text:
[XeTeX] Using the Unicode prime character (Was re. Single glyphs from a font)

Ross Moore ross at ics.mq.edu.au
Tue Jun 9 23:28:17 CEST 2009

Hi Joel,

On 10/06/2009, at 12:32 AM, Joel C. Salomon wrote:

> Ross,
>
> Thanks for all the help here.

You're welcome.

>>>> Now even this works:    \tag{$\ref{abc}'$}
>>>> using the ordinary ASCII single-quote/apostrophe.
>>> Not quite; $'$ has some superscripting magic associated with it.
>>> Try
>>> $\prime{}'$ and you’ll see the difference.  (Without the ‘{}’,
>>> you’ll
>>> get a weird primed prime.)
>>
>> That is TeX's  \mathcode"8000  trick, which allows ordinary
>> letters to become macros when in math-mode.
> <snip>
>> This makes it very convenient to do  f'' and f'''  etc.
>> But other complicated superscripting should not be done this way.
>
> I’d read about that trick.  The problem is that it want to treat $x'$
> like $x^\prime$ using the large CM Math prime character.

There is a discussion of these "pseudo-script" characters also
taking place on a MathML-related list.

Some software uses the Unicode character directly, while
other software (such as TeX) has a character that is required
to be superscripted; e.g., as $...^\prime ...$.
With these it is not normal to use the oversized character by itself.
It is only included in the font for the convenience of encoding,
and when it might be needed at \Large (etc.) sizes.

> The Unicode
> prime character is already at the correct size and position.  Try
> $\prime{}'$ and you’ll see that the second one is shifted up by some
> amount.  (With $\prime'$ it’s even worse.)

I don't see a difference in the placement. (see attached image)

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> I suppose the fix for this would be to redefine the \mathcode"8000
> macro
> for ['] so that it directly evaluates to [?].  Or just to avoid
> the $x'$
> form and directly use $x\prime$ or $x?$.

I'd not mess with the \mathcode of ['].
You'll surely break multiple primes. See my attached image.

If you want to mess with anything, do some tracing to look at how
$f''$ expands. Hook into the internal macros rather than change
the user-interface.

>
> —Joel Salomon

Hope this helps.

Ross

------------------------------------------------------------------------
Ross Moore                                       ross at maths.mq.edu.au
Mathematics Department                           office: E7A-419
Macquarie University                             tel: +61 (0)2 9850 8955
Sydney, Australia  2109                          fax: +61 (0)2 9850 8114
------------------------------------------------------------------------


text:
# Calculating current through an IR circuit

Hi, if I have a DC indoctor-resistor circuit with and inductance (L) of 700μH and a resistance (R) of 0.8Ω, I realise that I can use:

IL = (Vsource/R) * (1-e-t*(R/L))

to calculate the current through the inductor if I have a constant voltage source, but would it be possible to calculate how the current build (and also decay) might look in this or a similar method if my source voltage looked more like:

http://www.dransfield.talktalk.net/VoltageTrace.jpg [Broken]

So far I can't work out how to make this work with a changing voltage, as the equation appears to assume that you always start at 0

Also, on the decay side, the equation:

IL = (Vsource/R) * (e-t*(R/L))

Seems to assume that you always start at current I=V/R, which is not always necessarily true

I hope I've explained myself well enough, and thanks for any help in advance!

Last edited by a moderator:

tiny-tim
Homework Helper

LdI/dt + RI = V(t)

so use an integrating factor to simplify the LHS into a single integral, and then use area-under-a-graph

(btw, it's not called an IR circuit, it's an RL circuit)

Ah, thanks very much, I know it was something similar, but got the wrong letter!

In this case, which "I" would I use in RI? Would this also be the induced current, or just the dI/dt?

tiny-tim
Homework Helper
In this case, which "I" would I use in RI? Would this also be the induced current, or just the dI/dt?

uhh?

I is I

dI/dt is d/dt of I

(and the current through R is the same as the current through L … where else would it go?)

I was under the impression (from here and a few other places) that the total current was I = V/R , but there were different amounts in the inductor and resistor as the current has to build over time in the inductor (iL), but the current in the resistor (iR) was I-iL

I am expecting the current to rise sharply due to the high voltage, then remain a little more constant through the lower (hold) voltage

Simulation and testing shows that this happens, though creating a more one-dimensional approach is proving a little more tricky

tiny-tim
Homework Helper
I was under the impression (from here and a few other places) that the total current was I = V/R , but there were different amounts in the inductor and resistor as the current has to build over time in the inductor (iL), but the current in the resistor (iR) was I-iL

"once the Storage Phase has finished, the Current, iL, that flows through it is stable, iL = V / R, no Self Induced e.m.f. is produced and vL is zero. The Inductor acts like an ordinary connecting wire, its Resistance is zero."
… iL only = V/R when steady-state has been reached

until then, the total voltage = the voltage drop across the resistor (IR) plus the "induced" LdI/dt across the inductor.

It's the same I across both, for the same reason you have the same current through a water pipe that goes through various attachments …

water in = water out, and electric current in = electric current out.

In that case, how do I calculate the induced current at any time-step using a changing voltage input?

The equation you specified doesn't appear to calculate the induced current, which is what I'm after?

tiny-tim
Homework Helper
In that case, how do I calculate the induced current …

the current I is the current through the inductor

it's "induced voltage" (LdI/dt) across an inductor, not "induced current"

(eg if you look up "induced current" in wikipedia, you'll be redirected to a page which doesn't mention "induced current")

an inductor is a voltage source (emf source), not a current source

format: plain


text:
Timezone: »

Oral
Efficient Nonconvex Regularized Tensor Completion with Structure-aware Proximal Iterations
Quanming Yao · James Kwok · Bo Han

Thu Jun 13 11:35 AM -- 11:40 AM (PDT) @ Room 101

Nonconvex regularizers have been successfully used in low-rank matrix learning. In this paper, we extend this to the more challenging problem of low-rank tensor completion. Based on the proximal average algorithm, we develop an efficient solver that avoids expensive tensor folding and unfolding. A special sparse plus low-rank" structure, which is essential for fast computation of individual proximal steps, is maintained throughout the iterations. We also incorporate adaptive momentum to further speed up empirical convergence. Convergence results to critical points are provided under smoothness and Kurdyka-Lojasiewicz conditions. Experimental results on a number of synthetic and real-world data sets show that the proposed algorithm is more efficient in both time and space, and is also more accurate than existing approaches.


text:
SCIENCE CHINA Information Sciences, Volume 61 , Issue 1 : 012201(2018) https://doi.org/10.1007/s11432-016-9054-x

## Quasi-consistent fusion navigation algorithm for DSS

• AcceptedMar 15, 2017
• PublishedAug 30, 2017
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### Abstract

A fusion navigation algorithm for the distributed satellites system (DSS) utilizing relative range measurements is proposed in this paper.Based on the quasi-consistent extended Kalman filter (QCEKF), an on-line evaluation of the navigation precision can be provided by the fusion navigation algorithm.In addition, the upper bound for the estimation error obtained from the fusion navigation algorithm is lower than those with any groups of measurements,which indicates that the fusion navigation algorithm can automatically choose the suitable redundant measurements to improve the navigation precision.The simulations show the feasibility and effectiveness of the proposed fusion navigation algorithm.

### Acknowledgment

This work was supported by National Basic Research Program of China (973 Program) (Grant No. 2014CB845303) and National Center for Mathematics and Interdisciplinary Sciences, Chinese Academy of Sciences.

### Supplement

Appendix

Proof of Proposition sect. 3

Since the trackers are observable according to the inertial measurements obtained from the GPS. The observability analysis of the DSS is to investigate the observability of the users based on the relative range measurements. The observability matrix associated with the relative range measurement vector $H(X(t))$ can be denoted by $$D^{\rm relative}=\left[ (C^{\rm relative})^{\rm T}, (C^{\rm relative}A)^{\rm T} , (C^{\rm relative}A^2)^{\rm T}, \ldots \right]^{\rm T}, \tag{31}$$ where $A=\frac{\partial~F}{\partial~X^{\rm~T}},C^{\rm~relative}=\frac{\partial~H}{\partial~X^{\rm~T}}$. The fourth-order submatrix of the observability matrix $D^{\rm~relative}$ which is associated with the measurement $H_l=\rho_{ij}$, $l=\frac{(14-i)(i-1)}{2}+j-i$, can be partitioned as $$\left[ c_{ij}^{\rm T}, (c_{ij}A)^{\rm T},(c_{ij}A^2)^{\rm T}, (c_{ij}A^3)^{\rm T} \right]^{\rm T} =\left[ 0_{4\times 6(i-1)}, P_{ij}, 0_{4\times 6(j-i-1)}, Q_{ij}, 0_{4\times 6(n-j)} \right], \tag{32}$$ where $c_{ij}=\frac{\partial~\rho_{ij}}{\partial~X^{\rm~T}}$, and $$P_{ij}= \left[ \begin{array}{cc} \frac{\partial \rho_{ij}}{\partial r_i^{\rm T}}& 0\\ 0& \frac{\partial \rho_{ij}}{\partial r_i^{\rm T}} \\ \frac{\partial \rho_{ij}}{\partial r_i^{\rm T}} \frac{\partial f_i}{\partial r_i^{\rm T}} & 0\\0 &\frac{\partial \rho_{ij}}{\partial r_i^{\rm T}} \frac{\partial f_i}{\partial r_i^{\rm T}} \\ \end{array} \right], Q_{ij}= \left[ \frac{\partial \rho_{ij}}{\partial r_j^{\rm T}}& 0\\ 0& \frac{\partial \rho_{ij}}{\partial r_j^{\rm T}} \\ \frac{\partial \rho_{ij}}{\partial r_j^{\rm T}} \frac{\partial f_j}{\partial r_j^{\rm T}} & 0\\ 0 &\frac{\partial \rho_{ij}}{\partial r_j^{\rm T}} \frac{\partial f_j}{\partial r_j^{\rm T}} \\ \end{array} \right].$$ Since the $L_2$-norm of the matrix $\frac{\partial~f_i}{\partial~r_i^{\rm~T}}$, which is the maximum of the absolute value of its eigenvalues, satisfies $\|~\frac{\partial~f_i}{\partial~r_i^{\rm~T}}~\|~\leq~3.0824\times~10^{-6}$, the observability of the users can be analyzed by the submatrix $D^{\rm~relative}_2=[~(C^{\rm~relative})^{\rm~T},~~~(C^{\rm~relative}A)^{\rm~T}~]^{\rm~T}$. Therefore, the DSS is observable if and only if $${\rm{rank}} (D^{\rm relative}_2)=6(n-3). \tag{33}$$

Then a necessary condition for the DSS to be observable is that $${\rm{rank}} (D_i)=6, D_i=\left[ C_i^{\rm T}, (C_i A_i)^{\rm T} \right]^{\rm T}, i=4,\ldots,n, \tag{34}$$ where $A_i=\frac{\partial~F_i}{\partial~X_i^{\rm~T}}$, $C_i$ is composed of $\{\frac{\partial~\rho_{ij}}{\partial~X_i^{\rm~T}},~\rho_{ij}\in~\pi_i\}$.

