TITLE: Hardness of Vertex Separators
QUESTION [12 upvotes]: For a given graph $G$, the Separator Problem asks whether a vertex or edge set of small cardinality (or weight) exists whose removal partitions $G$ into two disjoint graphs of approximately equal sizes. This is called the Vertex Separator Problem when the removed set is a vertex set, and the Edge Separator Problem when it is an edge set. Both problems are NP-complete for general unweighted graphs. What is the best known hardness of approximating vertex separator ? Is a PTAS ruled out ? What are the best known hardness results in the directed setting ?
Correction : The following links and answers did not help me because I did not state my question correctly. My question is related to the following theorem of Leighton-Rao :
Theorem : There exists a polynomial time algorithm that, given a graph $G(V,E)$ and a set $W \subseteq V$, finds a $\frac{2}{3}$ vertex separator $S \subseteq V$ of $W$ in $G$ of size $O(w.{\log}n)$, where $w$ is the minimum size of a $\frac{1}{2}$-vertex separator of $W$ in $G$.
Given a graph $G(V,E)$ and a set $W \subseteq V$, I want to find a $\delta$-vertex separator (where $\frac{1}{2} \leq \delta \leq 1$ is a constant) of size $w$, where $w$ is the minimum size of a $\frac{1}{2}$-vertex separator of $W$ in $G$. What is the best known hardness of this problem ? The above theorem gives an $O({\log}n)$ approximation for this problem.
Note that I am allowing constant factor blow-up in the size of the resulting components after removing the separator, but I want to minimize the size of the separator itself. The links mentioned in the comments point to minimum b-vertex separator, in which we insist that the size of the resulting components is at most $|V|/2$.

REPLY [5 votes]: The approximability of the separator question in the sense you want is closely related to the approximability of the uniform sparsest cut problem. An $O(\sqrt{\log n})$-factor approximation was obtained by Arora-Rao-Vazirani improving the $O(\log n)$ of
Leighton and Rao; they did this for the edge case. Agrawal-Charikar-Makarychev-Makarychev used the result to obtain similar bound for directed sparsest cut (if one is interested in vertex bipartition cuts). Feige-Hajiaghayi-Lee at the same time obtained a similar bound again via ARV for vertex separators (and also pointed out that treewidth can be approximated within the same factor). One should note that there is another notion of sparsest cut in directed graphs for which Chuzhoy-Khanna showed hardness results in the non-uniform case but I am not sure about the uniform case.  I think super-constant hardness results are known for (uniform) sparsest cut under UGC but I am not sure.


TITLE: Linear approximation of time dilation. In what point is it?
QUESTION [1 upvotes]: I have watched a video about linear approximation and there was an example, exactly here: https://www.youtube.com/watch?v=BSAA0akmPEU&feature=youtu.be&list=PL590CCC2BC5AF3BC1&t=32m50s about linear approximation of time dilation. I have started to count and this is the result:
$$ f(t,v) = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}} $$
$$ \frac{df(t,v)}{dv} = \frac{tv}{c^2(1-\frac{v^2}{c^2})^{\frac{3}{2}}} $$
so combining those two things into linear approximation I have:
$$ f(t,v) \approx f(t,v_0) + f(t,v_0)'(v-v_0) $$
then putting real bodies of functions:
$$ f(t,v) \approx \frac{t}{\sqrt{1-\frac{v_0^2}{c^2}}} + \frac{tv_0}{c^2(1-\frac{v_0^2}{c^2})^{\frac{3}{2}}}(v-v_0)  \approx \frac{t}{\sqrt{1-\frac{v_0^2}{c^2}}} - \frac{tv_0^2}{c^2(1-\frac{v_0^2}{c^2})^{\frac{3}{2}}} + \frac{tv_0v}{c^2(1-\frac{v_0^2}{c^2})^{\frac{3}{2}}} $$
then I can imagine that:
$$ v_0 << c $$
So I think that I can do something like that:
$$ \frac{1}{\sqrt{1-\frac{v_0^2}{c^2}}} \approx 1 $$
then
$$ f(t,v) \approx t - \frac{tv_0^2}{c^2} + \frac{tv_0v}{c^2} $$
and then compering to result from video it should be equal to
$$ T' = T(1 + \frac{1}{2}\frac{v^2}{c^2}) $$
but unfortunately it is not equal. Can somebody tell me what mistakes I have made? What is wrong with my reasoning? I will be really appreciate. :)

REPLY [0 votes]: You effectively killed $f(t,v)$ when you set
$$\frac{1}{\sqrt{1-\frac{v_0^2}{c^2}}} \approx 1$$
The result can be obtained much simpler by using known approximation rules
$$\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} \approx\dfrac{1}{1-\dfrac{v^2}{2c^2}}\approx 1+\dfrac{v^2}{2c^2}$$


TITLE: Is the gravitational constant $G$ a rational number or an irrational number?
QUESTION [4 upvotes]: Is the gravitational constant $G$ a rational number or an irrational number?
Not this only, what about such similar questions about all the universal constants?

REPLY [2 votes]: I would like to add a few comments to this discussion.
Newton's gravitational constant has dimensions.  This is the point of Gandalf61's answer:  We can either define the dimension of $G_N$ by setting $G_N = 1$ or making some other convention, or we can try to compute $G_N$ in terms of another dimensionful parameter.  In the latter case, Firdous' original question makes sense.  If we have a dimensionful parameter $A$, we can try to prove an equation $G_N = C \cdot A$ and ask whether $C$ is an integer.
In quantum field theory, we usually work in "natural units" in which $c = 1$ and $\hbar = 1$.  Then $G_N$ has the dimensions of (mass)$^{-2}$.  If $M$ is some mass in the theory, we can ask if the theory can predict $G_N = C \cdot M^{-2}$, where $C$ is an integer.
There is actually a theory in which this is possible. There is a quantum field theory model called the "nonlinear sigma model" in which the fields take values a sphere instead of, as usual, a flat Euclidean space. The radius of the sphere is a quantity called $F_\pi$, which has the dimensions of mass.  Next, consider the supersymmetric generalization of this model.  In 1982, Edward Witten and Jonathan Bagger studied the coupling of this model to supergravity.  They found that there are topological constraints that can only be satisfied if  $G_N = m/F_{\pi}^2$, where $m$ is an integer! See Witten and Bagger, Physics Letters B115, 202 (1982).
This model is very artificial.  It also requires a lot of sophisticated theoretical baggage to understand. That is probably to be expected; finding an equation for Newton's constant is not an easy problem to solve.  Nevertheless, it provides a concrete existence proof that a solution to Firdous' problem may be possible.


TITLE: Equivalent normal vectors to a surface
QUESTION [0 upvotes]: I want to evaluate the following surface integral: $\iint_{\partial\Omega}(x^3,z,y)\cdot n\ ds$ where $\partial\Omega=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=4\}$ and the normal vector is directed outside.
I parametrized the surface like this: $x=2\sin\varphi\cos\theta, y=2\sin\varphi\sin\theta, z=2\cos\varphi$ so I get as a normal vector: $-r_{\theta}\times r_{\varphi}=-(-2\sin\varphi\sin\theta,2\sin\varphi)\times (2\cos\varphi\cos\theta, 2\cos\varphi\sin\theta,-\sin\varphi)=(4\sin^2\varphi\cos\theta,4\sin^2\varphi\sin\theta,4\sin\varphi\cos\varphi)$
thus $\iint_{\partial\Omega}(x^3,z,y)\cdot n\ ds=$
$\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2} (8\sin^3\varphi\cos^3\theta,2\cos\varphi,2\sin\varphi\sin\theta)\cdot(4\sin^2\varphi\cos\theta,4\sin^2\varphi\sin\theta,4\sin\varphi\cos\varphi)d\varphi d\theta =\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2}(32\sin^5\varphi\cos^4\theta+16\sin^2\varphi\cos\varphi\sin\theta)d\varphi d\theta=\frac{256}{20}\pi=\frac{64}{5}\pi $.
I have two questions about this:
1) Is this correct?
2) I thought of another way to compute this; since a normal vector is also (in cartesian coordinates): $n=(x,y,z)$ I thought about using it in spherical coordinates, i.e. $n=(2\sin\varphi\cos\theta, 2\sin\varphi\sin\theta,2\cos\varphi)$ and I get for the surface integral:
$\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2} (8\sin^3\varphi\cos^3\theta,2\cos\varphi,2\sin\varphi\sin\theta)\cdot (2\sin\varphi\cos\theta,2\sin\varphi\sin\theta,2\cos\varphi)d\varphi d\theta =\frac{9\pi^2}{4}$.
Can someone explain to me why I get a different result (i.e. why are the two normal vectors I've found not equivalent for the purpose of evaluting the surface integral)?

