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# Forecasting short time-series at scale

I am looking for a method of forecasting short time series. I need to make multiple such forecasts at parallel, so I need some simple method that scales to large data. My data looks like $n\times k+1$ (where $n$ is in thousands or more and $k$ is something between 3 and 10) matrix

$$\begin{array}{ccccc} x_{1,t-k} & x_{1,t-k+1} & \dots & x_{1,t-1} & x_{1,t} \\ x_{2,t-k} & x_{2,t-k+1} & \dots & x_{2,t-1} & x_{2,t} \\ \dots \\ x_{n,t-k} & x_{n,t-k+1} & \dots & x_{n,t-1} & x_{n,t} \\ \end{array}$$

The series may, or may not, be correlated with each other.

I need to make forecases for the $x_{i, t+1}, x_{i, t+2}, x_{i, t+3}$ points (the forecast horizon would hopefully be short) for each $i$-th series. In most cases I can assume some simple linear trends (I am not hoping for much more since the limited data for each of the series), so I could simply try something like a linear regression to approximate AR($k$) model, but in my data, from time to time after a steady linear upward trend I observe a random drop and I would like my method to be somehow sensitive to noticing such change (e.g. when there is a change between $x_{i,t-1}$ and $x_{i,t}$) and not suggesting linear upward trend after noticing it. Below I show a made-up example that looks similar to my data, that was created using the a10 dataset from fpp package by Rob Hyndman.

library("fpp")

s1 <-  seq(1, 100, by = 10)
s2 <- seq(10, 110, by = 10)

X <- NULL

for (i in seq_along(s1))
X <- rbind(X, a10[s1[i]:s2[i]])

All this means that the common simple methods like linear regression; random walk forecast that takes $x_{i,t}$ as a forecast for the $t+1, t+2, t+3$ values; random walk with drift; predicting mean of previous timepoints etc. do not seem as a good choice for me. On another hand, limited data and computational limitations make more advanced methods not feasible.

I was thinking of something like using simple exponential smoothing of the changes between the timepoints $\Delta_{i,t} = x_{i,t} - x_{i,t-1}$, i.e. taking

\begin{align} \ell_{i,t} &= (1-\alpha) \ell_{i,t-1} + \alpha \Delta_{i,t} \\ \hat x_{i,t+1} &= x_{i,t} + \ell_{i,t} \\ \hat x_{i,t+2} &= \hat x_{i,t+1} + \ell_{i,t} \\ \dots \end{align}

with some pretty hign $\alpha$ (this could be optimized), but I do not want to re-design the wheel, so I'm looking for some hints either on this approach, or suggesting something better then this. I'd need prediction intervals for my forecast, so I'd be grateful also for a comment on this.

Another problem is that this approach ignores the fact that I could also use the other series to build a general model. I believe that using the whole dataset could lead to improvements (some kind of shrinkage). Maybe I could use something like hierarchical forecasting described by Rob Hyndman (see e.g. those slides or the hts vignette) where the parameters to aggregate the individual series could be estimated using regression (this easily scales).

• Can you elaborate on why simple methods do not seem a good choice? – S. Kolassa - Reinstate Monica Nov 6 '17 at 10:32
• @StephanKolassa rw or mean are flat and in most cases I'll be seeing the upward trend (as in the data example); regression or rw + drift do not adapt to random drops. – Tim Nov 6 '17 at 10:33

A good method for forecasting short time series is double exponential smoothing, which I would prefer to applying single exponential smoothing to the differenced series. This will also allow for changes in trends, in contrast to a simple linear regression. Here is the relevant section in FPP, 2nd edition. The state space framework that, e.g., forecast::ets() uses will give you prediction intervals.

If you can meaningfully group your time series (or at least some of them) so they exhibit common patterns, you might be able to improve forecasts by forecasting aggregates (on which the signal is hopefully more easily visible) and adapting lower-level forecasts. Here is the relevant chapter in FPP2. I have repeatedly found this "optimal combination" approach to improve forecasts on all levels of the hierarchy.

Making optimal combinations play nice with prediction intervals (or even better, predictive densities), is an ongoing research topic. There was a presentation on "Coherent probabilistic forecasts for hierarchical time series" by Souhaib Ben Taieb, James Taylor and Rob Hyndman at this year's International Symposium on Forecasting. I'd recommend you ping the authors and ask for the presentation.

Of course, if you don't know a priori which time series will be positively correlated so you can hierarchically treat them, this is a problem - especially for short time series. You could simply cluster them, using one minus some correlation as a distance measure. But of course, if your series are short, then you might be clustering a lot of noise.