The condition $\left|\pi_i\right|~\geq~3$ in sect. 3.1 means that the number of the relative range measurements associated with $S_i$ is larger than 3, which is a necessary condition for 34. Moreover, $\left|\pi_4~\cup~\pi_5~\cup~\cdots~\cup~\pi_n\right|~\geq~3(n-3)$ indicts that the number of the components of the measurement vector $H(X(t))$ is larger than $3(n-3)$, which is a necessary condition for 33. Therefore, $\left|\pi_i\right|~\geq~3$ and $\left|\pi_4~\cup~\cdots~\cup~\pi_n\right|~\geq~3(n-3)$ are the necessary conditions for the DSS to be observable.

Proof of Proposition TH-obs

The submatrix of the observability matrix $D^{\rm~relative}$ in 31, which is associated with the users, can be denoted by $$D_{\rm user}=\left[ C_{\rm user}^{\rm T}, (C_{\rm user}A_{\rm user})^{\rm T} \right]^{\rm T}, \tag{35}$$ where $A_{\rm~user}=\frac{\partial~F_{\rm~user}}{\partial~X_{\rm~user}^{\rm~T}}$, $C_{\rm~user}=\frac{\partial~H}{\partial~X_{\rm~user}^{\rm~T}}$, $F_{\rm~user}=[F_4^{\rm~T},\ldots,F_n^{\rm~T}]^{\rm~T}$, $X_{\rm~user}=[X_4^{\rm~T},\ldots,X_n^{\rm~T}]^{\rm~T}$. Then, the observability matrix $D_{\rm~user}$ with the relative measurement vector $H_{\rm~nec}(X(t))$ in 9 can be partitioned as $$D_{\rm user}=\left[ D_i=\left[ d_i&0_{3\times3} \\ 0_{3\times3}&d_i \end{array}\right]_{6\times 6}, d_i=\left[ \dfrac{(r_i-r_1)}{\rho_{1i}} , \dfrac{(r_i-r_2)}{\rho_{2i}}, \dfrac{(r_i-r_3)}{\rho_{3i}} \end{array}\right]^{\rm T}, i=4,\ldots,n. \tag{36}$$

If the satellites $S_i$, $i=4,\ldots,n$, are not coplanar with the trackers, $S_1,S_2,S_3$, then ${\rm{rand}}~(D_i)=6$. Therefore, we can obtain ${\rm{rand}}~(D_{\rm~user})=6(n-3)$, which means that the DSS is observable.

Proof of Theorem TH_QCEKF

We use the mathematical induction to prove the quasi-consistency of the algorithm. Assume that at the $k$th sampling point, there is $E[(\hat~X_k-X_k)(\hat~X_k-X_k)^{\rm~T}]\leq~P_k$. Linearizing the process equations in 10 and 14 at $\forall\xi\in~\mathbb{R}^{6n}$ result in $$X_k+\Delta t F(X_k)=\xi+\Delta t F(\xi)+A_\xi( X_k- \xi)+\varphi_1, \hat X_k+ \Delta t F(\hat X_k)=\xi+\Delta t F(\xi)+A_\xi(\hat X_k- \xi)+\varphi_2, \tag{37}$$ where $A_\xi=I_{6n}+\Delta~t~\frac{\partial~F}{\partial~X^{\rm~T}}|_{~X=\xi}$, the linearization errors are $$\varphi_1=\frac{\Delta t^2}{2} (X_k- \xi)^{\rm T} \left.\frac{\partial^2 F}{\partial X^2}\right|_{\xi} (X_k- \xi)+\cdots, \varphi_2=\frac{\Delta t^2}{2} (\hat X_k- \xi)^{\rm T} \left.\frac{\partial^2 F}{\partial X^2}\right|_{\xi} (\hat X_k- \xi)+\cdots. \tag{38}$$ Thus the predicted error at the $(k+1)$th sampling point is \begin{align}\bar X_{k+1}-X_{k+1}=A_\xi(\hat X_k-X_k)+\varphi_k -d_k, \tag{39} \end{align} where $\varphi_k=\varphi_2-\varphi_1$. If $\Delta~Q_k$ satisfies the inequality: \begin{aligned} \Delta Q_k \geq & A_\xi E\left[(\hat X_k-X_k)(\hat X_k-X_k)^{\rm T}\right]A_\xi^{\rm T}+A_\xi E\left[(\hat X_k-X_k)(\varphi_k-d_k)^{\rm T}\right]+E\left[(\varphi_k-d_k)(\hat X_k-X_k)^{\rm T}\right]A_\xi^{\rm T} \\ &+E\left[(\varphi_k-d_k)(\varphi_k-d_k)^{\rm T}\right]-A_kE\left[(\hat X_k-X_k)(\hat X_k-X_k)^{\rm T}\right] A_k^{\rm T}, \\ \end{aligned} \tag{40} then the mean square error of the predicted value can be obtained as follows: \begin{aligned} E\left[(\bar X_{k+1}-X_{k+1})(\bar X_{k+1}-X_{k+1})^{\rm T}\right] =&A_\xi E\left[(\hat X_k-X_k)(\hat X_k-X_k)^{\rm T}\right]A_\xi^{\rm T}+A_\xi E\left[(\hat X_k-X_k)(\varphi_k-d_k)^{\rm T}\right] \\ &+E\left[(\varphi_k-d_k)(\hat X_k-X_k)^{\rm T}\right]A_\xi^{\rm T}+E\left[(\varphi_k-d_k)(\varphi_k-d_k)^{\rm T}\right] \\ \leq& A_k P_{k} A_k^{\rm T} + \Delta Q_k =\bar P_{k+1}, \end{aligned} \tag{41} which indicates that the prediction at the $(k+1)$th sampling point is quasi-consistent.

Similarly, the linearizations of the measurement functions $G(~X_{k+1})$ and $G(\bar~X_{k+1})$ at $\forall\eta~\in~\mathbb{R}^{6n}$ are $$G( X_{k+1})=G(\eta)+C_\eta(X_{k+1}-\eta)+\psi_1, G(\bar X_{k+1})=G(\eta)+C_\eta(\bar X_{k+1}-\eta)+\psi_2, \tag{42}$$ where $C_\eta=\frac{\partial~G}{\partial~X^{\rm~T}}|_{X=\eta}$, the linearization errors are $$\psi_1=\frac{1}{2} (X_{k+1}-\eta)^{\rm T} \left.\frac{\partial^2 G}{\partial X^2}\right|_{\eta} (X_{k+1}-\eta)+\cdots, \psi_2=\frac{1}{2} (\bar X_{k+1}-\eta)^{\rm T} \left.\frac{\partial^2 G}{\partial X^2}\right|_{\eta} (\bar X_{k+1}-\eta)+\cdots. \\ \tag{43}$$ Hence, the filtering error at the $(k+1)$th sampling point is $$\hat X_{k+1}-X_{k+1}=(I-K_kC_\eta)(\bar X_{k+1}-X_{k+1})-K_k\psi_{k+1}+K_k w_{k+1}, \tag{44}$$ where $\psi_{k+1}=\psi_2-\psi_1$. If $\Delta~R_{k+1}$ satisfies the inequality: \begin{aligned} \Delta R_{k+1}\geq &E\left[(I-K_kC_\eta)(\bar X_{k+1}-X_{k+1})(\bar X_{k+1}-X_{k+1})^{\rm T}(I-K_kC_\eta)^{\rm T}\right] -E\left[(I-K_k C_\eta)(\bar X_{k+1}-X_{k+1}) \psi_{k+1}^{\rm T} K_k^{\rm T}\right] \\ &-E\left[K_k\psi_{k+1}(\bar X_{k+1}-X_{k+1})^{\rm T}(I-K_kC_\eta)^{\rm T}\right] +E\left[K_k\psi _{k+1} \psi_{k+1}^{\rm T} K_k^{\rm T}\right] +E\left[K_kw_{k+1}w_{k+1}^{\rm T} K_k^{\rm T}\right] \\ &-K_kE\left[w_{k+1}w_{k+1}^{\rm T}\right] K_k^{\rm T} -(I-K_kC_{k+1})E\left[(\bar X_{k+1}-X_{k+1})(\bar X_{k+1}-X_{k+1})^{\rm T}\right](I-K_kC_{k+1})^{\rm T}, \end{aligned} \tag{45} then the mean square error of the filtering value satisfies \begin{aligned} &E\left[(\hat X_{k+1}-X_{k+1})(\hat X_{k+1}-X_{k+1})^{\rm T}\right] \\ & =E\left[(I-K_kC_\eta)(\bar X_{k+1}-X_{k+1})(\bar X_{k+1}-X_{k+1})^{\rm T}(I-K_kC_\eta)^{\rm T}\right] -E\left[(I-K_kC_\eta)(\bar X_{k+1}-X_{k+1}) \psi_{k+1}^{\rm T} K_k^{\rm T}\right] \\ & -E\left[K_k\psi_{k+1}(\bar X_{k+1}-X_{k+1})^{\rm T}(I-K_kC_\eta)^{\rm T}\right] +E\left[K_k\psi _{k+1} \psi_{k+1}^{\rm T} K_k^{\rm T}\right]+E\left[K_kw_{k+1}w_{k+1}^{\rm T} K_k^{\rm T}\right] \\ & \leq (I-K_kC_{k+1})\bar P_{k+1}(I-K_kC_{k+1})^{\rm T}+K_kR_{k+1}K_k^{\rm T}+ \Delta R_{k+1} =P_{k+1}, \end{aligned} \tag{46} which means that the filtering result at the $(k+1)$th sampling point is quasi-consistent. Therefore, when $\Delta~Q_k$ and $\Delta~R_{k+1}$ satisfy the inequalities 40 and 45, the results of the prediction, the filtering, as well as the estimation are all quasi-consistent.