REPLY [1 votes]: There are multiple questions. Roughly in the order they're asked:

Your parametrization covers the sphere if $0 \leq \varphi \leq \pi$. As written (with $0 \leq \varphi \leq \pi/2$) you get only the hemisphere $z \geq 0$.
The cross product $r_{\theta} \times r_{\varphi}$ is inward-pointing (as Triatticus notes).
I haven't checked the numerical evaluation of your integral, but using the correct limits of integration causes more cancellation than is currently present. (The integral of $2yz$ over the entire sphere vanishes.)
In your second attempt, using the Cartesian form of the normal, you've

Used a normal vector of length $2$ (rather than a unit normal);
Written $ds = d\theta\, d\varphi$, while the surface element is
$$
ds = \|r_{\theta} \times r_{\varphi}\|\, d\theta\, d\varphi.
$$


TITLE: Prove Inequality $\frac{\left(1-\alpha\right)\left(1+{\alpha}^{k}\right)}{\left(1+\alpha\right)\left(1-{\alpha}^{k}\right)}\geqslant\frac{1}{k}$
QUESTION [9 upvotes]: I would like to prove the following inequality:
$$ \frac{\left(1 - \alpha \right )\left(1 + {\alpha}^{k} \right )}{\left(1 + \alpha \right )\left(1 - {\alpha}^{k} \right )} \geqslant \frac{1}{k} \ \forall \alpha > 0 ,  k \in \mathbb{N} $$
Any other properties of:
$$ \frac{\left(1 - \alpha \right )\left(1 + {\alpha}^{k} \right )}{\left(1 + \alpha \right )\left(1 - {\alpha}^{k} \right )} $$
Would be great as well.

REPLY [11 votes]: Here's an answer in terms of hyperbolic functions. The (corrected) inequality to be shown is
$${(1 - \alpha)(1 + \alpha^k) \over (1 + \alpha)(1 - \alpha^k)} \geq {1 \over k}$$
This is the same as
$${{(\alpha^{-{1 \over 2}} - \alpha^{{1 \over 2}})(\alpha^{-{k \over 2}} + \alpha^{{k \over 2}})} \over {(\alpha^{-{1 \over 2}} + \alpha^{{1 \over 2}})(\alpha^{-{k \over 2}} - \alpha^{{k \over 2}})}}\geq {1 \over k}$$
For any $\alpha > 0$ there exists some $t$ for which $e^t = \alpha^{1 \over 2}$. The above can be written as
$${\tanh(t) \over \tanh(kt)} \geq {1 \over k}$$
Since ${\displaystyle {\tanh(t) \over \tanh(kt)}}$ is even, it suffices to prove this for $t > 0$. For $t > 0$ the above equation can be rewritten as
$$\tanh(kt) \leq k\tanh(t)$$
 The derivative of the left hand side is $k$ sech^2$(kt)$, while the derivative of the right hand side is $k$ sech^2$(t)$. Since sech is decreasing for $t \geq 0$ (this comes from the fact that $\cosh$ is increasing), the left hand side's derivative is less than that of the right. Since the left and right hand sides of this are equal at $t = 0$, the left hand side is always at most the right hand side and the inequality is proven.


TITLE: Computing the $x$-$y$ position of a vehicle for constant acceleration
QUESTION [0 upvotes]: I want to very simplistically plot the $x$-$y$ position of a vehicle in 2D that can accelerate/brake and drive a constant speed while turning etc.
The basic equations for determining the position for constant speed seem to be
$$x_{n+1} = x_n + vt\cos(\theta)$$
$$y_{n+1} = y_n + vt\sin(\theta)$$
where $\theta$ is the turning angle given in radians.
Am I correct in assuming that since
$$v_{n+1} = v_n + at$$
for a constant acceleration we have:
$$x_{n+1} = x_n + (v_n+at)t\cos(\theta)$$
$$y_{n+1} = y_n + (v_n+at)t\sin(\theta)$$
?

REPLY [0 votes]: So I did a little C# demo to check that my math is correct first.

The math behind each simulation step is as follows (assuming non-zero steering angle). Below $\Delta t$ is the time step, $v_0$ the initial speed, $\vec{p}_0$ is the initial position, $\varphi_0$ the initial orientation angle, and $\ell$ is the wheelbase.

Define the step steering angle $\theta$, as well as the acceleration $\dot{v}$.
Define the turn radius $R = \ell / \tan \theta$.
Get speed change $\Delta v = \dot{v} \Delta t$.
Get distance change $\Delta s = v_0 \Delta t + \tfrac{1}{2} \dot{v} \Delta t^2$.
Get orientation angle change $\Delta \varphi = \Delta s / R$
Get new orientation angle $\varphi_1 = \varphi_0 + \Delta \varphi$
Get initial normal direction $\hat{n}_0 = \pmatrix{-\sin (\varphi_0) & \cos (\varphi_0)}$
Get final normal direction $\hat{n}_1 = \pmatrix{-\sin (\varphi_1) & \cos (\varphi_1)}$
Get position change $\Delta \vec{p} = R\, (\hat{n}_0 - \hat{n}_1)$
New position $\vec{p}_1 = \vec{p}_0 + \Delta \vec{p}$
New speed $v_1 = v_0 + \Delta v$
Get tangent direction $\hat{e}_1 = \pmatrix{ \cos(\varphi_1) & \sin(\varphi_1)}$
Get new velocity vector $\vec{v}_1 = v_1 \hat{e}_1$

NOTE: All angles are in radians. The car position is defined by the center of the rear axle since a car turns about the rear wheels.
The key to this and the answer to your question is #9. It takes the current position, moves to the center of rotation (pivot point), and back into the new position by traversing an arc of length $\Delta s$.

The kinematics of a 2D car is described by the following picture

where $R$ is the radius of turn (and $1/R$ is the turn curvature). In pink is the arc the car tracks (the local coordinate of the car shown mid-rear axle) and it has an arc length of $\Delta s$. This arc sweeps an angle of $\Delta \varphi = \Delta s/R$.  The angle $\theta$ describing the big blue triangle is the steering angle, due to similar triangles in the steering geometry.
The initial and final position and orientation of the car coordinate system is shown at the ends of the pink arc, and you see they share the common pivot point which is always $R$ distance away, along the $\hat{n}$ direction.


TITLE: Calculating $\operatorname{Res} \left(\frac{f(z)}{g(z)}, z=a\right)$ with $a$ a double zero of $g$.
QUESTION [2 upvotes]: I have to show that for $f,g$ analytic on some domain and $a$ a double zero of $g$, we have:
$$\operatorname{Res} \left(\frac{f(z)}{g(z)}, z=a\right) = \frac{6f'(a)g''(a)-2f(a)g'''(a)}{3[g''(a)]^2}.$$
The problem is that direct calculation using the formula (for pole of order $2$):
$$\operatorname{Res}(h(z),z=a)=\lim_{z \to a} \frac{d}{dz}\left(  (z-a)^2h(z) \right)$$
is extremely ugly, given that we're dealing with a quotient. Is there some sort of trick to make the calculation more manageable?

REPLY [1 votes]: Because $a$ is a double zero of $g(z)$, write
$$g(z) = (z-a)^2 p(z)$$
where $p(a) \ne 0$ and is analytic, etc. etc.
Then
$$\operatorname*{Res}_{z=a} \frac{f(z)}{g(z)} = \left [\frac{d}{dz} \frac{f(z)}{p(z)} \right ]_{z=a}$$
Now,
$$\frac{d}{dz} \frac{f(z)}{p(z)} = \frac{f'(z) p(z)-f(z) p'(z)}{p(z)^2}$$
Also, note that
$$g(z) = \frac12 g''(a) (z-a)^2 + \frac16 g'''(a) (z-a)^3+\cdots = p(a) (z-a)^2 + p'(a) (z-a)^3+\cdots$$
Therefore
$$p(a) = \frac12 g''(a)$$
and
$$p'(a) = \frac16 g'''(a)$$
Thus
$$\operatorname*{Res}_{z=a} \frac{f(z)}{g(z)} = \frac{f'(a) \frac12 g''(a) - f(a) \frac16 g'''(a)}{\frac14 [g''(a)]^2}$$
which is equivalent to the formula you seek.


TITLE: Finding eigenvalues/vectors of a matrix and proving it is not diagonalisable.
QUESTION [1 upvotes]: I have got the following matrix.
$$\begin{pmatrix}
-7 &4 \\
-9 &5
\end{pmatrix}$$
I need to find the eigenvalues, eigenvectors and $\textbf{prove}$ that it is not diagonalisable.
I have managed to show that the only eigenvalue is $\lambda=-1$ (from the equation $det(A-\lambda I)=0$). Then I have calculated the only eigenvector, which is $(2,3)$. Now I need to prove that the matrix is not diagonalisable. I think it would be reasonable to prove that if eigenvalues are repeating, then the matrix is not diagonalisable.
Have anyone got any suggestions how to proceed? Thank you very much.

REPLY [5 votes]: An $n\times n$ matrix $A$ is diagonalizable if and only if there exists a basis $\{v_1,\dotsc,v_n\}$ for $\Bbb R^n$ consisting of eigenvectors of $A$. Here you have a $2\times 2$ matrix and you've shown that every eigenvector is a scalar multiple of $\vec v=(2,3)$. What can you conclude?
Note that your suggestion that "if eigenvalues are repeating, then the matrix is not diagonalizable" is false. For example, the identity matrix $I_n$ has only one eigenvalue $\lambda=1$ and this eigenvalue has algebraic multiplicity $n$. However, $I_n$ is clearly diagonalizable.