As to change point detection, this is of course also a hard topic for short series. The strucchange package for R contains a number of useful functions. You might want to apply these to differenced series, or to the residuals in-sample, so your breakpoint detector is not thrown off by the trend.

If you can cluster your series a priori (or an automatic clustering works well), then you might be able to detect the change points on the aggregates instead of the individual series, where once again, the signal will be stronger.

The problem of course then is what to do with detected change points. A state space framework should of course be able to include an external step change predictor, but unfortunately I believe that neither forecast::ets() nor the functions in the smooth package allow for this (but it might be good if you checked, I'm not entirely sure). Alternatively, you could use auto.arima() and feed a change point indicator into the xreg parameter. Here is Rob Hyndman explaining the difference between regression with ARIMA errors and ARIMAX, recommended reading.

EDIT - you write:

Maybe I could use something like hierarchical forecasting described by Rob Hyndman (see e.g. those slides or the hts vignette) where the parameters to aggregate the individual series could be estimated using regression

I agree that the hierarchical approach could definitely be helpful. However, if you estimate "optimal" coefficients for aggregating series, don't forget that this will be an additional source of variance, which could make your forecasts worse than just using flat weights of 1 (see Claeskens et al., IJF 2016, "The forecast combination puzzle: A simple theoretical explanation").


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# condition of orthogonality

## Primary tabs

\documentclass{article}
% this is the default PlanetMath preamble.  as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.

% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\begin{document}
Let two straight lines of the $xy$-plane have the slopes $m_1$ and $m_2$. \,The lines are at right angles to each other iff $m_1$ and $m_2$ are the {\em opposite inverses} of each other, i.e. \PMlinkname{iff}{Biconditional}
$m_1m_2 = -1.$

\textbf{Example.} \,The lines \,$y = (1+\sqrt{2})x$\, and \,$y = (1-\sqrt{2})x$\, are at right angles to each other.
\begin{center}
\includegraphics{orthogon}
\end{center}
%%%%%
%%%%%
nd{document}

format: plain


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# Prove that $\frac{1}{2h}\int_a^b\mu(A\cap(x-h,x+h))\,\text{d}x\le \mu(A)$

I'm preparing to the second mini-test in measure theory. Here is one of the problems I cannot deal with. I would appreciate any help, thank you.

Let $\mu$ be a Radon measure on $\mathbb{R}$, suppose that $A$ is a $\mu$–measurable subset of $[a,b]$ and let $h$ be a positive number. Prove that $$\frac{1}{2h}\int_a^b\mu(A\cap(x-h,x+h))\,\text{d}x\le \mu(A).$$

-

If you apply Fubini (or Tonelli) in the second equality below (everything is positive) you get $$\frac1{2h}\,\int_{[a,b]}\mu(A\cap(x-h,x+h))\,dx=\frac1{2h}\,\int_{[a,b]}\int_{(x-h,x+h)}\,1_A(t)\,d\mu(t)\,dx\\ \leq\frac1{2h}\int_{(a-h,b+h)}\int_{[t-h,t+h]}\,1_A(t)\,dx\,d\mu(t)=\int_{(a-h,b+h)}\,1_A(t)\,d\mu(t)=\mu(A\cap(a-h,b+h))=\mu(A).$$

-
So do we have in fact the equality? – Kuba Helsztyński Nov 19 '12 at 0:26
No, we don't. Now I see where my mistake was. When I'm doing Tonelli, the region on the right-hand-side is smaller than the one I'm using. So I'm changing that equal sign to a $\leq$. If that's not clear enough, let me know and I'll try to clarify. – Martin Argerami Nov 19 '12 at 0:32
Do you mean the reason for inequality sign is that you enlarge the region from $[a,b]$ to $(a-h,b+h)$? – Kuba Helsztyński Nov 19 '12 at 0:37
No. The region of integration is a parallelogram. When you swap the variables for Tonelli, its expression complicates a little. If I have to explain this to my Calculus II students, I would say that the region (in Stewart's Calculus language) is Type I but not Type II. So, for the second integrals I'm switching to a slightly larger parallelogram, hence the inequality. If you draw the region, you'll see it quickly. – Martin Argerami Nov 19 '12 at 0:40
Ah, ok, now I see. We are adding two small triangles to the initial parallelogram and integrating over slightly larger one. All is clear now, thank you for your help! – Kuba Helsztyński Nov 19 '12 at 1:22


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# Group Theory : Sum of order of elements of a group with finite and even order.

Consider a group $G$ which is finite and had even order. If we consider the sum of the order of the cyclic groups created by the elements of $G$, is this sum odd? How to prove this?