Next, we prove $\Delta~Q_k$ and $\Delta~R_{k+1}$ in 18 satisfy 40 and 45. Set $\Delta~Q_k$ and $\Delta~R_{k+1}$ as follows: \begin{aligned} &\Delta Q_k = (1+\alpha_{k})A_\xi P_k A_\xi^{\rm T} +\left(1+\frac{1}{\alpha_{k}}\right)\Delta Q_{\varphi,k} -A_k P_k A_k^{\rm T}, \\ &\Delta R_{k+1} = (1+\beta_{k}) (I-K_kC_\eta)\bar P_{k+1}(I-K_kC_\eta)^{\rm T} +\left(1+\frac{1}{\beta_{k}}\right) K_k \Delta R_{\psi,k+1} K_k^{\rm T} -(I-K_kC_{k+1})\bar P_{k+1}(I-K_kC_{k+1})^{\rm T}, \end{aligned} \tag{47} where \begin{aligned} &\alpha_{k}=\sqrt{\frac{\|\Delta Q_{\varphi,k}\|}{\|A_\xi P_k A_\xi^{\rm T}\|}}, \Delta Q_{\varphi,k}=6n {\rm{diag}} \left( \left[ \delta\varphi_{k,1}^2,\ldots, \delta\varphi_{k,6n}^2 \right] \right)+2d_k d_k^{\rm T}, \\ &\beta_{k}=\sqrt{\frac{\| K_k \Delta R_{\psi,k+1} K_k^{\rm T}\|}{\| (I-K_kC_\eta)\bar P_{k+1}(I- K_kC_\eta)^{\rm T} \|}}, \Delta R_{\psi,k+1}={m_{\rm nec}} {\rm{diag}} \left( \left[ \delta\psi_{k+1,1}^2,\ldots, \delta \psi_{k+1,{m_{\rm nec}}+9}^2 \right] \right), \\ \end{aligned} \tag{48} and $\delta~\varphi_{k,i}^2$, $\delta~\psi_{k+1,j}^2$ are the upper bounds of the linearization errors’ variances, that is $$\delta \varphi_{k,i}^2 \geq E(\varphi_{k,i}^2), \delta \psi_{k+1,j}^2 \geq E(\psi_{k+1,j}^2), i=1,\ldots,6n, j=1,\ldots,{m_{\rm nec}}+9. \tag{49}$$ Ignoring the effect of the third-order terms of the Taylor series expansions in 38 and 43, the linearization errors are \begin{aligned} &\varphi_{k,i}= \begin{cases} 0, i=6j-5,\ldots,6j-3, j=1,\ldots,n. \cr \frac{\Delta t}{2} (\hat X_k-X_k)^{\rm T} \left.\frac{\partial^2 f_j(l)}{\partial X^2}\right|_{\xi} (\hat X_k-X_k) + \Delta t(\hat X_k-X_k)^{\rm T} \left.\frac{\partial^2 f_j(l)}{\partial X^2}\right|_{\xi} (X_k-\xi), i=6j-3+l, l=1,2,3, j=1,\ldots,n. \end{cases} \\ &\psi_{k+1,j}= \begin{cases} 0, j=1,\ldots,9. \cr \frac{1}{2} (\bar X_{k+1}-X_{k+1})^{\rm T} \left.\frac{\partial^2 G(j)}{\partial X^2}\right|_{\eta} (\bar X_{k+1}-X_{k+1})+(\bar X_{k+1}-X_{k+1})^{\rm T} \left.\frac{\partial^2 G(j)}{\partial X^2}\right|_{\eta} (X_{k+1}-\eta), j=10,\ldots,{m_{\rm nec}}+9. \end{cases} \\ \end{aligned} \tag{50} Since the randomness of the gain matrix $K_k$ can be neglected for the navigation of the DSS studied, and the following inequality holds: $$AB^{\rm T}+BA^{\rm T}\leq \theta AA^{\rm T}+\frac{1}{\theta} BB^{\rm T}, \forall \theta >0, \tag{51}$$ where $A,B\in~\mathbb{R}^{n_1\times~n_2}$, $n_1,n_2\in~N_+$, $\Delta~Q_k$ and $\Delta~R_{k+1}$ in 47 with the constraints of $\delta~\varphi_{k,i}^2$ and $\delta~\psi_{k+1,j}^2$ in 49 satisfy the inequalities 40 and 45.

To be specific, if we choose $\xi=\hat~X_k$ and $\eta=\bar~X_{k+1}$, then $\Delta~Q_k$ and $\Delta~R_{k+1}$ in 18 with $\delta~\varphi_{k,i}^2$ and $\delta~\psi_{k+1,j}^2$ being chosen as in 19 satisfy 40 and 45. Therefore, the results of the prediction, the filtering and the estimation of the navigation algorithm 1417 are quasi-consistent.

Proof of Theorem TH_fus

We prove that for $\forall~\omega_j\in\Omega$, the corresponding estimate error satisfies the following result: $$E(\hat X_{{\rm fus},k}(j)-X_k(j))^2 \leq U_{{\rm fus},k}(j), \forall k, \forall j\in\{1,2,\ldots,6n\}. \tag{52}$$ Let $e_f=\hat~X_{{\rm~fus},k}(j)-X_k(j)$, $u_f=U_{{\rm~fus},k}(j)$, $e_i=\tilde~X_{i,k}(j)-X_k(j)$, $p_i=\tilde~P_{i,k}(j,j)$ and $\alpha_i=\omega_j(i)$, $i=1,\ldots,N$, $\forall~k$, $\forall~j$. Then the equivalent result of 52 is $E(e_f^2)\leq~u_f$ for $N=2^{m_{\rm~red}}$, which will be proved by the mathematical induction.

Firstly, we prove the inequality $E(e_f^2)\leq~u_f$ for $N=2$. From 26, we can obtain $e_f^2=u_f^2~(~\frac{\alpha_1}{p_1}~e_1~+~\frac{\alpha_2}{p_2}~e_2~)^2~$. Hence, it needs to be proved that $$\frac{1}{u_f} \geq E\left[ \left( \frac{\alpha_1}{p_1} e_1 + \frac{\alpha_2}{p_2} e_2 \right)^2\right]. \tag{53}$$ According to the quasi-consistency of the estimation $\tilde~X_{i,k}$, i.e., $E(e_i^2)~\leq~p_i$, there is $$\frac{1}{u_f}=\frac{\alpha_1}{p_1} + \frac{\alpha_2}{p_2} \geq \frac{\alpha_1}{p_1^2} E(e_1^2) + \frac{\alpha_2}{p_2^2} E(e_2^2). \tag{54}$$ Meanwhile, considering the relationship $\alpha_1+\alpha_2=1$, we can obtain $$\frac{\alpha_1}{p_1^2} e_1^2 + \frac{\alpha_2}{p_2^2} e_2^2 - \left( \frac{\alpha_1}{p_1} e_1 + \frac{\alpha_2}{p_2} e_2 \right)^2 =\alpha_1(1-\alpha_1) \left( \frac{\alpha_1}{p_1} e_1 - \frac{\alpha_2}{p_2} e_2 \right)^2 \geq 0. \tag{55}$$ Therefore, from 54 and 55, we can get the inequality 53, which means that 52 holds for $N=2$.

Then, assume for $N={N_0}$, Eq. 52 holds. For $N={N_0}+1$, from 26, there is $$\frac{1}{u_f}=\frac{\alpha_1}{p_1} +\cdots+ \frac{\alpha_{{N_0}+1}}{p_{{N_0}+1}}, e_f=u_f \left( \frac{\alpha_1}{p_1} e_1 +\cdots + \frac{\alpha_{{N_0}+1}}{p_{{N_0}+1}} e_{{N_0}+1} \right) . \tag{56}$$ Let $$\frac{1}{\bar{p}_1}=\frac{\beta_1}{p_1} +\cdots+ \frac{\beta_{N_0}}{p_{N_0}}, \bar{e}_1=\bar{p}_1 \left( \frac{\beta_1}{p_1} e_1 +\cdots+ \frac{\beta_{N_0}}{p_{N_0}} e_{N_0} \right), \tag{57}$$ where $\beta_i=\frac{\alpha_i}{\alpha_1+\cdots+\alpha_{N_0}},~i=1,2,\ldots,N_0$, then from the assumption for $N={N_0}$, we can get $E(\bar{e}_1^2)\leq~\bar{p}_1$. Similarly, let $$\frac{1}{\bar{p}_2}=\frac{\gamma_1}{\bar{p}_1} + \frac{\gamma_2}{p_{{N_0}+1}}, \bar{e}_2=\bar{p}_2 \left( \frac{\gamma_1}{\bar{p}_1} \bar{e}_1+ \frac{\gamma_2}{p_{{N_0}+1}} e_{{N_0}+1}\right), \tag{58}$$ where $\gamma_1=1-\alpha_{{N_0}+1},\gamma_2=\alpha_{{N_0}+1}$, then from the result of $N=2$, there is $E(\bar{e}_2^2)\leq~\bar{p}_2$. Moreover, we can obtain \begin{aligned} \frac{1}{\bar{p}_2}&=(1-\alpha_{{N_0}+1})\left( \frac{\beta_1}{p_1} +\cdots+ \frac{\beta_{N_0}}{p_{N_0}} \right) + \frac{\alpha_{{N_0}+1}}{p_{{N_0}+1}} =\frac{\alpha_1}{p_1} +\cdots + \frac{\alpha_{{N_0}+1}}{p_{{N_0}+1}}=\frac{1}{u_f}, \end{aligned} \tag{59} and \begin{aligned} \bar{e}_2=&\bar{p}_2 \left[ (1-\alpha_{{N_0}+1}) \left( \frac{\beta_1}{p_1} e_1 +\cdots+ \frac{\beta_{N_0}}{p_{N_0}} e_{N_0} \right) + \frac{\alpha_{{N_0}+1}}{p_{{N_0}+1}} e_{{N_0}+1}\right] =u_f \left( \frac{\alpha_1}{p_1} e_1 + \cdots + \frac{\alpha_{{N_0}+1}}{p_{{N_0}+1}} e_{{N_0}+1} \right)=e_f, \end{aligned} \tag{60} which means that Eq. 52 holds for $N=N_0+1$. Therefore, according to the mathematical induction, Eq. 52 holds for $\forall~N\in~\mathbb{R}$ and $\forall~\omega_j\in\Omega$.

Specifically, for $\omega_j=\omega_j^*$ and $N=2^{m_{\rm~red}}$, the corresponding results in property (i) can be obtained.

The results in property (ii) are obvious from the fusion scheme in 27 and 28.

### References

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[27] Simon D. Optimal State Estimation — Kalman, $H_\infty$, and Nonlinear Approaches. New Jersey: John Wiley $\&$ Sons Inc., 2006. Google Scholar

• Figure 1

Relative orbits of $S_1,S_2,\ldots,S_7$ in the virtual reference satellites LVLH frame.

• Figure 2

The mean square error and the diagonal elements of $P^{\rm~EKF}_k$ from the EKF based navigation algorithm of $S_4$. (a) Position errors; (b) velocity errors.

• Figure 3

The mean square error and the diagonal elements of $\tilde~P_k$ from the QCEKF based navigation algorithm of $S_4$. (a) Position errors; (b) velocity errors.

• Figure 6

The mean square error and the diagonal elements of $\tilde~P_k$ from the QCEKF based navigation algorithm of $S_7$. (a) Position errors; (b) velocity errors.

• Figure 7

The mean square error and the components of $U_{{\rm~fus},k}^*$ from the fusion algorithm of $S_4$. (a) Position errors; protectłinebreak (b) velocity errors.

• Figure 10

The mean square error and the components of $U_{{\rm~fus},k}^*$ from the fusion algorithm of $S_7$. (a) Position errors; (b) velocity errors.

• Figure 11

Results of the fusion algorithm and the algorithm without fusion of $S_4$. (a) Position errors; (b) velocity errors.

• Figure 12

Results of the fusion algorithm and the centralization algorithm of $S_4$. (a) Position errors; (b) velocity errors.