TITLE: Eilenberg's rational hierarchy of nonrational automata & languages -- where is it now?
QUESTION [31 upvotes]: In the preface to his very influential books Automata, Languages and Machines (Volumes A, B), Samuel Eilenberg tantalizingly promised Volumes C and D dealing with "a hierarchy (called the rational hierarchy) of the nonrational phenomena... using rational relations as a tool for comparison. Rational sets are at the bottom of this hierarchy. Moving upward one encounters 'algebraic phenomena,'" which lead to "to the context-free grammars and context-free languages of Chomsky, and to several related topics."
But Eilenberg never published volume C. He did leave preliminary handwritten notes for the first few chapters (http://www-igm.univ-mlv.fr/~berstel/EilenbergVolumeC.html) complete with scratchouts, question marks, side notes and gaps. But they do not reveal much beyond the beginnings of the well-known power series approach to grammars.
So, my actual question -- does anyone know of work along the same lines to possibly reconstruct what Eilenberg had in mind? If not, what material is likely closest to his ideas?
The site http://x-machines.net/ is about x-machines, one of Eilenberg's key innovations, but it deals mainly with applications of x-machines rather than further developing the theory as Eilenberg seemed to promise.
Also, anyone know why Eilenberg stopped before making much progress on Volume C? This was the late 70's, and he lived until 1998, though he did not appear to have published any math after Volume B. Yet he seemed to have the math for Volumes C and D largely done, at least in his mind.
(Same question asked on math.stackexchange -- https://math.stackexchange.com/questions/105091/eilenbergs-rational-hiererchy-of-nonrational-automata-languages -- apologies if this is considered cross-posting.)

REPLY [4 votes]: An accepted answer to this question was given by J.-E. Pin at Mathematics Stack Exchange.


TITLE: Extensions of beta-theory of lambda calculus
QUESTION [16 upvotes]: The beta-eta-theory of the lambda-calculus is Post-complete.  Can additional rules be added to extend the beta-theory of the lambda-calculus to get confluent theories other than the beta-eta theory?
Postscript
This question violated my own rule that questions should explain why the questioner cares.
It struck me one night, not long before this site went into private beta, that because extensionality and the principle of the excluded middle are related, the eta rule is some sort of extensionality rule, and there are intermediate logics between intuitionistic and classical logic, then it would be interesting if there was such a thing as "intermediate eta" theories.
If I has done so, it would have been obvious that Evgenij's answer raises an obvious problem in the way I'd formulated the question, rather than being what I was after.

REPLY [8 votes]: Yes. There is for example beta + the rule {s = t | s and t are closed unsolvable terms}. This is as far as I remember not equal to beta-eta, and is consistent. See mathgate for a short description and reference to Barendregt.


TITLE: Can the average of an unbounded sequence of positive numbers be 0?
QUESTION [2 upvotes]: Looking for verification of a proof. Find $\{s_n\}$ a sequence of positive numbers such that $\limsup_{n\to\infty}s_n=\infty$, and $\lim_{n\to\infty}\sigma_n=0$ where $\sigma_n=\frac1{n+1}\sum_{k=0}^ns_n$.
Let
$$s_n=
\begin{cases}
m+1, & \text{if $n=2^m$}\\
\frac1{2^n}, & \text{otherwise}
\end{cases}$$
If $2^m\le n<2^{m+1}$,
$$\begin{align}
\sigma_n & =\frac1{n+1}\left(\sum_{k=0}^n\frac1{2^k}+\sum_{k=0}^mk-\sum_{k=0}^m\frac1{2^{2^k}}\right)\\
& =\frac1{n+1}\left(2-\frac1{2^n}+{(m+2)(m+1)\over 2}-\sum_{k=0}^m\frac1{2^{2^k}} \right).
\end{align}$$
Now for $2^m\le n<2^m-1$,
$$\begin{align}
\sigma_n-\sigma_{n+1} & =\frac1{n+1}\left(2-\frac1{2^n}\right)-\frac1{n+2}\left(2-\frac1{2^{n+1}}\right)\\
& +\frac1{(n+1)(n+2)}\left({(m+2)(m+1)\over 2}-\sum_{k=0}^m\frac1{2^{2^k}}\right).
\end{align}$$
Note that the last term is clearly positive, and
$$\begin{align}
& \frac1{n+1}\left(2-\frac1{2^n}\right)-\frac1{n+2}\left(2-\frac1{2^{n+1}}\right)\\
= & {2\over (n+1)(n+2)}+{1\over 2^{n+1}(n+2)}-{1\over 2^n(n+1)}\\
= &{2^{n+2}-(n+3)\over 2^{n+1}(n+1)(n+2)}.
\end{align}$$
This difference is also positive because $2^{n}>n+1$ for $n\ge2$. So,
$$\max\{\sigma_n:2^m\le n\le 2^{m}-1\}=\sigma_{2^m}.$$
Therefore to prove $\lim_{n\to\infty}\sigma_n=0$, it suffices to prove $\lim_{m\to\infty}\sigma_{2^m}=0$.
$$\begin{align}
\lim_{m\to\infty}\sigma_{2^n} & \le \lim_{m\to\infty}\frac1{2^m+1}\left(2-\frac1{2^{2^m}}+{(m+2)(m+1)\over 2}\right)\\
& =\lim_{m\to\infty}\frac2{2^m+1}-\lim_{m\to\infty}{1\over (2^m+1)2^{2^m}}+\lim_{m\to\infty}{(m+1)(m+2)\over 2(2^m+1)}\\
& = 0.
\end{align}$$

REPLY [2 votes]: Your example $s_n$ works, and the proof looks reasonable, although I haven't gone through every computation. However, I strongly suggest refactoring it as follows.

Consider the sequence defined by $a_n = k$ when $n=2^k$ and $a_n=0$. It is unbounded. The  partial sums $s_n=a_1+\dots+a_n$ are given by $k(k+1)/2$ where $k\le \log_2 n$. Thus, $s_n/n\to 0$.
Consider the sequence $b_n=1/n$. It converges to zero, therefore its means also converge to zero: (standard fact, proved here).
Conclude from 1 and 2 that the sequence $s_n=a_n+b_n$ has the desired properties.


TITLE: How do I adjust values for consistency?
QUESTION [1 upvotes]: What's a fair way to adjust value for consistency?
For example, Machine A pumps out 10 lollipops every day.  Machine B pumps out 0 lollipops one day and 20 lollipops the next, which averages 10 lollipops a day (you just don't know when you'll get 0 or 20).  I want to adjust the 10 lollipops per day numbers in this example for consistency, giving a better value for the more consistent machine.  How should I do this?  I have logs of all machine output.
Don't know how to tag this.

REPLY [0 votes]: You want the standard deviation.
First, define the variance, which is the average of the square of the offset.  In your example, the first machine is always at 10, so the offset is zero, and the variance is zero.  In the second machine, the offset is either 20-10=10 or 0-10=-10.  The square in both cases is $10^2=100$ or $(-10)^2=100$, so the variance is 100.
The standard deviation is the square-root of the variance.  In this case it is 0 for the first machine and 10 for the second machine.
Usually, you keep both numbers - the mean and the standard deviation - rather than combine them.  So you know the (10,0) machine is more reliable than the (10,10) machine, but both give the same output in the long run.


TITLE: If $f$ is a function that can be differentiated two times on $[a;b]$
QUESTION [1 upvotes]: I asked this question before, but I quickly deleted it so that I can think about it more. the question was how can I show that:
$\exists c \in [a,b]$ such that $ f(b)=f(a)+(b-a)f'(a)+\frac{(b-a)^2}{2}f''(c)$ knowing that $f$ is a function that can be differentiated two times on $[a,b]$
Here is what I did:
I used the mean value theorem on $f$ in the interval $]a;b[$ then I got:
$$\exists c_0 \in ]a;b[:f(b) - f(a) = f'(c_0)(b-a)$$
rearranging gives:   (1)
$$\exists c_0 \in ]a;b[:f'(c_0) = \frac{f(b)-f(a)}{b-a}$$
Then I used the theorem on $f'$ but in the interval $]a;c_0[$ which gives me
$$\exists c \in ]a;c_0[:f'(c_0) - f'(a) = f''(c)(c_0-a)$$
by replacing $f'(c_0)$ in the last equation with (1) and multiplying by $(b-a)$ I got
$$\exists c \in ]a;c_0[:f(b)-f(a)-(b-a)f'(a)=f''(c)(c_0-a)(b-a)$$ and rearranging this gets me really close to what I am supposed to get:
$$\exists c \in ]a;c_0[:f(b)=f(a)+(b-a)f'(a)+f''(c)(c_0-a)(b-a)$$
and because $]a;c_0[ \subset]a;b[$ I can write : $$\exists c \in ]a;b[:f(b)=f(a)+(b-a)f'(a)+f''(c)(c_0-a)(b-a)$$
If only that $(c_0-a)$ was $\frac{(b-a)}{2}$ or something like that.
Am I on the right track? I'd appreciate any help on this, I am open to any suggestion.
note: I asked my teacher about this problem and he told me to write I new function in terms of $f(x)$, maybe that will work, but I don't know about that, it seemed to me that he doesn't know the answer either, he could be wrong.

