetnikov didn't show a complete solution. Just the final answer. – Zaid Alyafeai Jan 3 '17 at 5:56
• @ZaidAlyafeai: Yes, but it's a great final answer. He used an $\alpha$ that turns out to be the quartic factor of the quintic in this post. How he came across the correct root and the correct form is remarkable. I just found the $p=7$ by analogy. – Tito Piezas III Jan 3 '17 at 6:04
• It looks like one can solve the general problem here for an arbitrary $k$. If $$I_k(n) = \int_{0}^{1}\arctan\left._k\right.F_{k-1}\left(\frac{1}{k+1},\frac{2}{k+1}\cdots,\cdots,\frac{k}{k+1},\frac{k+1}{k}; \frac{nx}{k^k}\right){\rm d}x$$ then we should have something like $$I_k(n) = \arctan\left(\frac{1}{z_0}\right) + \frac{(k+1)^{k+1}}{n}\int_{z_0}^1\frac{w^{k+1}-w^k}{1+w^2}{\rm d}w$$ where $z_0$ is a solution to $z_0^{k+1} - z_0^k + \frac{n}{(k+1)^{k+1}} = 0$. – Winther Jan 3 '17 at 22:38

## 2 Answers

I think you should start by the following general form

$${}_{k}F_{k-1}\left(\frac{1}{k+1} ,\cdots ,\frac{k}{k+1};\frac{2}{k} \cdots ,\frac{k-1}{k},\frac{k+1}{k};\left( \frac{m(1-m^k)}{f_k}\right)^k \right) = \frac{1}{1-m^k}$$

Where

$$f_k \equiv \frac{k}{(1+k)^{1+1/k}}$$

Put $k=4$

$${}_{4}F_{3}\left(\frac{1}{5} ,\cdots ,\frac{4}{5};\frac{2}{4} \cdots ,\frac{3}{4},\frac{5}{4};\left( \frac{m(1-m^4)}{f_4}\right)^4 \right) = \frac{1}{1-m^4}$$

The argument simplifies to

$$\left(\frac{m(1-m^4)}{\frac{4}{5^{1+1/4}}}\right)^4 = \frac{5^5 \,m^4(1-m^4)^4}{4^4}$$

Hence we have

$${}_{4}F_{3}\left(\frac{1}{5} ,\cdots ,\frac{4}{5};\frac{2}{4} \cdots ,\frac{3}{4},\frac{5}{4};\frac{5^5 (1-m)m^4}{4^4} \right) = \frac{1}{m}$$

Now suppose we want to find

$$\int \arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{n}{4^4}\,x\right)\,\mathrm dx$$

Use $nx = 5^5m^4(1-m)$

$$\frac{5^5}{n}\int (4m^3(1-m)-m^ 4)\arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{5^5 (1-m)m^4}{4^4}\right)\,\mathrm dm$$

This simplifies to

$$\frac{5^5}{n}\int (4m^3(1-m)-m^ 4)\arctan\,\frac{1}{m}\,\mathrm dm$$

The anti-derivative of this is elementary.

Note that when we use substittution we will need the roots of

$$m^5-m^4+\frac{n}{5^5}= 0$$

As conjectured by the OP.

Using the value $n=4$ we can verify Reshetnikov result

\begin{align}\int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx &= \frac{5^5}{4}\int^{\alpha}_1 (4x^3(1-x)-x^ 4)\arctan \left(\frac{1}{x} \right)\,\mathrm dx\\&=0.7857194\dots \end{align}

Where $\alpha$ is the largest positive root of

$$m^5-m^4+\frac{4}{5^5}= 0$$

• Ok, but perhaps you should change your variable $x$ to the more generic $m$, as it will confuse/conflate the use of $F(x)dx$ in the post. – Tito Piezas III Jan 3 '17 at 6:51
• @TitoPiezasIII, yes sure. – Zaid Alyafeai Jan 3 '17 at 6:56
• Ah, beautiful! Mathematica can integrate $$\int^{\alpha}_1 (4x^3(1-x)-x^ 4)\arctan\left(\frac{1}{x} \right)\,\mathrm dx$$ for generic $\alpha$. I can now see why using $\ln(1+\alpha^2)$ and the other expressions make sense. – Tito Piezas III Jan 3 '17 at 12:29
• do you have a proof for your first eqn.? – tired Jan 3 '17 at 13:24
• @tired, I never worked with the generalized hyper geometric function. I will try to provide a proof for k=2. – Zaid Alyafeai Jan 3 '17 at 18:36

The result does generalize. We will derive a closed formula for the integral $$I_k(n) = \int_0^1\arctan\left[_{k}F_{k-1}\left(\frac{1}{k+1},\cdots, \frac{k}{k+1},\frac{2}{k},\cdots,\frac{k+1}{k}; \frac{nx}{k^k}\right)\right]\,{\rm d}x$$ valid for $|n| \leq k^k$.