• Table 1   The choices of the redundant measurements for the fusion navigation algorithm
 $t_1=161.08$ s $t_4=1928.66$ s $\rho_{45}$ $\rho_{46}$ $\rho_{47}$ $\rho_{56}$ $\rho_{57}$ $\rho_{67}$ $\rho_{45}$ $\rho_{46}$ $\rho_{47}$ $\rho_{56}$ $\rho_{57}$ $\rho_{67}$ $X_4(1)$ $\star$ $\star$ $X_4(1)$ $\star$ $\star$ $\star$ $X_4(2)$ $\star$ $\star$ $\star$ $X_4(2)$ $\star$ $\star$ $X_4(3)$ $\star$ $\star$ $X_4(3)$ $\star$ $X_4(4)$ $\star$ $\star$ $X_4(4)$ $\star$ $\star$ $\star$ $X_4(5)$ $\star$ $X_4(5)$ $\star$ $\star$ $\star$ $X_4(6)$ $\star$ $X_4(6)$ $\star$ $\star$ $\star$ $X_5(1)$ $\star$ $X_5(1)$ $\star$ $\star$ $\star$ $\star$ $\star$ $\star$ $X_5(2)$ $\star$ $\star$ $X_5(2)$ $\star$ $\star$ $\star$ $\star$ $\star$ $\star$ $X_5(3)$ $\star$ $\star$ $\star$ $X_5(3)$ $\star$ $X_5(4)$ $\star$ $\star$ $\star$ $X_5(4)$ $\star$ $X_5(5)$ $\star$ $\star$ $X_5(5)$ $\star$ $\star$ $X_5(6)$ $\star$ $\star$ $X_5(6)$ $\star$ $\star$ $\star$ $X_6(1)$ $\star$ $\star$ $X_6(1)$ $\star$ $X_6(2)$ $\star$ $\star$ $\star$ $X_6(2)$ $\star$ $X_6(3)$ $\star$ $\star$ $\star$ $X_6(3)$ $\star$ $\star$ $\star$ $X_6(4)$ $\star$ $\star$ $\star$ $X_6(4)$ $\star$ $\star$ $\star$ $X_6(5)$ $\star$ $\star$ $X_6(5)$ $\star$ $\star$ $\star$ $X_6(6)$ $\star$ $\star$ $X_6(6)$ $\star$ $\star$ $\star$ $X_7(1)$ $\star$ $\star$ $X_7(1)$ $\star$ $\star$ $\star$ $X_7(2)$ $\star$ $\star$ $X_7(2)$ $\star$ $\star$ $X_7(3)$ $\star$ $\star$ $\star$ $X_7(3)$ $\star$ $\star$ $\star$ $X_7(4)$ $\star$ $\star$ $\star$ $X_7(4)$ $\star$ $\star$ $\star$ $X_7(5)$ $\star$ $X_7(5)$ $\star$ $\star$ $\star$ $X_7(6)$ $\star$ $\star$ $X_7(6)$ $\star$ $\star$ $\star$

Citations

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text:
# What's a convenient/appropriate way to override my document class' \paragraph style?

The document class I'm using (ieeetran) typesets paragraphs as in the following:

-------
[space here] a) The Paragraph Title In Italics:[space here]The quick brown fox jumped
over the lazy dog. The quick brown fox jumped over the lazy dog. The quick brown fox
jumped over the lazy dog.
-------

(actually, it doesn't matter exactly how they're typeset, since...)

I want paragraphs like in the article document class - with no indentation before the title, with boldface instead of italics, and with no numbering, thank you very much!

Besides creating a modified document class, what's a reasonable way to achieve this?

• When submitting to IEEE, the chance is high that they revert your changes. A journal should have a consistent look, throughout. – Johannes_B Mar 28 '15 at 13:36
• @Johannes_B: But I don't want their 'structural' paragraph, I want an 'unstructued paragraph' – einpoklum Mar 28 '15 at 13:52
• A theoretical answer can be given in a comment, copy the definition of paragraphs from the cls and redefine it in the document preamble. Or use package titlesec to do the redefinition with a nice interface. With a MWE, a ready to use solution can be presented. – Johannes_B Mar 28 '15 at 14:05
• The most reasonable way, barring other information, is: simply use article if you want it to look like article. – Paul Gessler Mar 28 '15 at 14:28
• If you aren't submitting to IEEE, you are better off using a standard class (e.g. article). If you are submitting to IEEE, modifying the paragraph style is certainly pointless: journals don't allow authors to determine the way their articles are typeset. Content, sure. Format is the journal's business and they impose a consistent style for all articles. – cfr Mar 29 '15 at 3:37

\makeatletter


text:
# Asymmetric Cournot competition with linear demand

Posted on Jul 19, 2020

Consider the following setting: $n$ firms simultaneously choose their individual output level $q_i$. The market clearing price is determined by the total sum of outputs and the firms aim to maximize their profit which is given by revenue minus total costs. (Think of the market price being determined through an auction.) What is the Nash equilibrium of outputs if marginal costs are not necessarily equal?
In this note, I will present a matrix formulation for this problem. This note shows how matrices can be used to find solutions to otherwise tenacious optimization problems in economics.

Let aggregate demand be given by

$$p^{D}(\pmb{q}) = A-b\sum_{i} q_{i},$$

where $A, b > 0$ and $\pmb{q}$ is a $n\times1$ vector of quantities. Each firm $i$ maximizes its profit

$$\pi_{i} = \underbrace{p^{D}(\pmb{q}) q_{i}}_{\text{revenue}} - \underbrace{c_{i}q_{i}}_{\text{total costs}},$$

where $c_i$ is the firm's marginal cost of production. There are no fixed costs of production.

Each firm can only choose $q_i$, but it has to take all the other quantities as given. The simultaneous maximum of profits is the Nash equilibrium of the Cournot oligopoly. This gives rise to the following set of $n$ first order conditions:

\begin{align*} 0 &= \frac{\partial p^{D}(\pmb{q})}{\partial q_{i}}q_{i} + p^{D}(\pmb{q}) - c_{i} \hspace{0.5cm}&\forall i\\ &= A-b\sum_{i} q_{i} - b q_{i} - c_{i} &\forall i \end{align*}

Let us now investigate these first order conditions more closely. Each firm's quantity, $q_i$, appears twice in its first order condition, but only once in all of the other firm's FOCs. Thus, the full set of FOCs is as follows:

\begin{align*} \begin{array}{rrrrrrrrrrrr} 0 &= A&-&2bq_{1}&-&bq_{2}&-&\cdots&-&bq_{n} &-&c_{1}\\ 0 &= A&-&bq_{1}&-&2bq_{2}&-&\cdots&-&bq_{n} &-&c_{2}\\ &&&&\vdots\\ 0 &= A&-&bq_{1}&-&bq_{2}&-&\cdots&-&2bq_{n} &-&c_{n} \end{array} \end{align*}

or, equivalently,

\begin{align*} \begin{array}{rcrcrcrrrr} b(2q_{1}&+&q_{2}&+&\cdots&+&q_{n}) &= A-c_{1}\\ b(q_{1}&+&2q_{2}&+&\cdots&+&q_{n}) &= A-c_{2}\\ &&&&\vdots\\ b(q_{1}&+&q_{2}&+&\cdots&+&2q_{n}) &= A-c_{n} \end{array} \end{align*}

By recognizing the sums and products, we can assemble these FOCs in matrix form:

\begin{align*} \underbrace{b \begin{pmatrix} 2 & 1 & \cdots & 1\\ 1 & 2 & \ddots & \vdots\\ \vdots & \ddots & \ddots & 1\\ 1 & \cdots & 1 & 2 \end{pmatrix}}_{\pmb{M}} \underbrace{\begin{bmatrix} q_{1}\\ q_{2}\\ \vdots\\ q_{n}\\ \end{bmatrix}}_{\pmb{q}} &= \underbrace{\begin{bmatrix} A-c_{1}\\ A-c_{2}\\ \vdots\\ A-c_{n}\\ \end{bmatrix}}_{\pmb{R}} \end{align*}

Note that our aim is to find $\pmb{q}$. Now that our FOCs can be expressed in matrix form,

$$\pmb{M} \pmb{q} = \pmb{R},$$

we may use matrix operations to solve for $\pmb{q}$ (which is then the Cournot-Nash equilibrium, denoted by $\pmb{q}^C$). What immediately comes to mind is to pre-multiply the inverse of $\pmb{M}$. It would then follow that

\begin{align*} \pmb{M}^{-1}\pmb{M}\pmb{q} &= \pmb{M}^{-1}\pmb{R},\\ \pmb{q}^{C} &= \pmb{M}^{-1}\pmb{R}. \end{align*} Using the fact that the inverse of a scalar times an invertible matrix is just the reciprocal of the scalar times the inverse of the matrix, we have \begin{align*} \pmb{M}^{-1} &= \dfrac{1}{b} \begin{pmatrix} 2 & 1 & \cdots & 1\\ 1 & 2 & \ddots & \vdots\\ \vdots & \ddots & \ddots & 1\\ 1 & \cdots & 1 & 2 \end{pmatrix}^{-1}. \end{align*}

It can be shown that

\begin{align*} \pmb{M}^{-1} &= \frac{1}{b(n+1)}\begin{pmatrix} n & -1 & \cdots & -1\\ -1 & n & \ddots & \vdots\\ \vdots & \ddots & \ddots & -1\\ -1 & \cdots & -1 & n \end{pmatrix} \end{align*}

is the required inverse. (I found this matrix by thinking about the structure of $\pmb{M}$ and experimenting a little bit. However, it is trivial to directly prove that the inverse is correct, and therefore the proof is left to you, the reader.)

Now that we have found $\pmb{M}^{-1}$, the Cournot-Nash equilibrium follows as

\begin{align*} \pmb{q}^{C} &= \pmb{M}^{-1}\pmb{R},\\ &= \frac{1}{b(n+1)}\begin{pmatrix} n & -1 & \cdots & -1\\ -1 & n & \ddots & \vdots\\ \vdots & \ddots & \ddots & -1\\ -1 & \cdots & -1 & n \end{pmatrix} \begin{bmatrix} A-c_{1}\\ A-c_{2}\\ \vdots\\ A-c_{n}\\ \end{bmatrix}\\ &= \frac{1}{b(n+1)}\begin{bmatrix} n(A-c_{1}) - (A-c_{2}) - (A-c_{3}) - \cdots - (A-c_{n})\\ n(A-c_{2}) - (A-c_{1}) - (A-c_{3}) - \cdots - (A-c_{n})\\ \vdots\\ n(A-c_{n}) - (A-c_{1}) - (A-c_{2}) - \cdots - (A-c_{n-1})\\ \end{bmatrix}\\ &= \frac{1}{b(n+1)}\begin{bmatrix} n(A-c_{1}) - (n-1)A + \sum_{j \neq 1} c_{j}\\ n(A-c_{2}) - (n-1)A + \sum_{j \neq 2} c_{j}\\ \vdots\\ n(A-c_{n}) - (n-1)A + \sum_{j \neq n} c_{j}\\ \end{bmatrix}\\ &= \frac{1}{b(n+1)}\begin{bmatrix} A -nc_{1} + \sum_{j \neq 1} c_{j}\\ A -nc_{2} + \sum_{j \neq 2} c_{j}\\ \vdots\\ A -nc_{n} + \sum_{j \neq n} c_{j}\\ \end{bmatrix}\\ &= \frac{1}{b(n+1)}\begin{bmatrix} A -(n+1)c_{1} + \sum_{i} c_{i}\\ A -(n+1)c_{2} + \sum_{i} c_{i}\\ \vdots\\ A -(n+1)c_{n} + \sum_{i} c_{i}\\ \end{bmatrix}\\ &= \frac{1}{b} \begin{bmatrix} \frac{A + \sum_{i} c_{i}}{n+1} - c_{1}\\ \frac{A + \sum_{i} c_{i}}{n+1} - c_{2}\\ \vdots\\ \frac{A + \sum_{i} c_{i}}{n+1} - c_{n}\\ \end{bmatrix} \end{align*}

Thus, we find that for each firm $i = 1, \dots, n$ it is optimal to choose the following output level:

$$q_i = \frac{1}{b} \left( \frac{A + \sum_{j} c_{j}}{n+1} - c_i \right),$$

which is equal to the (perhaps well known) solution to the asymmetric Cournot oligopoly.