REPLY [2 votes]: Given $f\in C([a,b])$ twice-differentiable and $x\in(a,b]$, consider
$$
F(t) = f(t)+f'(t)(x-t), \qquad G(t) = (t-x)^2.
$$
By Cauchy's mean value theorem (a simple consequence of Rolle's theorem), we have
$$\tag{1}
\frac{F(x)-F(a)}{G(x)-G(a)} = \frac{F'(c)}{G'(c)}
$$
for some $c\in(a,x)$. But
$$
F'(t)=f'(t)+f''(t)(x-t)-f'(t)=f''(t)(x-t)
$$
and
$$
G'(t) = 2(t-x),
$$
so $(1)$ translates to
$$
\frac{f(x)-f(a)-f'(a)(x-a)}{-(a-x)^2} = \frac{f''(c)(x-c)}{2(c-x)} = -\frac{f''(c)}2.
$$
Rearranging the terms leads to
$$
f(x) = f(a)+f'(a)(x-a)+\frac{f''(c)}2(x-a)^2.
$$


TITLE: What is the range of values of the random variable $Y = \frac{X - \min(X)}{\max(X)-\min(X)}$?
QUESTION [3 upvotes]: Suppose $X$ is an arbitrary numeric random variable. Define the variable $Y$ as
$$Y=\frac{X-\min(X)}{\max(X)-\min(X)}.$$
Then what is the range of values of $Y$?

REPLY [2 votes]: If $X$ takes values over any finite (closed) interval, then the range of $Y$ is $[0,1]$.


TITLE: How to show this random variable equals its conditional expectation?
QUESTION [3 upvotes]: Suppose $(\Omega,\mathcal F,\mathbb{P})$ is a probability space and that $\mathcal G$ is a sub-sigma-algebra of $\mathcal F$. If $X$ is an integrable, non-negative random variable with the same distribution as $\mathbb{E}[X|\mathcal G]$, how does one show that $X=\mathbb{E}[X|\mathcal G]$ a.s.?
Thank you.
(I can do the case where $X$ has finite variance, but the method doesn't seem to extend to this...)
Edit
I've just found a hint for this question (but remain stuck):

Show that $f(X)=f(\mathbb{E}[X|\mathcal{G}])$, almost surely, where $f(x)$ equals $1$ if $x>0$, $0$ if $x=0$ and $-1$ otherwise.

REPLY [1 votes]: By conditional Jensen $ e^{-E(X\vert \mathcal G)} \le E(e^{-X} | \mathcal G) $, by taking expectations equality must hold.  I would like to claim here that your result follows by characterization of equality in conditional Jensen, but I'm not sure about that. However, the lhs has the same distribution as $e^{-X}$  , and now you have your original condition with bounded r.v.s


TITLE: How prove this $p\equiv 1\pmod 8$
QUESTION [4 upvotes]: let $m,n$ are positive integer numbers,and such $m<n$,if

$$p=\dfrac{n^2+m^2}{\sqrt{n^2-m^2}}$$
  is prime number.

show that
$$p\equiv 1 \pmod 8$$
My try:
let $$p=\dfrac{n^2+m^2}{\sqrt{n^2-m^2}}=\dfrac{n^2-m^2+2m^2}{\sqrt{n^2-m^2}}=\sqrt{n^2-m^2}+\dfrac{2m^2}{\sqrt{n^2-m^2}}$$
Then I can't

REPLY [1 votes]: If $p$ divided either of $n$ and $m$, then it must also divide the other, so dividing by $p$ yields (with $n = p\nu,\, m = p\mu$)
$$1 = \frac{\nu^2 + \mu^2}{\sqrt{\nu^2-\mu^2}},$$
from which it follows that $\nu = 1,\, \mu = 0$ contradicting the positivity of $n$ and $m$. So $p$ divides neither $n$ nor $m$.
Now, for
$$\frac{n^2+m^2}{\sqrt{n^2-m^2}}$$
to be an integer, we must have
$$n^2 - m^2 = k^2,$$
and hence
$$n^2 + m^2 = k^2 + 2m^2.$$
$p = 2$ is impossible, since $\dfrac{n^2+m^2}{\sqrt{n^2-m^2}} > \sqrt{n^2+m^2}$, which would force $n = m = 1$ to have $\sqrt{n^2+m^2} < 2$. So $p$ is an odd prime.
$p \mid n^2 + m^2$ implies that $-1$ is a quadratic residue modulo $p$, so $p \equiv 1 \pmod{4}$.
$p \mid k^2 + 2m^2$ implies that $-2$ is a quadratic residue modulo $p$, so $p \equiv 1 \pmod{8}$ or $p \equiv 3 \pmod{8}$. Since we have $p \equiv 1 \pmod{4}$ from above, it follows that $p \equiv 1 \pmod{8}$.


text:
\begin{document}
\newpage
\doublespacing
\doublespacing
\title {\Large {  Optimum Production for a heaped stock  dependent breakable item through variational principle}}
\author{{J. N. Roul $^{a},$ K. Maity  $^{b}, $   S. Kar $^{c},$ and M. Maiti $^{d}$}\\\\
  $^{a}$Patha Bhavana, Visva-Bharati, Santiniketan-731235, West Bengal, India,
\\   Email$-$jotin2008@rediffmail.com \\\\
  $^{b}$ Department of Mathematics, Mugberia
 Gangadhar Mahavidyalaya, Bhupatinagar,\\   Purba
Medinipur-721425, W.B., India, Email$-$kalipada\_maity@yahoo.co.in\\\\
   $^{c}$ Department of Mathematics, National Institute of
Technology, Durgapur-713209\\ W.B., India, Email$-$kar\_s\_k@yahoo.com\\\\
  $^{d}$Department of Applied Mathematics with Oceanology and Computer
 Programming,\\
 Vidyasagar University, Midnapore-711201,W.B., India,
Email$-$mmaiti2005@yahoo.co.in}

\maketitle
\begin{abstract}
Breakability rate of fragile item depends on the accumulated stress of heaped stock level. So   breakablility rate can be considered as dependent parameter  of stock variable.  The unit production cost is a function of production rate and also dependent on raw material cost, development cost  and wear-tear cost. The holding cost is assumed to be non-linear, dependent on time. Here optimal control problem  for a fragile  item under finite  time horizon is considered.  The profit  function which consists of  revenue, production  and holding costs is formulated as a Fixed-Final Time and Fixed State System(cf. Naidu (2000)[\ref{nai2000}]) optimal control problem with finite time horizon. Here  production rate is unknown and considered as a control variable and  stock level is taken as a state variable. It  is formulated to optimize the production rate so that total profit is  maximum.  As particular cases,  models are evaluated with and without  breakability. The models are  solved by using conventional Variational Principle  along with the non-linear optimization technique-Generalised Reduced Gradient Method (LINGO 12.0).  The optimum results are illustrated  both  numerically and graphically. Some sensitivity analysis on breakability coefficient are presented.
\end{abstract}
\noindent ~~~~~~{\bf Keywords:}  Variational Principle, Finite time horizon, Breakability, Optimal Control Problem.\\
\section{Introduction}

Available items  in the market can be broadly classified into two categories-damageable items and non-damageable items. Damageable items again can be classified into two sub-categories -deteriorating and breakable items. Deteriorating items are deteriorated with time.  Normally, seasonal goods like fruits, vegetables, X-mas cake, etc., are deteriorating in nature. Demands of these items normally exist in the market for a finite time and obviously these types of demands are dependent on time.\\

   Fashionable/decorating items are made of glass, ceramic, etc., and demand of these types of items continues over a fixed duration only. Sale of these fashionable products increases with the showing of stock. But manufacturers  face a conflicting situation in their business as breakability increases with the increase of piled stock.   Inventory practitioners  are not interested with this types of items for their breakability. Few articles in this direction have been focused by Mandal and Maiti 2000 [\ref{mandal2000}], Guchhait et al. (2010,2013) [\ref{guchhait2010},\ref{guchhait2013}].\\
Almost all inventory models are formulated with constant  holding cost ( Sarkar et al., 2011 [\ref{sar2011}; Roul, et al. 2017 [\ref{rou2017}] ). In reality, due to inflation, bank interest, rental charges, preservation cost, etc., it increases with time. Thus some factors are contributed to the holding cost change with time (Giri et al., 1996 [\ref{gir1996}], Alfares 2007 [\ref{alfares2007}],  Urban 2008 [\ref{urban2008}] ) and others remain constant. Also set-up cost depends on production rate as high production rate require sophisticated modern machineries. In this paper, holding and set-up costs are considered as functions of time and production rate respectively as a particular case. This is very simple application of breakability  in production inventory problem via variational principle. \\

In this paper, a production-inventory model with imperfect production process is considered for a breakable  item over a finite time horizon. The production rate varies with time.  Set-up cost is partially production rate dependent and holding cost is also partially time dependent. The unit production cost is a function of production rate, raw material cost, labour charge, wear and tear cost. The
models are formulated as optimal control problems for the maximization of total profits over the planning horizon and solved using variational principle. The profit is maximized using the optimization technique-Generalised Reduced Gradient method(LINGO 12.0) (cf. Gabriel and Ragsdell 1977)[\ref{gab1977}]. The models are numerically illustrated and optimum results are presented in tabular form and graphically . Some sensitivity analysis on breakability coefficient are given.
\newpage
\section{Proposed Optimal Control Models }
For  the optimal control problem under finite time horizon, following assumptions and notations are used.