• i amnot professional in group theory but i think it is because of order of any element divide order of group thus all elements except unit have even order thus sum will be odd – R.N Sep 12 '15 at 11:33
• @RaziehNoori what if an element has order 3, so it could still divide let's say the group order 6...that means the order of the elements don't have to be all even – user190080 Sep 12 '15 at 11:36
• I'm not sure what you mean - I don't think your question makes it clear what you are supposed to be adding. In relation to one possible interpretation, the cyclic group of order $4$ has cyclic subgroups of size $1,2,4$ sum $7$. The cyclic group order $6$ has subgroups of size $1,2,3,6$ sum $12$. – Mark Bennet Sep 12 '15 at 11:37
• @user190080 oh, silly mistake. sorry – R.N Sep 12 '15 at 11:37
• The way I read the question: In the cyclic group of order $6$, additive group of $\frac {\mathbb Z}{6 \mathbb Z}$ say, the group generated by $0$ has order $1$, the groups generated by $1$ and $5$ have order $6$, the groups generated by $2$ and $4$ have order $3$, the group generated by $3$ has order $2$...so the sum in question is: $1+2^*6+2^*3+2=21$. – lulu Sep 12 '15 at 11:44

Note that this is true if the group is cyclic (of order $n$). To see this, note that your sum is $$\sum_{d|n} d\varphi(d)$$. Look at each term in that sum. If $d$ is even the term is even, but if $d>1$ is odd then $\varphi(d)$ is even so the only odd term in the sum is $1^*\varphi(1)=1$.
For a general group, with elements $\{e=g_1,g_2, ..., g_n\}$ First look at the group generated by $g_2$. Computing your sum for the elements in that group you get an odd number by our first result. But after that you are summing over the elements in cyclic groups without adding in the identity element (as you already included that in your first sum), hence you always get an even number.
• sure. the sum is over all the divisors of $n$. $\varphi(n)$ means the euler phi function. Thus, for $n=6$ the divisors are $d=1,2,3,6$ and the sum is $$1^*\varphi(1)+2^*\varphi(2)+3^*\varphi(3)+6^*\varphi(6)=1+2+6+12=21$$ – lulu Sep 12 '15 at 14:38


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This is a problem from Kaplan Practice Test : A TV

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This is a problem from Kaplan Practice Test : A TV [#permalink]

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This is a problem from Kaplan Practice Test :

A TV salesman makes a profit of 25 percent on all sales. How many TVs will he have to sell for $325 each to make a totl profit of atlest$200 ?

Kaplan argues that

since profit is 25% of 325 ( little over $80 per tv) he has to sell 3 tv sets to cross$200 mark ..

But can we calculate profit on selling price ?

If he sells a tv set for $325 and makes a 25% profit, he must have bought it for (100/125) * 325 =$ 260

that means , he makes a profit of $65 per tv he sells .. So to cross$200 mark, he must sell at least 4 TV sets ..

_________________

Brainless

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Joined: 14 Aug 2003
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25 Aug 2003, 13:30
to me, making a 25% profit in a sale of 100$is to make a profit of 25$
25 Aug 2003, 13:30
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# Find rational numbers a and b such that

Question:

Find rational numbers a and b such that

(i) $\frac{\sqrt{2}-1}{\sqrt{2}+1}=a+b \sqrt{2}$

(ii) $\frac{2-\sqrt{5}}{2+\sqrt{5}}=a \sqrt{5}+b$

(iii) $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=a+b \sqrt{6}$

(iv) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}$

Solution:

(i)

$\frac{\sqrt{2}-1}{\sqrt{2}+1}$

$=\frac{\sqrt{2}-1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}$

$=\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2})^{2}-1^{2}}$

$=\frac{2+1-2 \sqrt{2}}{2-1}$

$=3-2 \sqrt{2}$

$\therefore \frac{\sqrt{2}-1}{\sqrt{2}+1}=3+(-2) \sqrt{2}=a+b \sqrt{2}$

$\Rightarrow a=3, b=-2$

(ii)

$\frac{2-\sqrt{5}}{2+\sqrt{5}}$

$=\frac{2-\sqrt{5}}{2+\sqrt{5}} \times \frac{2-\sqrt{5}}{2-\sqrt{5}}$

$=\frac{(2-\sqrt{5})^{2}}{(2)^{2}-(\sqrt{5})^{2}}$

$=\frac{4+5-4 \sqrt{5}}{4-5}$

$=\frac{9-4 \sqrt{5}}{-1}$

$=-9+4 \sqrt{5}$

$\therefore \frac{2-\sqrt{5}}{2+\sqrt{5}}=4 \sqrt{5}+(-9)=a \sqrt{5}+b$

$\Rightarrow a=4, b=-9$

(iii)