Note that the firm will only produce a positive amount if $\frac{A + \sum_{i} c_{i}}{n+1} > c_i$; we will assume that this is the case. The firm may choose not to produce anything if its marginal cost is too high. But then again, the firm will have to rely on its competitors to follow the Nash equilibrium, as otherwise the price may be pushed so far down that losses are incurred after the good is auctioned off. But the other firms will follow the Nash equilibrium, as not doing so would not be profit-maximizing.

For example, for $n = 2$ and $c \equiv c_1 = c_2$, this reduces to

$$q_i = \frac{1}{b} \left( \frac{A + 2 c}{3} - c \right) = \frac{A-c}{3b},$$

which is known to be the solution for the symmetric Cournot duopoly. It is, I am willing to posit, one of the few age-old adages of economics, known since time immemorial.

We can find total output by summing over all $q_i$,

\begin{align*} Q^C &= \sum_i q_i\\ &= \sum_i \frac{1}{b} \left( \frac{A + \sum_{j} c_{j}}{n+1} - c_i \right)\\ &= \frac{1}{b} \sum_i \left( \frac{A + \sum_{j} c_{j}}{n+1} - c_i \right)\\ &= \frac{1}{b} \left(\sum_i \frac{A + \sum_{j} c_{j}}{n+1} - \sum_i c_i \right)\\ &= \frac{1}{b} \left(\frac{n}{n+1} \left( A + \sum_{j} c_{j} \right) - \sum_i c_i \right)\\ &= \frac{1}{b} \left(\frac{n A - \sum_{j} c_{j}}{n+1}\right). \end{align*}

The market clearing price is found by

\begin{align*} p^{D}(\pmb{q}^C) &= A-b\sum_{i} q_{i}^C\\ &= A - b Q^C\\ &= A-\frac{n A - \sum_{j} c_{j}}{n+1}\\ &= \frac{A + \sum_{i} c_{i}}{n+1}, \end{align*}

which for large $n$ has a very nice approximate interpretation as $p \approx a + \overline{c}$ with $a = A/n$ and $\overline{c} = {\sum_{j} c_{j}}/n$, i.e. average marginal cost.


text:
## org.extex.unit.tex.math.fraction Class Over

java.lang.Object
org.extex.interpreter.type.AbstractCode
org.extex.unit.tex.math.AbstractMathCode
org.extex.unit.tex.math.fraction.Over

All Implemented Interfaces:
java.io.Serializable, Code

public class Overextends AbstractMathCode

This class provides an implementation for the primitive \over.

### The Math Primitive \over

The math primitive \over arranges that the material in the math group before it is typeset above the material after the primitive. The two parts are separated by a line of the default width which is taken from the font parameters. If the width is less than 0pt then no rule is drawn but the given height is left blank.

If several primitives of type \above, \abovewithdelims, \atop, \atopwithdelims, \over, or \overwithdelims are encountered in the same math group then the result is ambiguous and an error is raised.

If the primitive is used outside of math mode then an error is raised.

#### Syntax

The formal description of this primitive is the following:
    ⟨over⟩
→ ⟨math material⟩ \over ⟨math material⟩

#### Examples

    {a \over b}

Version:
$Revision: 4732$
Author:
Gerd Neugebauer
Serialized Form

Field Summary
protected static long serialVersionUID
The constant serialVersionUID contains the id for serialization.

Constructor Summary
Over(CodeToken token)
Creates a new object.

Method Summary
 void execute(Flags prefix, Context context, TokenSource source, Typesetter typesetter)
This method takes the first token and executes it. The result is placed on the stack. This operation might have side effects. To execute a token it might be necessary to consume further tokens.

Methods inherited from class org.extex.unit.tex.math.AbstractMathCode
getListMaker

Methods inherited from class org.extex.interpreter.type.AbstractCode
getLocalizer, getName, getToken, isIf, isOuter, readResolve, toString, toText, toText

Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, wait, wait, wait

Field Detail

### serialVersionUID

protected static final long serialVersionUID
The constant serialVersionUID contains the id for serialization.

Constant Field Values
Constructor Detail

### Over

public Over(CodeToken token)
Creates a new object.

Parameters:
token - the initial token for the primitive
Method Detail

### execute

public void execute(Flags prefix,
Context context,
TokenSource source,
Typesetter typesetter)
throws HelpingException,
ConfigurationException,
TypesetterException
This method takes the first token and executes it. The result is placed on the stack. This operation might have side effects. To execute a token it might be necessary to consume further tokens.

Specified by:
execute in interface Code
Overrides:
execute in class AbstractCode
Parameters:
prefix - the prefix controlling the execution
context - the interpreter context
source - the token source
typesetter - the typesetter
Throws:
HelpingException - in case of an error
ConfigurationException - in case of an configuration error
TypesetterException - in case of an error in the typesetter
AbstractCode.execute( org.extex.interpreter.Flags, org.extex.interpreter.context.Context, org.extex.interpreter.TokenSource, org.extex.typesetter.Typesetter)


text:
# Gaussian Elimination

## Main Question or Discussion Point

Solve the following system of equations using Gaussian elimination:
3w+10x-2y+3z=55
w+12x-11y-z= 69
-3w-6x-5y-10z=-47
-3w+9x+5y+4z=-1

Here's what I did so far:
swap R1 and R2 (row 1 and row 2)
-3R1+R2 (multiply row 1 by -3 and add row 2 to that product)
3R1+R3
3R1+R4

Now, after that I think I may have made a mistake because I eventually get some very odd fractions that could not possibly be correct. Can somebody tell me what the steps are???


text:
## Hooke's Law

Do you see where it says

m(dv/dt)v = (d/dt)(.5mv^2)

If it's an integration which I don't think it is then I would think it should be

(d/dt)((mv^3)/3) because you're taking the two v's and adding an additional power. I don't think it is an integration because it says right there in the book that you can't integrate, so if it's now then where does the 1/2 come from?
 Hi g.lemaitre It is indeed an integration and it is correct just derive (1/2mv²) and you will see that you get back to mvv' or mv dv/dt Cheers...
 It's a case of the chain rule: $\frac{d}{dt}\left(\frac{1}{2}mv^2\right)\:=\:\frac{d}{dv}\left(\frac{1} {2}mv^2\right)\:\times\frac{dv}{dt}\:=\:mv\frac{dv}{dt}$. Your textbook isn't very clear. You could integrate the left hand side of the equation wrt t, before multiplying by v. It's the right hand side that you can't integrate wrt to t until you've multiplied by v. That's why you need to multiply through by v. Then the left hand side integrates (wrt t) to give $\frac{1}{2}mv^2$. The right hand integrates to give $-\frac{1}{2}kx^2$ + constant. Try differentiating this wrt t using the chain rule!

## Hooke's Law

The spring constant 'k' is not a measure of the STRENGTH of the spring.
It is a measure of the STIFFNESS.... units are N/m
The strength is measured by ultimate tensile stress
edit....is this question about simple harmonic motion?
or hookes law (elasticity)


text:
Exponential Regression Model

I need to model my data ($(x,y)$ pairs) using the following exponential function:

$$f(x) = \exp((x + a)/b) - c$$

So, I need to find $a, b, c$ coefficients that are the best fit for my data. What is the algorithm that can solve this problem?

-
It depends on the probability model for the errors. Typically the observed value of $f(x)$ differs from the true $f(x)$; otherwise all the data points would fit the curve perfectly. You'd have a probability density $g(y)$ depending on $a$, $b$, $c$ and possible an error variance $\sigma^2$ or the like. Plug the data points into that function, and then view it as a function of $a,b,c,\sigma$. Find the values of $a,b,c,\sigma$ that maximize that function. – Michael Hardy Sep 12 '12 at 16:23

Assume that somehow you are able to find the value of $c$, and only $a$ and $b$ have to be found. Let $y=\log(f(x)+c)$. Then $y=a\,x+b/a$, so that simple linear regression will give the value of $a$ and $b$.
But how do we find $c$? One possibility is to try different values of $c$ until you get the best fit. Another one, is to observe that $$-c=\begin{cases}\lim_{x\to-\infty}f(x) & \text{if }a>0,\\ \lim_{x\to+\infty}f(x) & \text{if }a<0\end{cases}.$$ If the data decrease as $x$ increases, you can guess $c$ looking at your data for large values of $x$. If the data increase with $x$, look at large negative values of $x$.
I think if $|a|$ is relatively larger than $c$ linear term will be dominant and the solution that you suggested will be a good one. The other way around is another thing. – Seyhmus Güngören Sep 13 '12 at 13:49


text:
# Greedy Algorithms

(Algorithms chapter 5)

Greedy strategies take a very natural approach to solve a problem. Given the choices that we have made so far, make the choice that provides the most benefit going forward. We can think of a greedy algorithm as taking the choice that looks locally optimal. However, generally our goal is to find a global optimum. When do these locally optimal moves produce a global optimum? We start by seeing some examples.

Def 1. (Matching) Given a graph $G=(V, E)$, a subset of edges $M \subset E$ is a matching if every two edges in $E$ has disjoint endpoints.

### Maximum Matching Problem

Given a graph $G=(V, E)$, find a matching of maximum cardinality.

A greedy approach:  Pick any available edge that extends the current matching.

Does the above algorithm always output a maximum matching of the graph? No. Consider the example below which is a path of length three. If the green edge is chosen as the first edge by the algorithm, the algorithm stops there. The optimal solution though is a matching of size 2 (red edges).

### Shortest Path Problem

Given a graph $G=(V, E)$ with weights on the edges,  and vertices $(s,t)$, find a path between $s, t$ with minimum total weight.

A greedy approach: Start with $s$ and at each step choose the neighbour that gives the shortest path to $t$.

Does the above algorithm always find a shortest path? No. Consider the example below. At the first step, the algorithm chooses the neighbour $c$ cause the distance of $c$ to $t$ is 6 versus the distance of $a$ to $t$ which is 8. Therefore the greedy algorithm outputs the path $(s, c, d, t)$. However, the shortest path is $(s, a, b, t)$.

### Minimum Spanning Tree (MST) Problem

The next example is a problem (computing the minimum spanning tree of a graph) where a greedy algorithm finds the optimal solution. Before getting into the problem, we need to introduce some definitions of a graph $G=(V,E)$.

Def 2. (Connected Component) A connected component is a subset of vertices $M \subset V$ such that there is a path between any two vertices in $M$.

Def 3. (Spanning Subgraph) A subgraph that includes every vertex in $V$.

Def 4. (Tree) A tree is a connected component of $G$ with no cycles.

Claim 5: A minimal weight spanning connected component  $F\subseteq E$ is a spanning tree. That is,

1. $F$ has no cycles
2. $F$ has $|V|-1=n-1$ edges; i.e. $|F|=n-1$.

Proof: (1) Suppose $F$ had a cycle. Then, we can remove any edge on the cycle while maintaining connectivity between every pair of vertices. Therefore, there can be no cycles if $F$ is minimal.