\subsection{Assumptions}
\begin{itemize}
\item It is a single period production inventory model with finite time horizon $t\epsilon[0,T]$;
\item  Break-ability rate is function of stock;
\item Shortages are not allowed;
\item There is no repair or replacement of breakable units over whole time period;
\item Unit production cost depends on produced-quantity, raw material, wear-tear and development costs;
\item  Deterministic  variational principle is  used for the model.
\end{itemize}

\subsection{Notations}
{$T :$} total time for the system;\\
{$u(t):$} production rate at time $t$ which is taken as a control variable;\\
{$x(t):$} stock level at  time t which is a state variable;\\
{$h(t) :$}=$a+b t^n( a\,\,\,\, \mbox{and} \,\,\,\,b \,\,\,\,\mbox{are constants })$ holding cost per unit;\\
{$L :$} fixed cost like labour, energy, etc;.\\
{$N :$} cost of technology, design, complexity, resources, etc;.\\
{$c_{10}:$} constant material cost;\\
{$d_{1}:$} constant demand at initial stage;\\
{$d(t)=d_{1}+d_2 t+d_3 t^2:$} demand function with $d_2$ and $d_3$ as constants.\\
{$c_d(l)=N+L: $} development cost to improve the quality of the product;\\
{$\beta_{10}:$} wear-tear cost for the system.\\
{$p:$} is the selling price of unit quantity. \\
$ c_{u}(t)=c_{10} +\frac{c_{d}(l)}{u(t)}+\beta_{10} u(t)$ is the unit production cost.

\subsection{Model-1: Formulation of optimal control  models for quadratic  demand with holding cost $h(t)=a+bt^n$}
\indent Let us consider  a production system with production rate $ u(t)$, demand $d(t)$ and stock dependent breakability rate $B(x)$. In this model,  the differential equation for stock level
{$x(t)$} regarding above system
during a fixed time-horizon, T is
\begin{eqnarray} \frac{d}{dt}\bigg(x(t)\bigg)= u (t)-d(t)-B(x)
\end{eqnarray}
where $ d(t)=d_1 +d_2 t+d_3 t^2$ and $B(x)\equiv$ $B(x(t))$\\
The units made of China-clay, glass, ceramic, mud, etc., are
kept in showrooms as heaped stocks. Due to this, units at
the bottom are under stress and as a result these units, break or get cracked. Thus breakability
rate of such an item depends on the accumulated stress of
stock level. So here  breakablility rate is dependent on stock variable x(t). So the breakability rate $B(x)$ is of the form
\begin{eqnarray}
B(x)=b_1 x^{\gamma} \,\,\, \mbox{and} \,\,\,\,\, b_1 \mbox{=constant  with} \,\,\,\,\, \gamma>0;
\end{eqnarray}
 \noindent The unit
production cost is considered as a function of produced-quantity,
raw material cost, wear-tear and development costs(cf. Khouja [\ref{Khouja}]). So the total
production cost  is
\begin{eqnarray}
 c_{u} (t)u(t)=\bigg(c_{10}u(t) +{c_{d} (l
)} +\beta _{10} u^2 (t)\bigg)\end{eqnarray} \noindent
\noindent  The total holding cost over the finite time interval
$[0,\,T]$
 for the stock $x(t)$ is
\begin{eqnarray}
 \int _{0}^{T} h(t) x (t)\,dt \nonumber\
 \end{eqnarray}
 where $ h(t)=a+b t^n $ and $ a,b $ are being constants. \\
For a production system set-up cost is taken as normally constant. But, if dynamic
production rate is considered, some machineries, etc., are to be
set-up and maintained in such a way that the production
system can stand with the pressure of increasing demand.
Thus  a part of set up cost per unit time is linearly proportional to production rate and
hence the form is of the form   $\frac{s_1 +s_2 u(t)}{T}$. Here $s_1$ and $ s_2$  are constants.
Therefore the set up cost $s(t)$ for the model is taken as
\begin{eqnarray}
s(t)=\frac{s_1 +s_2 u(t)}{T}
\end{eqnarray}
 Let $p$  be the selling price of an unit.  Then the revenue from market demand is
$p d(t)$. Thus the problem reduces to maximization of the profit function $J$
subject to the  constraint satisfying the dynamic
production-demand relation.
\begin{eqnarray}
&&\mbox{Max} J=\int _{0}^{T}\bigg(p d(t) - h(t) x(t)-c_{u }
 (t) u (t)-s(t)\bigg) dt\nonumber\\
&&\mbox{sub\, to}\,\,
 \frac{d}{dt}\bigg(x(t)\bigg)= u (t)-d(t)-B(x)\label{5}\\
  &&\mbox{ where \,} u(t)\ge 0\,\,  ,x(t)\ge0\label{6}\\
\nonumber\end{eqnarray}
\noindent Then the expression (\ref{5}) is written as
\begin{eqnarray}
 J_{max}\bigg({x}(t),\dot{x}(t),t\bigg)
 &=&\int _{0}^{T}\bigg(pd(t)-h(t)\, x(t)-\bigg(c_{10}+\frac{ s_2}{T}\bigg)
 \bigg({\dot{x}(t)+d(t)+B(x)}\bigg) \nonumber\\&-&{ \beta_{10}}\bigg({\dot{x}(t)+d(t)+B(x)}\bigg)^2 -\bigg({c_{d}(l)}+\frac{ s_1}{T}\bigg)\bigg) dt   \,\, \mbox{ where \,} \,  x(t)\ge0\label{7}
 \end{eqnarray}
 \noindent The above problem (\ref{7}) is defined as an optimal
control problem with state variable $x(t)$. Here, (\ref{7}) contains $u(t)$ implicitly.\\
\noindent Using Euler-Lagrange's equation, Fixed-Final Time and Fixed State System (i.e, here final
time $T$
is specified and $x(t)\equiv x, \dot x(t)\equiv \dot x $, $x(0)=0$ , $ x(T)=0$); we
have
\begin{eqnarray}\frac{\partial
J}{\partial {x}}&-&\frac{d\ }{dt}\bigg(\frac{\partial
J}{\partial \dot{x}}\bigg)=0\label{8}\\
\nonumber\end{eqnarray}
\noindent Using equation (\ref{8}) in (\ref{7}), we get \\
\begin{eqnarray}
-h(t)-\bigg(c_{10}+\frac{s_2}{T}\bigg)b_1 \gamma x^{\gamma-1}-2 \beta_{10}\bigg(\dot x(t)+d(t)+b_1 x^{\gamma}\bigg)b_1 \gamma x^{\gamma-1}+2 \beta_{10}\frac{d}{dt}\bigg(\dot x(t)+d(t)+b_1 x^{\gamma}\bigg)=0\end{eqnarray}

Representing $x(t)\equiv x$,
 \begin{eqnarray}
\frac{d^2 }{dt^2}\bigg(x\bigg)-b^2_1 \gamma x^{2 \gamma-1}-\bigg(c_{10}+\frac{s_2}{T}+2 \beta_{10} d(t)\bigg)b_1 \gamma x^{\gamma-1} =\frac {1}{2\beta_{10}}\bigg(h(t)-{2\beta_{10}}(d_2+2 d_3 t)\bigg)
\end{eqnarray}
This is a second order first degree differential equation of the independent variable t.\\

\noindent Therefore, solving the above we can get values of
\begin{eqnarray}
x(t)=\mbox{Complementary Function + Particular Integral }
\end{eqnarray}
 Here the initial condition  $x(0)=0$  and final condition
  $x(T)=0$. Using these conditions, we can get the constants of integration. From(1), we have the production rate
 \begin{eqnarray}
 u(t)=\frac{d}{dt}\bigg(x(t)\bigg) +d_1 +d_2 t+d_3 t^2 +b_1 x(t)  \end{eqnarray}

  \noindent Then the  profit functional $J$ takes the form $J_{max}$
 \begin{eqnarray}
  &=&\int _{0}^{T}\bigg(p(d_1 +d_2 t+d_3 t^2)- h(t) \, x(t)-c_{10} u(t)-c_d(l)-\beta_{10} u^2(t)-\frac{s_1}{T}-\frac{s_2}{T} u(t)\bigg) dt\nonumber\\
 &=&\int _{0}^{T}\bigg(p(d_1 +d_2 t+d_3 t^2)- h(t) \, x(t)-\bigg(c_{10}+\frac{s_2}{T}\bigg) u(t)-\beta_{10} u^2(t)-\bigg(c_d(l)+\frac{s_1}{T}\bigg)\bigg) dt
\end{eqnarray}