For this particular combination of arguments the hypergeometrical function simplifies greatly, see property $(25)$ here$^{(1)}$. Using this we can perform a substitution $\frac{nx}{(k+1)^{k+1}} = y^k(1-y^k)^k$ and the integrand simplifies to $\frac{(k+1)^{k+1}}{n}\frac{1}{1-y^k}\frac{d}{dy}\left(y^k(1-y^k)^k\right)$. Now performing integration by parts and the substitutions $y^k\to y \to 1-y$ we arrive at

$$I_k(n) = \frac{\pi}{2} - \arctan(z_0) + \frac{(k+1)^{k+1}}{n}\int_{z_0}^1\frac{w^{k+1}-w^k}{1+w^2}{\rm d}w$$

where $z_0$ is the largest positive root of $z_0^{k+1} - z_0^k + \frac{n}{(k+1)^{k+1}} = 0$. The integral above can be evaluated in closed form using

$$\matrix{\int \frac{w^{2k}}{1+w^2}{\rm d}w &=& (-1)^k\left[\arctan(w) + F_k(w)\right] + C\\\int \frac{w^{2k+1}}{1+w^2}{\rm d}w &=& (-1)^k\left[\frac{1}{2}\log(1+w^2) + G_k(w)\right] + C}$$ where $F_k(x) = \sum_{i=1}^{k}\frac{(-1)^i x^{2i-1}}{2i-1}$ and $G_k(x) = \sum_{i=1}^{k}\frac{(-1)^i x^{2i}}{2i}$. These relations are easily proven by differentation and summing a geometrical series. This leads to

$$I_{2k}(n) = \frac{\pi}{2} - \arctan(z_0) + \frac{(-1)^k(2k+1)^{2k+1}}{n}\left[\frac{1}{2}\log\left(\frac{2}{1+z_0^2}\right) - \frac{\pi}{4} + \arctan(z_0)\right] + \frac{(-1)^k(2k+1)^{2k+1}}{n}\left[G_k(1) - G_k(z_0) - F_k(1) + F_k(z_0)\right]$$ $$I_{2k+1}(n) = \frac{\pi}{2} - \arctan(z_0) + \frac{(-1)^{k}(2k+2)^{2k+2}}{n}\left[-\frac{1}{2}\log\left(\frac{2}{1+z_0^2}\right) - \frac{\pi}{4} + \arctan(z_0)\right] + \frac{(-1)^{k}(2k+2)^{2k+2}}{n}\left[- G_k(1) + G_k(z_0) -F_{k+1}(1) + F_{k+1}(z_0)\right]$$

For the particular case $k = 2$ ($p=3$ in the question) you asked for this reduces to

$$I_2(1) = \frac{27}{2}\log \left(\frac{z_0^2+1}{2}\right) - \frac{27}{2}(z_0-1)^2 - 28\arctan(z_0) + \frac{29 \pi}{4} \simeq 0.795296 \\ \text{where } z_0 = \frac{1}{3} \left(1+2 \cos \left(\frac{\pi }{9}\right)\right)$$

Finlly since the expression has many terms we should check that it gives the correct result. Here is some code used to check this in Mathematica:

(* Working precision *)
ndigits = 100;

Block[{$MinPrecision = ndigits}, (* Pick k and n *) k = 4; n = 1; (* Solve for z0 *) z0 = Max[N[x /. Solve[{x^(k + 1) - x^k + n/(k + 1)^(k + 1) == 0, x > 0}, x], ndigits]]; (* Define the integral *) ak = Table[i/(k + 1), {i, 1, k}]; bk = Table[If[i == k, (i + 1)/k, i/k], {i, 2, k}]; exact = NIntegrate[ArcTan[HypergeometricPFQ[ak, bk, (x n)/k^k]], {x, 0, 1}, WorkingPrecision -> ndigits]; (* Our solution *) result = If[Mod[k, 2] == 0, (* For even k *) \[Pi]/2 - ArcTan[z0] + ((-1)^(k/2) (k + 1)^(k + 1))/ n (1/2 Log[2/(1 + z0^2)] - \[Pi]/4 + ArcTan[z0] + Sum[(-1)^i ((1 - z0^(2 i))/(2 i) - (1 - z0^(2 i - 1))/(2 i - 1)), {i, 1, k/2}]) , (* For odd k *) \[Pi]/2 - ArcTan[z0] + ((-1)^((k - 1)/2) (k + 1)^(k + 1))/n (-(1/2) Log[2/(1 + z0^2)] - \[Pi]/4 + ArcTan[z0] - Sum[(-1)^i ((1 - z0^(2 i))/(2 i)If[i == (k - 1)/2 + 1, 0, 1] + (1 - z0^(2 i - 1))/(2 i - 1)), {i, 1, (k - 1)/2 + 1}]) ]; (* Difference between integral and our formula *) Print["Difference: ", N[result - exact]]; ];  This demonstrates that we get the same result (to numerical precision) for a wide range of$k$and$n$.$^{(1)}$It's easy to check that this holds to arbitrary precisison (checked to a few hundred digits) at least for the smallest values of$k$. Marty Cohen found this reference discussing a very similar type of property which might be useful if anyone wants to try to derive it (which likely is a very messy affair). • +1 Thanks for this highly detailed explanation. If I could accept two answers, I would do so. :) – Tito Piezas III Jan 4 '17 at 5:07 • Unfortunately, the only justification for that amazing equation (25) is "(M. L. Glasser, pers. comm., Sept. 26, 2003)". So, we do not have a proof, just as assertion. – marty cohen Jan 4 '17 at 5:52 • @martycohen, Interesting. Maybe I will post a question about it later. – Zaid Alyafeai Jan 4 '17 at 5:55 • Doing a Google search for "M. L. Glasser hypergeometric functions" results in a number of interesting hits including this one: researchgate.net/profile/Larry_Glasser/publication/… – marty cohen Jan 4 '17 at 6:07 • It's more complicated than I though (and far beyond what I'm interested in doing) to derive the formula. To show it one needs to establish $$\frac{1}{1-x^k} = \sum_{n=0}^\infty \frac{\left(\frac{1}{k+1}\right)_n\cdots \left(\frac{k}{k+1}\right)_n}{\left(\frac{2}{k}\right)_n\cdots \left(\frac{k-1}{k}\right)_n\left(\frac{k+1}{k}\right)_n}\frac{x^{nk}(1-x^k)^{nk}(k+1)^{n(k+1)}}{k^{nk} n!}$$ where$(a)_n\$ is the falling factorial. – Winther Jan 4 '17 at 6:37