(2) We prove by induction that $|F|=n-1$. It’s clearly true for the base cases $n=1,2$. Now, for $n>2$, remove any edge in $F$. Then, $F \backslash \{e\}$ has $2$ components (otherwise $F$ was not minimal), and each component has $n_1, n_2 vertices with $n_1 + n_2 = n$. By induction,

$|F| = (n_1-1) + (n_2-1) + 1 = n-1$.

Minimum Spanning Tree (MST) Problem. Given a graph $G=(V, E)$ and weights on the edges, find a spanning tree of $G$ with minimum total weight.

A greedy approach: One possible approach is to simply select the lightest weight edge. Locally, this seems like the best edge to select. However, there is a potential problem: what if the lightest weight edge creates a cycle? Well, then we simply ignore that edge and proceed to the next lightest edge. We have essentially described Kruskal’s Algorithm:

Kruskal’s Algorithm for solving MST:

KRUSKAL$(V,E,w)$:

• Sort edges in increasing order of weight; so have an ordering $e_1, \ldots, e_m$ such that $w(e_i) \le w(e_{i+1})$.
• Set $T = \{\}$
• for $i=1$ to $m$
• Add $e_i$ to $T$ if it doesn’t create a cycle.
• Return $T$.

Theorem 6 (Kruskal): Kruskal’s algorithm produces a minimum weight spanning tree of $G=(V,E)$.

Before proving this theorem, we need to first introduce some definitions and lemmas.

Def 7. (Cut) A cut $(S,\bar S)$ is the set of edges between a subset of vertices $S \subseteq V$ and its complement $\bar S = V \backslash S$. That is,

$(S,\bar S) = \{(u_i, v_j) \in E | u_i \in S, v_j \in \bar S\}$.

Lemma 8. Fix any edge $e$ whose weight is the minimum in some cut $(S,\bar S)$. Then, there exists an MST $T$ containing $e$.

Proof: Suppose $e$ is not in an MST. Consider an MST $T$ and add $e$ to $T$. Then, $e$ induces a cycle $C$ (since $T \cup \{e\}$ has two paths between endpoints of $e$).

Consider the cut $(S,\bar S)$ for which $e$ is a minimum weight edge. Then $C$ must cross $(S,\bar S)$ in some other edge $f$ with $weight(e) \le weight(f)$.

Set $T' = T \cup \{e\} \backslash \{f\}$. Note that $weight(T') \le weight(T)$, which means $T'$ is an MST (since it connects every pair of vertices and is minimum weight).

Lemma 9 (cut property). Let $X$ be a subset of edges of an MST of a graph $G$. For any edge $e$ such that $e$ is a minimum-weight edge of a cut $(S, \bar{S})$ and $X \cap (S, \bar{S}) = \phi$, there is an MST $T$ s.t. $X \cup \{e \} \subseteq T$.

Proof. Take some MST $T$ that contains $X$.  Let $(S, \bar{S})$ be a cut in which $e$ is a minimum cost edge and assume that $e \notin T$. Add $e$ to $T$. This creates a cycle. At least one other edge of the cycle $e'$ must be in $(S, \bar S)$. Since $e$ was the edge of the cut with the smallest weight, $T \cup \{e\}\setminus \{e'\}$ has lower weight than $T$. This contradicts $T$ being an MST.

Proof (of Kruskal Theorem): Using Lemmas 8 and 9, we can now prove that Kruskal’s algorithm always produces an MST. When the next edge

When the next edge $e$ is added, we can identify a cut $(S, \bar S)$ such that $e$ is the first edge across this cut. Let $X$ be the edges that has been added so far. According to Lemma 9, $e$ belongs to some MST. We add $e$ to $X$ and repeat the process.

Running time of Kruskal’s algorithm: $O(m \log n)$.

Sorting the edges once up front takes $O(m \log n)$.

However, we also have to keep track of whether an added edge creates a cycle, i.e. when an edge connects two vertices in the same connected component. We can do this via a simple data structure.

For each vertex, introduce a component number, which is initially just the index of that vertex. When Kruskal’s algorithm adds an edge which connects two vertices with the same component number, we skip that edge because it creates a cycle. If the component numbers of the two vertices are different, then we relabel the vertices in the smaller component to the label of the larger component. Since we relabel the smaller component, the size of a vertex’s component at least doubles each time it is relabeled. Therefore a vertex’s label changes at most $O( \log n)$ times. Thus the total cost of this data structure is $O(n \log n)$.

Prim’s Algorithm:

Another algorithm to find the MST of a graph is Prim’s algorithm.

• Add the lightest weight edge from some visited vertex to an unvisited vertex; repeat until all vertices are visited.

As with Kruskal’s algorithm, its correctness follows from the cut property given above.

Running time of Prim’s algorithm: $O(m \log n)$.

### Matroids

Can we define a more general class of problems that could be solved by the above greedy algorithm?

Def 10. (Matroid) A matroid consists of a universe of elements $U$ and a collection $I$ of independent subsets of $U$ with the following properties

• $A \subseteq B, B\in I \Rightarrow A \in I$
• $|A|<|B|, A,B \in I \Rightarrow \exists e \in B$ s.t. $A\cup \{e\} \in I$

A Greedy approach. Sort elements of $U$ in increasing order of weights. Add the next element in the list that preserves the independency.

Theorem 11. The above greedy algorithm finds the maximum weight independent set.

Proof. We prove by contradiction. Let $x_1 \geq x_2 \geq X_3 \ldots$ be the elements in the output greedy algorithm in descending weight order and let $y_1, y_2, \ldots$ be the elements in the maximum independent set in descending weight order. Since the total weight of the optimal is higher than the total weight of the greedy solution, consider $k$ to be the first index that $w(x_k) < w(y_k)$. Then $|\{x_1, \ldots, x_{k-1}\}| < |y_1, \ldots, y_k|$ and both are independent. By the matroid property $\exists y_j \in \{ y_1, \ldots , y_k \}$ such that $\{ x_1, \ldots, x_k \} \cup \{y_j\}$ is an independent set. But $w(y_j) \geq w(y_k) > w(x_k)$ and this contradicts the fact that the greedy algorithm chose $x_k$ and not $y_j$.

Observation 12. Spanning trees form a matroid. Let’s define $U = E$ and $I$ as the subset of edges with no cycles (forests). It is simple to see that this satisfies the two matroid properties. Given this, Theorem 11 states that the greedy algorithm stated above finds the maximum spanning tree. This can be used to solve the MST problem by simply defining the weights $M_{ij} = W - w_{ij}$. Then finding an MST on a graph with weights $w$ is equivalent to finding the maximum spanning tree on the same graph with weights $M$.


text:
# Ian Stark

## The Continuous π-Calculus: A Process Algebra for Biochemical Modelling

In Computational Methods in Systems Biology: Proceedings of 6th International Conference CMSB 2008, Rostock, Germany, October 12–15, 2008. Lecture Notes in Computer Science 5307, pages 103–122. Springer-Verlag, 2008.

Fetch the paper or visit the publisher's page. Download the slides from Marek's conference talk, or from my later seminar presentation. Go to other papers, talks or my home page.

### Abstract

We introduce the continuous π-calculus, a process algebra for modelling behaviour and variation in molecular systems. Key features of the language are: its expressive succinctness; support for diverse interaction between agents via a flexible network of molecular affinities; and operational semantics for a continuous space of processes. This compositional semantics also gives a modular way to generate conventional differential equations for system behaviour over time. We illustrate these features with a model of an existing biological system, a simple oscillatory pathway in cyanobacteria. We then discuss future research directions, in particular routes to applying the calculus in the study of evolutionary properties of biochemical pathways.

@InProceedings{kwiatkowski/stark:continuous-pi,
author =       {Marek Kwiatkowski and Ian Stark},
title =        {The Continuous $\pi$-Calculus: A Process Algebra for
Biochemical Modelling},
booktitle =    {Computational Methods in Systems Biology: Process of the
Sixth International Conference CMSB~2008},
pages =        {103--122},
year =         2008,
series =       {Lecture Notes in Computer Science},
number =       5307,
publisher =    {Springer-Verlag},
url =          {http://homepages.ed.ac.uk/stark/continuous-pi.html},
pdf =          {http://homepages.ed.ac.uk/stark/continuous-pi.pdf},
doi =          {10.1007/978-3-540-88562-7_11}
}

This research was supported by an EPSRC studentship, the Laboratory for Foundations of Computer Science, and The University of Edinburgh School of Informatics.


text:
How would you define diffusion in terms of the particles of a gas?

Aug 25, 2016

Diffusion of gases is the process of two gases mixing together.

Explanation:

Diffusion of gases is the process of two gases mixing together.

The diffusion ratio could be measured by the distance traveled by each particles, which depends on the velocity of the gas.

The root mean square velocity (${U}_{r m s}$) of a gas depends on temperature ($T$) and its molar mass ($M M$) and it is given by the following expression:

${U}_{r m s} = \sqrt{\frac{3 R T}{M M}}$

The diffusion rate of two gases $A \text{ and } B$ can be written as:

("Distance traveled by A")/("Distance traveled by B")=sqrt(((color(red)(cancel(color(black)(3RT))))/(MM_A))/((color(red)(cancel(color(black)(3RT))))/(MM_B)))=sqrt((MM_B)/(MM_A)


text:
# Does a topological hypercover always have free degeneracies?

This question arises when I am reading Dugger and Isaksen's "Hypercovers in topology". According to Definition 4.1 in that paper, A hypercover of a space $X$ is an augmented simplicial space $U_*\to X$ such that the maps $U_n\to M^X_nU$ are open covering maps for all $n \geq 0$. Here $M^X_nU$ denotes the nth matching object of $U_*$ computed in the category $s(\mathcal{Top}\downarrow X)$.

On the other hand by Definition A.4 in the same paper, a simplicial space $X_*$ is said to be split, or to have free degeneracies, if there exist subspaces $N_k\hookrightarrow X_k$ such that the canonical map $$\coprod_{\sigma}N_{\sigma}\to X_k$$ is an isomorphism. Here the variable σ ranges over all surjective maps in $\bf{\Delta}$ of the form $[k]\to [n]$, $N_{\sigma}$ denotes a copy of $N_n$, and the map $N_{\sigma}\to X_k$ is the one induced by $\sigma^*: X_n\to X_k$. The idea is that the spaces $N_k$ represent the ‘non-degenerate’ part of $X_k$, sitting inside of $X_k$ as a direct summand.

Now Theorem 4.3 of the paper claims that if $U_*\to X$ is a a hypercover then the maps $\text{hocolim}U_*\to |U_*|\to X$ are all weak equivalences. In the proof it says that "the simplicial object $U_*$ is Reedy cofibrant since it has free degeneracies,"

It is clear that the Cech nerve of an open cover has free degeneracies. My question is: is it also true for any hypercover in the topological setting?