\subsection{Models with particular values of $n\,\,\, \mbox{and}\,\,\, \gamma $:}
\subsubsection{Model-1a: Formulation of optimal control  models for dynamic demand with holding cost $h(t)=a+bt$ for $n=1$ and linear breakability $B(x,t)=b_1 x(t) $ for $\gamma=1$ }
\indent In this model,  the differential equation for stock level
{$x(t)$} regarding above system
during a fixed time-horizon, T is
\begin{eqnarray} \frac{d}{dt}\bigg(x(t)\bigg)= u (t)-d(t)-b_1 x(t) \,\,\,\,\,\,\mbox{where} \,\,\,\,\, d(t)=d_1 + d_2 t+d_3 t^2 \label{14}
\end{eqnarray}

Proceeding as before,
 we have
\begin{eqnarray}
 J_{max}\bigg({x}(t),\dot{x}(t),t\bigg)
 &=&\int _{0}^{T}\bigg(pd(t)-h(t)\, x(t)-\bigg(c_{10}+\frac{ s_2}{T}\bigg)\bigg({\dot{x}(t)+d(t)+ b_1 x(t)}\bigg) \nonumber\\
 &-&{ \beta_{10}}\bigg({\dot{x}(t)+d(t)+ b_1 x(t)}\bigg)^2 -\bigg({c_{d}(l)}+\frac{ s_1}{T}\bigg)\bigg) dt   \,\, \mbox{ where \,} \,  x(t)\ge0 \label{15}
 \end{eqnarray}
 \noindent The above problem (\ref{15}) is defined as an optimal
control problem with state variable $x(t)$. Here, (16) contains $u(t)$ implicitly.\\
\noindent Using Euler-Lagrange's equation, Fixed-Final Time and Fixed State System (i.e, here final
time $T$
is specified and $x(t)\equiv x, \dot x(t)\equiv \dot x $, $x(0)=0$ , $ x(T)=0$); we
have
\begin{eqnarray}\frac{\partial
J}{\partial {x}}&-&\frac{d\ }{dt}\bigg(\frac{\partial
J}{\partial \dot{x}}\bigg)=0\label{16}\\
\nonumber\end{eqnarray}
\noindent Using equation (\ref{16}) in (\ref{15}), we get \\
\begin{eqnarray}
-h(t)-\bigg(c_{10}+\frac{s_2}{T}\bigg)b_1-2 \beta_{10}\bigg(\dot x(t)+d(t)+b_1 x(t)\bigg)b_1
+2 \beta_{10}\frac{d}{dt}\bigg(\dot x(t)+d(t)+b_1 x(t)\bigg)=0\end{eqnarray}
It can be written as
\begin{eqnarray}
\frac{d^2 }{dt^2}\bigg(x(t)\bigg)-b^2_1  x(t)= \frac{1}{2 \beta_{10}}\bigg(a_{11} +a_{22} t + a_{33} t^2\bigg)
\end{eqnarray}
 where $a_{11}=(a+b_1 c_{10}+ b_1 \frac{s_2}{T} +2 \beta_{10} b_1 d_1-2\beta_{10} d_2); \,\,\,
a_{22}=(b-4 \beta_{10} d_3 +2 \beta_{10} b_1 d_2);   \,\,\,
a_{33}=2 \beta_{10} b_1 d_3 $\\
Then we have
\begin{eqnarray}
\frac{d^2 }{dt^2}\bigg(x(t)\bigg)-b^2_1  x(t)=f(t)  \,\,\,\,\mbox{where}\,\,\,f(t)=\frac{1}{2 \beta_{10}}\bigg(a_{11} +a_{22} t + a_{33} t^2\bigg)
\end{eqnarray}
This is a second order first degree differential equation of the independent variable t.\\
Therefore,
\begin{eqnarray}
x(t)&=& A_1 e^{b_1 t} +  B_1 e^{-b_1 t}+ \frac{f(t)}{D^2-b^2_1 };  D\equiv \frac{d}{dt}\nonumber\\
&=& A_1 e^{b_1 t} +  B_1 e^{-b_1 t}-\frac{f(t)}{b^2_1}-\frac{a_{33}}{b^4_1 \beta_{10}}\label{21}
 \end{eqnarray}
 where $A_1$ and $B_1$ are constants.
 Therefore  $A_1$ and $B_1$  is given by the following relation. \\
  $A_1 +  B_1 = \frac{f(0)}{b^2_1}+\frac{a_{33}}{b^4_1 \beta_{10}}
$ and
$ A_1 e^{b_1 T} +  B_1 e^{-b_1 T}=\frac{f(T)}{b^2_1}+\frac{a_{33}}{b^4_1 \beta_{10}}
$.\\
 From(\ref{14}), we have the production rate
 \begin{eqnarray}
 u(t)= A_1 b_1 e^{b_1 t}-B_1  b_1 e^{-b_1 t} -\frac{a_{22}+2 a_{33}t }{2  b^2_1  \beta_{10}} +d_1 +d_2 t+d_3 t^2 +b_1 x(t) \end{eqnarray}
 It can be written as
  \begin{eqnarray}
 u(t)&=& 2 A_1 b_1 e^{b_1 t} -\bigg(\frac{a_{22} }{2  b^2_1  \beta_{10}}+\frac{a_{11}}{2 b_1 \beta_{10}}+\frac{a_{33}}{b^3_1 \beta_{10}}-d_1\bigg) -\bigg(\frac{ a_{33}}{  b^2_1  \beta_{10}} +\frac{a_{22}}{2 b_1 \beta_{10}}-d_2\bigg) t \label{22}
 \end{eqnarray}

  Then the  profit functional $J$ takes the form $J_{max}$
 \begin{eqnarray}
  &=&\int _{0}^{T}\bigg(p(d_1 +d_2 t+d_3 t^2)- h(t) \, x(t)-c_{10} u(t)-c_d(l)-\beta_{10} u^2(t)-\frac{s_1}{T}-\frac{s_2}{T} u(t)\bigg) dt\nonumber\\
 &=&\int _{0}^{T}\bigg(p(d_1 +d_2 t+d_3 t^2)- h(t) \, x(t)-\bigg(c_{10}+\frac{s_2}{T}\bigg) u(t)-\beta_{10} u^2(t)-\bigg(c_d(l)+\frac{s_1}{T}\bigg)\bigg) dt\nonumber\\
  &=&p(d_1 T +d_2 \frac{T^2}{2}+d_3 \frac{T^3}{3})-\bigg(c_d(l)+\frac{s_1}{T}\bigg)T\nonumber\\
  &-&a\bigg(\frac{A_1}{b_1} (e^{b_1 T}-1)+\frac{B_1}{b_1} (1-e^{-b_1 T})-\frac{1}{2 {b^2_1} \beta_{10}}\bigg(a_{11}T +a_{22} \frac{T^2}{2} + a_{33} \frac{T^3}{3}\bigg)-\frac{a_{33}}{b^4_1 \beta_{10}} T\bigg)+\frac{a_{33} b T^2}{b^4_1 2 \beta_{10}}\nonumber\\
  &-&b\bigg( A_1(\frac{T}{b_1} e^{b_1 T}+\frac{1}{b^2_1}(1-e^{b_1 T})) +  B_1 (-\frac{T}{b_1} e^{-b_1 T}+\frac{1}{b^2_1}(1-e^{-b_1 T}))-\frac{1}{2 {b^2_1} \beta_{10}}\bigg(a_{11}\frac{T^2}{2} +a_{22} \frac{T^3}{3} + a_{33} \frac{T^4}{4}\bigg)\bigg)\nonumber\\
  &-&C_{11}\bigg(2 A_1  (e^{b_1 T}-1) -\bigg(\frac{a_{22} }{2  b^2_1  \beta_{10}}+\frac{a_{11}}{2 b_1 \beta_{10}}+\frac{a_{33}}{b^3_1 \beta_{10}}-d_1\bigg)T -\bigg(\frac{ a_{33}}{  b^2_1  \beta_{10}} +\frac{a_{22}}{2 b_1 \beta_{10}}-d_2\bigg) \frac{T^2}{2}\bigg)\nonumber\\

 &-& \beta_{10} \bigg(M^2_1 T +M^2_2 \frac{T^2}{2}+M^2_3  \frac{(e^{2 b_1 T}-1)}{2b_1}+ M_1 M_2 T^2-2M_2 M_3 (\frac{T}{b_1} e^{b_1 T}+\frac{1}{b^2_1}(1-e^{b_1 T}))-2M_1 M_3 \frac{(e^{ b_1 T}-1)}{b_1} \bigg)\nonumber\\ \label{23}
 \end{eqnarray}
 \begin{eqnarray}
  &&\mbox{Where}\nonumber\\
  && M_3=-2 A_1 b_1 e^{b_1 T} ; M_1=\frac{a_{22} }{2  b^2_1  \beta_{10}}+\frac{a_{11}}{2 b_1 \beta_{10}}+\frac{a_{33}}{b^3_1 \beta_{10}}-d_1;
  M_2=\frac{ a_{33}}{  b^2_1  \beta_{10}} +\frac{a_{22}}{2 b_1 \beta_{10}}-d_2 \nonumber\\
  &&C_{11}=c_{10}+\frac{s_2}{T};
 \nonumber\\ \label{24}
\end{eqnarray}
\subsubsection{Model-1b: Formulation of optimal control  models for dynamic demand with holding cost $h(t)=a+bt$ with linear breakability $B(x,t)=b_1 x(t) $, $ b_1=0$}
If we take $b_1= 0$ in the
above model (Model-1a), then the present model represents an
inventory model without deterioration. As $b_1$ appears in the
denominator of the expression (\ref{21}), (\ref{22}), (\ref{23}) and (\ref{24}), it is not possible to put
$b_1$ directly  to obtain that expressions and the profit expression also. Thus, in
this case, the corresponding profit function is calculated omitting
$B(x,t)=b_1 x(t) $  and then proceeding in the same way as in Model-1a.
\begin{eqnarray}
a+bt- 2 \beta_{10} \frac{d\ }{dt} \bigg( \dot{x}(t)+d(t)\bigg)=0
\end{eqnarray}
Representing $x(t)\equiv x$,
 \begin{eqnarray}
\frac{d^2 x}{dt^2}  =f(t)  \,\,\,\,\mbox{where}\,\,\, f(t)=-2 d_3 t+\frac{ b t}{2\beta_{10}}  +\frac{ a}{2\beta_{10}}-d_2
\end{eqnarray}
This is the second order first degree differential equation of the independent variable t.\\
\noindent Therefore, solving the above we can get
\begin{eqnarray}
x(t)&=&A+Bt -(2d_3 -\frac{ b }{2\beta_{10}})\frac{t^3}{6}+(\frac{ a}{2\beta_{10}}-d_2) \frac{t^2}{2}
\end{eqnarray}
 Where $A$ and $B$ are constants. Here the initial condition  $x(0)=0$  and final condition
  $x(T)=0$. Using these conditions, we can get the values of  $A=0$  and $B=(2d_3 -\frac{ b }{2\beta_{10}})\frac{(T)^2}{6}+(\frac{ a}{2\beta_{10}}-d_2) \frac{T}{2}
$. \\
Then, we have the production rate
 \begin{eqnarray}
 u(t)=B+ d_1 +\frac{ a}{2\beta_{10}} t+\frac{ b }{2\beta_{10}}\frac{t^2}{2}
   \end{eqnarray}
\noindent Then the  profit functional $J$ takes the form $J_{max}$
 \begin{eqnarray}
 &=&\int _{0}^{T}\bigg(p(d_1 +d_2 t+d_3 t^2)- h(t) \, x(t)-\bigg(c_{10}+\frac{s_2}{T}\bigg) u(t)-\beta_{10} u^2(t)-\bigg(c_d(l)+\frac{s_1}{T}\bigg)\bigg) dt\nonumber\\
 &=&p(d_1 T +d_2 \frac{T^2}{2}+d_3 \frac{T^3}{3})-\bigg(c_d(l)+\frac{s_1}{T}\bigg)T-a\bigg( A T+B \frac{T^2}{2} -(2d_3 -\frac{ b }{2\beta_{10}})\frac{T^4}{24}+(\frac{ a}{2\beta_{10}}-d_2) \frac{T^3}{6}
\bigg)\nonumber\\
 &-&b\bigg( A \frac{T}{2}+B \frac{T^3}{3} -(2d_3 -\frac{ b }{2\beta_{10}})\frac{T^5}{30}+(\frac{ a}{2\beta_{10}}-d_2) \frac{T^4}{8}\bigg)-\bigg(c_{10}+\frac{s_2}{T}\bigg)\bigg( B T+ d_1 T +\frac{ b }{2\beta_{10}}\frac{T^3}{6}+\frac{ a}{2\beta_{10}} \frac{T^2}{2}\bigg)\nonumber\\
 &-&\beta_{10}\bigg( (B+ d_1)^2 +(\frac{ b }{2\beta_{10}})^2\frac{T^5}{20}+(\frac{ a}{2\beta_{10}})^2 \frac{T^3}{3}+\frac{ b(B+d_1) }{\beta_{10}}\frac{T^3}{6}+\frac{ a(B+d_1)}{2 \beta_{10}} T^2 +\frac{ a b}{4\beta^2_{10}} \frac{T^4}{8}\bigg)
\end{eqnarray}
\newpage
\section{Numerical Experiments}
\noindent To illustrate the models, we assume the following input data for Models -1a and 1b. For Model-1a, $b_1$ has taken as $0.02$. and the other inputs are given in the following Table-\ref{t1}
\subsection{Input:}
 Inputs are given in the following Table-\ref{t1}
\begin{table}[h]
\begin{center} \caption{Input Values}
\scriptsize
\begin{tabular}{|cc c  c ccc c |} \hline
$L$   &$ N$&$  cd(l)$ & $ c_{10}$& $ \beta_{10} $&$p \mbox{(in \$)}$& $ s_1$ &
\\\hline\label{t1}
$ 40$ & $60$    & $100$
&$0.7$ & $0.5$& $200$&  $ 10$ &\\\hline
$a$   &$ b $&$ d_1$ & $ d_2$& $ d_3 $&$T$& $ s_2$ &
\\\hline
$ 3$ & $0.2$        & $7$
&$4$ & $2$& $12$&  $ 3$ &\\\hline
\end{tabular}
\end{center}
\end{table}
\subsection{Output:}
\noindent Using the above input data, the profit functions of Model$-1a$ and $1b$ are maximized and
with the help of GRG (LINGO-11.0), the unknowns, $x(t),u(t)$ and maximum profits etc. are
evaluated. Here  profit is $\$180913.30$ for Model-1a when breakability $b_1=0.02$ and profit is          $ \$ 247007.30$ for Model-1b when  breakability $b_1=0.00$(i.e the word breakability is not applicable). The numerical values of $ u(t), d(t), x(t)$
for the Model-1a (where  breakability is only dependent on stock and linear holding cost) and Model-$1b$ are given in Tables-\ref{t2} and -\ref{t3}. These are also graphically depicted in Figure-\ref{f1} and \ref{f2}. For the value of $b_1=0.11$, the unknowns $x(t),u(t)$ and $d(t)$ are evaluated which are given in Table-\ref{t4} and these are depicted in Figure-\ref{f3}. For the different values of breakability co-efficient $b_1$, profits are given in Table-\ref{t5}.
\begin{table}[h]
\begin{center}
\caption{Values of $u(t), d(t)$ and $x(t)$;  for Model-1a}
\resizebox{\linewidth}{!}{{\scriptsize
\begin{tabular}{|c|c c c c c c c c c c c c c|} \hline
$t$   & $0$ & $1$ & $2$ & $3$ & $4$ &$5$ & $6$ & $7$& $8$& $9$ & $10$ & $11$ & $12$
\\\hline\label{t2}
$u(t)$ & $94.98$ & $100.05$ & $105.43$ &$111.16$ & $117.17$& $123.44$ &$130.09$ & $137.08$& $144.41$  & $152.10$& $160.14$ &$168.54$  & $177.32$
\\\hline
 $d(t)$ & $7.00$ & $13.00$ & $23.00$ & $37.00$ & $55.00$ & $77.00$  & $103.00$  & $133.00$  & $167.00$    & $205.00$  &$247.00$& $293.00$ & $343.00$
\\\hline
 $x(t)$ & $0$ & $86.95$&  $169.44$ &$243.86$ & $306.76$ & $354.70$ & $384.35$ & $392.44$ & $375.77$  & $331.22$& $255.72$ &$146.28$  & $0 $
\\\hline
\end{tabular}
}}
\end{center}
\end{table}
\begin{table}
\begin{center}\scriptsize
\caption{Values of $u(t), d(t)$ and $x(t)$;  for Model-1b}
\begin{tabular}{|c|c c c c c c c c c c c c c|} \hline
$t$   & $0$ & $1$ & $2$ & $3$ & $4$ &$5$ & $6$ & $7$& $8$& $9$ & $10$ & $11$ & $12$
\\\hline\label{t3}
$u(t)$ & $104.20$ & $107.30$ & $110.60$ &$114.10$ & $117.17$& $121.70$ &$125.80$ & $130.10$& $134.60$  & $139.30$& $144.20$ &$149.30$  & $154.60$
\\\hline
 $d(t)$ & $7.00$ & $13.00$ & $23.00$ & $37.00$ & $55.00$ & $77.00$  & $103.00$  & $133.00$  & $167.00$    & $205.00$  &$247.00$& $293.00$ & $343.00$
\\\hline
 $x(t)$ & $0$ & $96.06$&  $187.33$ &$270.00$ & $340.26$ & $394.33$ & $428.66$ &$438.66$& $421.33$ & $372.60$  & $288.66$& $165.732$&   $0 $
\\\hline
\end{tabular}
\end{center}
\end{table}
\begin{figure}
\centering
\includegraphics[width=7.5cm, height=6.5cm]{fig01}
\caption{Time Vs Production, Stock Level and Demand for Model-1a.}\label{f1}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=7.5cm, height=6.5cm]{fig05}
\caption{Time Vs Production, Stock Level and Demand for Model-1b.}\label{f2}
\end{figure}