• Don't know the answer, but: if I'm understanding the definition of hypercover, it seems to induce a simplicial object in the topos generated by open Sets of $X$, with an epimorphism property at every level; D+I have arranged their definition the other way around, starting with sSpaces over X and then paring down what kind of Spaces are in each level. Aug 4, 2015 at 20:51
• @JesseC.McKeown That is correct. In topos-theoretic language, my argument amounts to the observation that a sheaf whose espace étalé is Hausdorff is a decidable object, i.e. $A \times A \cong \Delta_A \amalg B$, where $\Delta_A$ is the diagonal; a simplicial sheaf that is degreewise decidable will then have free degeneracies. Aug 4, 2015 at 23:48

Here is a proof for the case where $X$ is a Hausdorff space. Note that each $U_n$ is also Hausdorff in this case.
A standard argument shows that the face operators of $U_\bullet$ are (surjective) local homeomorphisms, so the degeneracy operators of $U_\bullet$ are open embeddings. On the other hand, degeneracy operators are split monomorphisms, and any split monomorphism in the category of Hausdorff spaces is a closed embedding. Thus, the image of any degeneracy operator of $U_\bullet$ is a "direct summand" (i.e. clopen subspace). It more or less follows that $U_n$ decomposes according to the formula you stated.


text:
1. ## trig substitution

suggest an appropriate trig substitution for x in

int(x^5/(x^2-5)^3/2)dx
and
int((3-x^2)^5/2)dx

2. Originally Posted by samdmansam
suggest an appropriate trig substitution for x in

int(x^5/(x^2-5)^3/2)dx
$x = \sqrt{5} \sec \theta$

int((3-x^2)^5/2)dx
$x = \sqrt{3} \sin \theta$

3. The first one is also solvable by standard algebraic techniques.

4. Originally Posted by Krizalid
The first one is also solvable by standard algebraic techniques.
you mean a regular substitution of $u = x^2 - 5$, or something else?

yes, that regular substitution will do it, but the poster asked for a trig substitution, so ...

[off-topic]

[/off-topic]

5. Yeah, but observe that I said "also."


text:
# Other realms for studying symbolic dynamics

I hope to find an online version of accessible texts in symbolic dynamics. Marcus and Lind have a text I hope to get online. What I don't know is if any text yet exists that considers symbolic dynamics over alternative domains or realms.

Much of the current theory studies what I will call one-dimensional shifts, which are essentially translations of strings of symbols, and so (to some extent) involve the combinatorics of infinite words, or certain subsets of $F^Z$ with $F$ often a finite set or alphabet and $Z$ being the integers, so the elements are bi-infinite words. But what of dynamics on $F^{Z \times Z}$, or on more exotic structures, say, where the exponent might be a sufficiently rich directed graph? Has the theory advanced to the point where symbolic dynamics in these realms can be as clearly understood as in the one-dimensional realm?

An ideal answer would come from a student or researcher who can fully address questions like "In order to get ready for studying SD in such realms, which portions of Marcus and Lind (or other accessible texts) should I read first?" or like "What are the one or two papers that illumined the subject of exotic SD for you?". I also welcome other references or suggested lines of research.

I am quite at the beginning of such studies; it won't bother me if you assume I am totally ignorant of dynamics in supplying an answer. My motivation is to find a few undecidable (in the sense of Turing computable) problems that I can use to gauge the decidability of certain problems in some resticted systems of second-order logic. A possible line of attack involves seeing how much symbolic dynamics in one-dimension carries over to more exotic realms. I welcome commments about such connections, too.

Gerhard "Time To Learn Something New" Paseman, 2011.04.29

• What do you mean by dynamics on $F^{\mathbb{Z} \times \mathbb{Z}}$? Are you still studying a single shift operator or are you studying both the horizontal and vertical shifts? – Qiaochu Yuan Apr 30 '11 at 10:43
• For that example, studying ( what I call ) a pair of shifts even like north and northwest would make sense. I use $Z \times Z$ more as an example, though. I'm hoping that a directed graph or a partially ordered algebraic structure might serve as exponent. Gerhard "Ask Me About System Design" Paseman, 2011.04.30 – Gerhard Paseman Apr 30 '11 at 15:52

I believe what you're referring to here are called $\mathbb Z^d$-actions (with $d=2$ in your setting). This is a pretty large area of (algebraic) dynamics with people like Klaus Schmidt, Doug Lind, Thomas Ward and Manfried Einsiedler (and many others) actively working in it.


text:
# where

Copies of indexes of a list or keys of a dictionary

where x    where[x]

Where x is a:

## Vector of non-negative integers

returns a vector containing, for each item of x, that number of copies of its index.

q)where 2 3 0 1
0 0 1 1 1 3
q)raze x #' til count x:2 3 0 1
0 0 1 1 1 3

Where x is boolean, the result is the indices of the 1s. Thus where is often used after a logical test:

q)where 0 1 1 0 1
1 2 4
q)x:1 5 6 8 11 17 20 21
q)where 0 = x mod 2        / indices of even numbers
2 3 6
q)x where 0 = x mod 2      / select even numbers from list
6 8 20

## Dictionary whose values are non-negative integers

returns a list of keys repeated as many times as the corresponding value.

q)d:amribmmsft!2 3 1
q)where d
amramribmibmibmmsft
q)where 2 3 0 1               / usual operation on integer list
0 0 1 1 1 3
q)where 0 1 2 3 ! 2 3 0 1     / same on dictionary with indices as keys
0 0 1 1 1 3
`

Insight

If a list is viewed as a mapping from indexes to entries, than the definition for the integer list above is merely a special case.


text:
# Vertical interference in double-slit

Why is no interference pattern in the vertical axis in a double-slit ? Isn't a vertical slit just thousands of small slits one next to the other ? So shouldn't this create an interference pattern also along the vertical axis ?

• The vertical dimension is usually too big. Had it been comparable to the slit width you would have obtained an interference along that dimension too. – Raziman T V Jan 8 '17 at 3:44

We can construct such a one-dimensional function as a square function that represents the slit convolved by a comb function $$h(y/w) \otimes C_d(y)$$ where $h(y/w)$ is the is the square function of width $w$ and $C_d(y)$ is a comb function consisting of impulse functions separated by $d$.
The Fourier transform of this is given by the product of the Fourier transform of the square function, which is a sinc function $${\rm sinc}(wu) = \frac{\sin(\pi wu)}{\pi wu}$$ and the Fourier transform of the comb function, which is again a comb function, but where the impulses are separated by $1/d$. Note that the sinc function has zeros where $u=n/w$ for any integer $n$ not equal to zero.


text:
# Need help on integral question

1. Mar 19, 2007

### Mag|cK

Hello, i have a question. This is integral question, but i dont have any program to write the equation with:

integral (sin5x)^(-2) dx= ?

Thank you..

2. Mar 19, 2007

### cristo

Staff Emeritus
This is one of those you should really know, since it can be rewritten cosec2(5x).

However, if you don't know the antiderivative, then it can be worked out by a few substitutions, the first being u=sin(5x)

3. Mar 19, 2007

### Mag|cK

yes i have done using substitution method but can't solve the problem. Can you step by step show me how please? thx

4. Mar 19, 2007

### cristo

Staff Emeritus
It's your homework, not mine! Show how you did the first substitution, and we'll go from there.

5. Mar 19, 2007

### Mag|cK

ok sorry
if i substitute with sin5x,

integral (sin5x)^(-2) d(sin5x)/5cos5x

this is how far i can get if i substitute with sin5x as i cant eliminate the cos5x.
I already tried substituting with cosec5x and sec5x, but still no solution.

6. Mar 19, 2007

### jakcn001

With -d[cot(x)]=(sinx)^(-2) dx ,you can solve it easily.Have a try!

7. Mar 19, 2007

### cristo

Staff Emeritus
If you've substituted, why do you still have functions of x in the integral? If you let u=sin(5x), then du=5cos(5x)dx=5[sqrt(1-u2)]. This transforms your integral to $$5 \int \frac{du}{u^2\sqrt{1-u^2}}$$.

There you go, I've done one step for you. Now, try the substitution v=1/u.
If you look back, I said you'll need a few substitutions!

I did allude to that in my first post.

8. Mar 20, 2007

### Mag|cK

ok thx guys i understand now


text:
# abresler/MLNotes forked from johnmyleswhite/MLNotes

Very concise notes on machine learning and statistics.
This branch is 2 commits behind johnmyleswhite:master.
Fetching latest commit…
Cannot retrieve the latest commit at this time.
 Failed to load latest commit information. README.md compile.R compile.sh mcmc.html mcmc.rmd norms.html norms.rmd optimization.html optimization.rmd outline.html outline.rmd regularization.html regularization.rmd

# MLNotes

Notes written for myself to keep ML/stats theory ideas organized. Will hopefully be useful to others.

To best take advantage of these notes, you should download the HTML files and view them in any browswer that supports MathJax. Eventually I'll organize them more clearly and maybe even make a PDF, but for now I'm just keeping ideas in a file named based on the central theme of that file.

# Compile Notes

To make sure the notes you're using are up-to-date, it's probably a good idea to compile the notes using the knitr package in R. Once it's installed, you can just run:

bash compile.sh

which will generate the appropriate HTML files.