\section{Sensitivity Analysis}
For production system, a general belief that the system is more and more perfect as breakability of the system decreases.  From Table-\ref{t4}, it is seen that profit decreases as breakability rate increases. This is as per our expectation. The breakability constant $b_1$ has taken in the input as $0.02$. For this value of $b_1$, the profit for Model-1a is 180913.30 $\$$. With $b_1=0.11$ , the value of profit for Model-1a is 153447.70 $\$$. For that value of $b_1$, the unknowns, $x(t),u(t)$ and $d(t)$ are evaluated which are given in Table-\ref{t4} and these are depicted in figure-\ref{f3}.
Taking the value of $b_1$ as $b_1=0.11$  and$b_1=0.02$ for Model-1a, the comparison  diagram for production and stock  has been drawn in figures-\ref{f4} and \ref{f5}.
\begin{table}
\begin{center}
\caption{Values of $u(t), d(t)$ and $x(t)$;  for Model-1a, [The value of $b_1$ as $b_1=0.11$ , the value of profit for Model-1a is 153447.7 $\$$]}
\resizebox{\linewidth}{!}{{\scriptsize
\begin{tabular}{|c|c c c c c c c c c c c c c|} \hline
$t$   & $0$ & $1$ & $2$ & $3$ & $4$ &$5$ & $6$ & $7$& $8$& $9$ & $10$ & $11$ & $12$
\\\hline\label{t4}
$u(t)$ & $48.96$ & $58.04$ & $68.38$ &$80.15$ & $93.48$& $108.58$ &$125.65$ & $144.92$& $166.63$  & $191.09$& $218.56$ &$249.45$  & $284.18$
\\\hline
 $d(t)$ & $7.00$ & $13.00$ & $23.00$ & $37.00$ & $55.00$ & $77.00$  & $103.00$  & $133.00$  & $167.00$    & $205.00$  &$247.00$& $293.00$ & $343.00$
\\\hline
 $x(t)$ & $0$ & $41.44$&  $80.15$ &$113.90$ & $140.83$ & $159.44$ & $168.59$ & $167.44$ & $155.48$  & $132.49$& $98.59$ &$54.18$  & $0 $
\\\hline
\end{tabular}
}}
\end{center}
\end{table}
\begin{figure}
\centering
\includegraphics[width=7.5cm, height=6.5cm]{fig04}
\caption{Time Vs Production, Stock Level and Demand for Model-1a when breakability $b_1=0.11$ }\label{f3}
\end{figure}
\begin{table}
\begin{center}
 \caption{The values of $b_1$ Vs Profit of Model-1a }
 \resizebox{\linewidth}{!}{{\scriptsize
\begin{tabular}{|c|cc c c c c cc|} \hline
$b_1$  &$0.01$ & $0.02$ & $0.03$ & $0.04$ & $0.05$ & $0.06$ &$0.07$&$0.08$
\\\hline\label{t5}
$\mbox{Profit} \,\,\, J\,\,\, in \,\,\, $\$$$ &$185131.50$& $180913.30$ & $176871.90$ & $173036.20$ &$169431.00$ & $166076.60$& $162988.70$&$160178$
\\\hline
 \end{tabular}
 }}
\end{center}
\end{table}
\begin{figure}
\centering
\includegraphics[width=7 cm, height=6.5 cm]{fig02}
\caption{Comparison of Time Vs Stock Level for Model-1a for the values of $b_1=0.11$  and $b_1=0.02$.}\label{f4}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=7 cm, height=6.5 cm]{fig03}
\caption{Comparison of Time Vs Production rate for Model-1a for the values of $b_1=0.11$  and $b_1=0.02$.}\label{f5}
\end{figure}
\newpage
\section{Discussion}
 For production system, a general belief is that cost of the produced item increases as breakability
units of the system increase. Consequently, Min cost of the production system increases. i.e,  total profit of the production system decreases.  These are as per our expection. In the Figures-\ref{f1}and \ref{f2}\,\,  for the models, initially production is more than demand and hence stock is built up after satisfying the demand. As demand is here dynamic which  rapidly increases with time, after some time,  surpasses the production and then excess demand is satisfied from the stock. So, the deficit in demand occurs, so the stock starts to decrease at this point and gradually decreases up to the end of the cycle
and finally becomes zero. It is observed that  profit is $ \$ 247007.30$ for Model-1b without  breakability which is more than the profit  $\$180913.30$ for Model-1a when breakability $b_1=0.02$. When the value of $ b_1$  increases, the value of profit decreases. This is clear in Table-4. For particular value of $b_1$ as $b_1=0.11$ , the value of profit for Model-1a is 153447.7 $\$$. This result is very natural and  is as per  our expectation.
\section{Conclusion}
Here a production-inventory model for a breakable item is developed.  Here time dependent production rate fetches more profit for  time-dependent demand. From the present model it can be concluded that optimal control of production rate reduces holding cost as well as damageability which in turn increases profit separately for breakable/deteriorating items.  The models are also solved taking some of the inventory costs. The present models can be extended to the rough, fuzzy-rough, random, fuzzy-random environment taking constant part of holding cost, set-up cost, etc. as uncertain in nature. As the formulation and solution are quiet general, the results can be obtained for other forms of dynamic demand such as linearly increasing demand, ramp demand etc. The model can be extended to include multi-item fuzzy inventory problem with fuzzy space and budget constraints.

config: arxiv


TITLE: Proving the ratios
QUESTION [1 upvotes]: Now, I tried this by the usual determinant way of doing it. However, I somehow always end up with the result: $a=b=c$. And then, I get thoroughly baffled at what to do next and how to generate square terms out of the variables. So, how is it to be done really? What should be the approach?

If $$a(y+z)=x,~~b(z+x)=y,~~c(x+y)=z,$$
prove that $$\frac{x^2}{a(1-bc)}=\frac{y^2}{b(1-ca)}=\frac{z^2}{c(1-ab)}$$

REPLY [1 votes]: You can write the system of first three identities as
$$
\left( {\begin{array}{*{20}c}
   { - 1} & a & a  \\
   b & { - 1} & b  \\
   c & c & { - 1}  \\
 \end{array} } \right)\left( {\begin{array}{*{20}c}
   x  \\
   y  \\
   z  \\
 \end{array} } \right) = \left( {\begin{array}{*{20}c}
   0  \\
   0  \\
   0  \\
 \end{array} } \right)
$$
and for it to have solutions you shall impose that the determinant be null
$$
\left| {\left( {\begin{array}{*{20}c}
   { - 1} & a & a  \\
   b & { - 1} & b  \\
   c & c & { - 1}  \\
 \end{array} } \right)} \right| = 0\quad  \Rightarrow \quad ab + ac + bc + 2abc = 1
$$
so that, actually, the three parameters $a,b,c$ are not independent.
Solving for the null vector, with the condition above, you get (eliminating for instance $a$):
$$
\frac{x}
{{1 - bc}} = \frac{y}
{{b(1 + c)}} = \frac{z}
{{c(1 + b)}}
$$
To pass to the squares, and get a more symmetrical formulation, you can follow the answer given by @Robert Z.


TITLE: Any chances to find the following sum analytically?
QUESTION [4 upvotes]: Consider the following sum:
\begin{equation}
S(x, k, N, \alpha_1, \alpha_2) = \sum\limits_{m=1}^{N} \dfrac{ e^{i \phi_m k}}{x - \Omega(m, \alpha_1, \alpha_2, N)},
\end{equation}
where $x$ - complex valued parameter, $k, N$ - positive integers ($k<N$), $\phi_m = \dfrac{2 \pi (m-1)}{N}$, $\Omega (m, \alpha_1, \alpha_2, N) = i \alpha_1 \dfrac{\xi + \xi^{1/N} e^{-i \phi_m}}{1 + \xi^{1/N} e^{-i \phi_m}} - i \alpha_2$, and $\alpha_1, \alpha_2$ - real-valued constants, and $0<\xi<1$.
I really doubt that for any $N$ this can be written as a simple formula in terms of several known functions, but may there be any possible way to simplify it in certain limits? Like $N \to \infty$, or saying that $\alpha_2 \gg \alpha_1$.

REPLY [3 votes]: This sum has a closed form. After some simplification (multiplying the numerator and the denominator of the summand by $1+\xi^{1/N}e^{-i\phi_m}$, replacing $m$ with $m+1$, etc.), it is a linear combination of sums of the form (here $z\in\mathbb{C}$) $$G(n,k,z)=\sum_{m=0}^{n-1}\frac{\omega^{km}}{1-z\omega^m},\qquad\omega=e^{2\pi i/n}.$$
This can be computed like I did here (or here), or as follows: if $|z|<1$ then $$G(n,k,z)=\sum_{m=0}^{n-1}\omega^{km}\sum_{j=0}^\infty z^j\omega^{mj}=\sum_{j=0}^\infty z^j\sum_{m=0}^{n-1}\omega^{m(k+j)},$$ and the inner sum is $0$ unless $k+j$ is a multiple of $n$; then, if $1\leqslant k\leqslant n$, $$G(n,k,z)=n\sum_{q=1}^\infty z^{nq-k}=\frac{nz^{n-k}}{1-z^n}.$$
Using $G(n,k,z)=G(n,n+k,z)$, and analytic continuation (out of $|z|<1$), $$G(n,k,z)=\frac{nz^{(-k)\bmod n}}{1-z^n}.$$