text:
# Recent questions tagged graphs

1 vote
1
Revenue obtained by a publishing house by selling books, magazines, and journals The number of years in which there was an increase in revenue from at least two categories is 1 2 3 4
1 vote
2
Revenue obtained by a publishing house by selling books, magazines, and journals If ’93 were to show the same growth as ’92 over ’91, the revenue in ’93 must be 194.5 186.6 172.4 176.7
1 vote
3
Revenue obtained by a publishing house by selling books, magazines, and journals In ’92, what percent of the total revenue came from books? 45% 55% 35% 25%
1 vote
4
Revenue obtained by a publishing house by selling books, magazines, and journals. The growth in total revenue from ‘89 to ‘92 is 21% 28% 15% 11%
1 vote
5
Revenue obtained by a publishing house by selling books, magazines, and journals Which year shows the least change in revenue obtained from journals? ’89 ’90 ’91 ’92
1 vote
6
Which of the following statements can be inferred from the graph? 1. In all the years shown in graph, the trade deficit is less than the export earning. 2. Export earnings increased in every year between 89-90 and 91-92. 3. In all the years shown in the graph, the earning by exports is less than the expenditure on imports in the preceding year. 1 only 2 only 3 only 1 and 3 only
1 vote
7
In the last three years the total export earnings have accounted for how many percent of the value of the imports? 80 83 95 89
1 vote
8
In how many years was the trade deficit less than the trade deficit in the succeeding year? 1 2 3 4
1 vote
9
Export earning in 90-91 is how many percent of imports in 91-92? 82 85 92 15
1 vote
10
In which year was the trade deficit greatest? 87-88 88-89 89-90 90-91
–1 vote
11
Use the following data: A manufacturer can choose from any of the three types of tests available for checking the quality of his product. The graph gives the relative costs for each of these tests for a given percentage of defective pieces. If $p <0.2$, then the best alternative will be:. Test-2 Test-3 Test-1 Not Test-3
–1 vote
12
Use the following data: A manufacturer can choose from any of the three types of tests available for checking the quality of his product. The graph gives the relative costs for each of these tests for a given percentage of defective pieces. When is Test-$1$ feasible? $P < 0.05$ $0.0 < p < 0.2$ $0.1 < p < 0.2$ $0.05$ to $0.2$
–1 vote
13
use the following data: A manufacturer can choose from any of the three types of tests available for checking the quality of his product. The graph gives the relative costs for each of these tests for a given percentage of defective pieces. When will Test-$3$ be feasible? $P > 0.2$ $0.1 < p < 0.2$ $0.05 < p < 0.1$ $P < 0.05$
–1 vote
14
use the following data: A manufacturer can choose from any of the three types of tests available for checking the quality of his product. The graph gives the relative costs for each of these tests for a given percentage of defective pieces. If p is equal to 0.2, then which test will be feasible? either 1 or 2 2 only 3 only either 2 or 3
15
use the following data: A manufacturer can choose from any of the three types of tests available for checking the quality of his product. The graph gives the relative costs for each of these tests for a given percentage of defective pieces.
16
These questions are based on the following graph. Referring to the above question, in which year was the per capita consumption of this nutrient the highest? 1993 1994 1995 1996
17
These questions are based on the following graph. If one gallon of milk contains $120$ gm of a particular nutrient and one ton of food grains contains $80$ gm of the same nutrient, in which year was the availability of this nutrient the maximum? 1993 1994 1995 1996
18
These questions are based on the following graph. In which year was the difference between the increase in the production of food grains and milk the maximum? 1993 1994 1995 1996
19
These questions are based on the following graph. When was the per capita production of food grains the most? 1992 1993 1994 1995
20
These questions are based on the following graph. When was the per capita production of milk the least? 1990 1992 1994 1996
21
Answer the question on the basis of following information. Shown below is a layout of major streets in a city. Two days (Thursday and Friday) are left for campaigning before a major election, and the city administration has received requests from five political parties ... Thursday. The BJP procession can only take place on Friday. Congress and BSP can take out their processions on the same day.
22
Answer the question on the basis of following information. Shown below is a layout of major streets in a city. Two days (Thursday and Friday) are left for campaigning before a major election, and the city administration has received requests from five political ... Congress procession can be allowed only on Thursday. only on Friday. on either day. only if the religious procession is cancelled.
23
Answer the question based on the information given below: A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in ... commuter travelling from point S to point T under this policy will be Rs $7$ Rs $9$ Rs $10$ Rs $13$ Rs $14$
24
Answer the question based on the information given below: A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for travelling along a street is indicated ... $0, 5, 4, 1$ $0, 5, 2, 2$ $1, 5, 3, 3$ $1, 5, 3, 2$ $0, 4, 3, 2$
25
Answer the question based on the information given below: A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for travelling along a street is indicated ... $0, 5, 2, 2$ $0, 5, 4, 1$ $1, 5, 3, 3$ $1, 5, 3, 2$ $0, 4, 3, 2$
26
Answer the question based on the information given below: A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for travelling along a street is indicated ... $1, 5, 3, 3$ $1, 4, 4, 3$ $1, 5, 4, 2$ $0, 5, 2, 3$ $0, 5, 2, 2$
27
The following sketch shows the pipelines carrying material from one location to another. Each location has a demand for material. The demand at Vaishali is $400,$ at Jyotishmati is $400,$ at Panchal is $700,$ and at Vidisha is $200.$ Each arrow indicates the direction of ... of each pipeline is $1000.$ What is the free capacity available in the Avanti-Vidisha pipeline? $300$ $200$ $100$ $0$
28
The following sketch shows the pipelines carrying material from one location to another. Each location has a demand for material. The demand at Vaishali is $400,$ at Jyotishmati is $400,$ at Panchal is $700,$ and at Vidisha is $200.$ Each arrow indicates the direction of ... capacity of each pipeline is $1000.$ The free capacity available at the Avanti-Vaishali pipeline is $0$ $100$ $200$ $300$
29
The following sketch shows the pipelines carrying material from one location to another. Each location has a demand for material. The demand at Vaishali is $400,$ at Jyotishmati is $400,$ at Panchal is $700,$ and at Vidisha is $200.$ Each arrow indicates the direction of ... met. The capacity of each pipeline is $1000.$ The quantity moved from Avanti to Vidisha is $200$ $800$ $700$ $1000$
30
The figure below shows the network connecting cities A, B, C, D, E and F. The arrows indicate permissible direction of travel. What is the number of distinct paths from A to F? $9$ $10$ $11$ None of these
31
There are three cities A, B and C, each of these cities is connected with the other two cities by at least one direct road. If a traveller wants to go from one city (origin) to another city (destination), she can do so either by traversing a road connecting the two cities directly, or ... $6$ $3$ $5$ $10$
A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with $3$ ... satisfy the condtion $11 \leq e \leq 66$ $10 \leq e \leq 66$ $11 \leq e \leq 65$ $0 \leq e \leq 11$
The graph $y-x$ against $y +x$ is as shown as below. (all graphs in this question are drawn to scale and the same scale has been used on each axis). Then, Which of the options given shows the graph of $y$ against $x.$
Answer the question based on the information given below: A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for travelling along a street is indicated ... $2, 5, 3, 2$ $0, 5, 3, 1$ $1, 5, 3, 2$ $2, 3, 5, 1$ $1, 3, 5, 1$


text:
Dirac Notation!

1. Aug 6, 2015

leonmate

I've been working through some dirac notation and I'm stuck...

Here's where I'm at:

I understand that an expectation value: <x> = ∫ ψ* x ψ dx = <ψ|xψ> = <ψ|x|ψ>
Also, we can say H|ψ> = E|ψ> where E is an eigenvalue of the operator H and |ψ> represents a state your acting on.
I get that you can represent a vector 'in Hilbert space' instead of a wavefunction inside the ket and these can be operated on to transform your vector into another vector, Q|a> = |b>
|a> represents a column vector, <a| represents a row vector

It's taking some time to get used to using Dirac notation, and I've come across this in some angular momentum notes:

Lz|m,l> = mħ|m,l> .. So Lz is the operator, mħ is the corresponding eigenvalue. But what does having two values in the ket mean?
Does, L|m,l> = L|m> + L|l>

Moreover, what if we have:

<l,m|L+L-|l,m>

Should this be written in more classical notation as ∫ (l,m)* L+L- (l,m)* dx

Any guide or text book with a good section on Dirac formalism would be great, I've been looking but can't find anything to explain this trickier stuff. I've attached a page of my notes so that it's more clear with what I'm working with. I've read through Griffith's introduction to quantum mechanics chapter on Dirac formalism but I'm still getting undone by it!

Thanks,
Leon

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2. Aug 6, 2015

blue_leaf77

The fact that you can write H|ψ> = E|ψ>, implies that the state |ψ> is an eigenstate of H, it cannot be any state.
$|l,m \rangle$ is eigenstate of both $L^2$ and $L_z$, this can be so because $L^2$ and $L_z$ commute - $[L^2,L_z] = 0$. If you have studied linear algbera you may remember that if two operators commute and no other operators can be added into this group of operators without violating the mutual commutativity, they must share the same set of eigenstates and it will be uniquely specified by two indices, in this case $l$ and $m$. The former corresponds to the operator $L^2$ and the latter to $L_z$.
No it doesn't.
You can do the calculation in position representation like you wrote there, but if you know the property of $L_+$ and $L_-$ and how they act on $|l,m \rangle$, it saves a lot of time working in Dirac notation.

3. Aug 6, 2015

Staff: Mentor

The ket $|l,m\rangle$ is a state that is both an eigenstate of $L_z$ with eigenvalue $m$ and an eigenstate of $L^2$ with eigenvalue $l(l+1)$. This is possible only because $L_z$ and $L^2$ commute.

4. Aug 6, 2015

leonmate

When we have two commuting operators we can have two values inside the ket, both are eigenstates of my operators?
Ok, I see that I can, I'm stuck on the how. Also, I'm confused at how eigenvalues seem to be plucked out of thin air in every example I look at. I can find how to find eigenvalues from matrices, but in examples like this is there some way to work out what the eigenvalue should be?

E.g. Lz|l,m> = mħ|l,m> ... where has that mħ come from?!

Is it easier to just learn all the eigenvalues for each operator? There's a way I can work out what they should be, my notes, other course notes and text books (that I've looked at) don't seem to show this. What if Lz|ψ> where ψ is some different eigenstate, are the eigenvalues the same or different?

Another example from my notes is this:

L2 |l, m >= l(l + 1)ħ2|l, m >

but,

L22 , m >= β2ħ22 , m >

These are two different eigenstates and two different eigenvalues that are just stated in my notes, you can see in the images I attached above. How, why? ... in the same example changing the eigenstate Lz is acting on doesn't change the eigenvalue. Very confused :(

5. Aug 6, 2015

blue_leaf77

At some point in your textbook you should find relation like
$$L_+|l,m \rangle = C_+ |l,m+1 \rangle$$
for $m \neq l$, and
$$L_-|l,m \rangle = C_- |l,m-1 \rangle$$
for $m \neq -l$. The constants $C_+$ and $C_-$ are found using normalization condition.
Working out the eigenvalues of $L^2$ and $L_z$ in Dirac notation does require some elaborate works, for this purpose I suggest Modern Quantum Mechanics by Sakurai sub-chapter 3.5. But honestly as I remember Griffith must also provide the same material, I think you just need to be persistent enough to read through that chapter in Griffith's book patiently.

If |ψ> is not an eigenstate of $L_z$, Lz|ψ> won't be eigenvalue equation anymore. In such a case, you can represent |ψ> as a linear combination of the eigenstates of $L_z$.

6. Aug 6, 2015

kith

Yes. You could start with the definition of the orbital momentum operator and plug in the linear momentum operator: $\mathbf{L} = \mathbf{r} \times \mathbf{p} = -i \hbar(\mathbf{r} \times \nabla)$. Now write down the eigenvalue equation for the operators, for example $$L_z u(\vec r) = \mu u(\vec r)$$ This partial differential equation can be solved for $u(\vec r)$ if you go over to spherical coordinates. So in the continuous case, you don't calculate eigenvalues of matrices but solve differential equations.

Now the case of angular momentum is special because there exists a simpler and more powerful technique to calculate the eigenvalues. The idea is to start with abstract equations and use only the commutation relations of the angular momentum operators to obtain the possible eigenvalues (it is quite remarkable that this works!).

Since $\mathbf{L^2}$ and $L_z$ commute, we can find states which are eigenstates to both operators. So instead of two independent eigenvalue equations
$$\mathbf{L^2} |\lambda \rangle = \lambda |\lambda \rangle\\ L_z |\mu \rangle = \mu |\mu \rangle$$
we have
$$\mathbf{L^2} |\lambda, \mu \rangle = \lambda |\lambda, \mu \rangle\\ L_z |\lambda, \mu \rangle = \mu |\lambda, \mu \rangle$$

Now we could try to use the commutation relations in order to determine the possible values of $\lambda$ and $\mu$ but usually, the derivation uses additional information. First, we know that we are dealing with angular momentum in Quantum Mechanics, so it seems sensible to write the eigenvalues as multiples of $\hbar$ which has the correct unit and expected magnitude. Also some authors write the eigenvalues of $\mathbf{L^2}$ in the special form $l(l+1) \hbar^2$ from the beginning, because they already know that the result will involve an integer $l$. I don't like the pedagogy of this but mathematically, it is of course valid.

Now check if you understand the derivation of the eigenvalues in a textbook. It should be included in Griffiths. If not, check a couple of other textbooks, as blue_leaf77 wrote, Sakurai definitely has it. It is a really common derivation.

Last edited: Aug 6, 2015