$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=\frac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

$=\frac{3+2+2 \times \sqrt{3} \times \sqrt{2}}{3-2}$

$=5+2 \sqrt{6}$

$\therefore \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=5+2 \sqrt{6}=a+b \sqrt{6}$

$\Rightarrow a=5, b=2$

(iv)

$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}$

$=\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}$

$=\frac{35-20 \sqrt{3}+14 \sqrt{3}-24}{(7)^{2}-(4 \sqrt{3})^{2}}$

$=\frac{11-6 \sqrt{3}}{49-48}$

$=11-6 \sqrt{3}$

$\therefore \frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=11+(-6) \sqrt{3}=a+b \sqrt{3}$

$\Rightarrow a=11, b=-6$


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# Almost sure convergence of sequences that are equal in distribution

I got two sequences of random variables $$(X_n)_{n \in \mathbb N}$$ and $$(Y_n)_{n \in \mathbb N}$$ and know that

1) $$X_n \to 0, \, n \to \infty,$$ almost surely 2) For all $$n \in \mathbb N$$ the sequences are equal in distribution: $$X_n \stackrel{d}{=} Y_n$$

Now I'm wondering, does this imply that also $$Y_n \to 0, n \to \infty,$$ almost surely?

I don't have a simple counterexample at the moment but you can argue using Skorohod's Theorem: let $$(Y_n)$$ tend to $$0$$ is probability but not almost surely. Since $$Y_n \to 0$$ in distribution Skorohod's Theorem tells us that there exits random variable $$X_1,X_2,...$$ such that $$X_n \to 0$$ almost surely and $$X_n$$ has same distribution as $$Y_n$$ for each $$n$$.
A better example: consider $$[0,1)$$ with Lebesgue measure. Arrange the intervals $$[\frac {i-1} {2^{n}},\frac i {2^{n}})$$, $$1 \leq i \leq 2^{n}$$, $$n \geq 1$$ in as sequence using the 'natural' ordering. Let $$Y_1,Y_2,..$$ be the indicator functions of these intervals. Now form $$X_1,X_2,...$$ by replacing $$[\frac {i-1} {2^{n}},\frac i {2^{n}})$$ by $$[0,\frac 1 {2^{n}})$$ for each $$i$$. Then $$X_n \to 0$$ almost surely, $$Y_n$$ does not tend to $$0$$ almost surely and $$X_n$$ has the same distribution as $$Y_n$$ for each $$n$$.
• Thanks. So if I get this right, this means the statement does not hold on the original probability space that $(X_n)_n$ and $(Y_n)_n$ are defined on, but using Skorohods theorem I can define those sequences on some other space in which the almost surve convergence of $(Y_n)$ holds? – Bazzan Jul 19 at 10:14


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# Interpreting precision and recall graphs

i have a CNN for sentiment analysis whose precision and recall for validation data over 10 training epochs is (average:macro):

The dataset contains more positive samples than negatives.I have problem interpreting the results.It seems like it has higher recall and lower precision. One way is to say that since there are so many positive samples, there are more examples that can become false negatives leading to smaller recall and high precision. Or if i try to interpret my own results it can be that since number of positive samples is bigger,the classifier is biased towards positive class in beginning leading to more false positives hence lower Precision and higher recall. How are these graphs interpreted actually? Thanks

Firstly, note that $Precision = \frac{TP}{TP+FP}$, whereas $Recall = \frac{TP}{TP+FN}$, where $TP$ is the number of true positives, $FP$ is the number of false positives, and $FN$ is the number of false positives.

Since you have mentioned that your data is skewed towards the positive case, (depending on how you have trained your model and chosen your probability cutoff) your predictions will hence likely be biased towards the positive case also.

This means your model will be more likely to predict an observation as being positive than negative, leading to more frequent false positives occurring. Since $Precision$ is inversely proportional to the number of false positives, $FP$, this increase in $FP$ leads to a larger decrease in $Precision$ relative to the decrease in $Recall$.

This happens as the bias in the data towards the positive class means on average, the negative class is less likely to be predicted, and hence less 'negatives' are predicted compared to what we would expect with a more balanced dataset. Less negative predictions explain the relative decrease in less false negative predictions, and why $Recall$ is higher, by a similar logic to the above.

Side note: For your graphs, are they zero-indexed (0-9) when they should be 1-10? Not a big issue but it may make it harder to read results off the graph if you constantly have to shift the x-index forward by one, and so forth.

• Thanks. Thats what i had thought as well so I calculated scores for each class separately which made me confused a bit. e.g. the precision graph shows high values for positive class which suggests lowest FP (incase of positive class meaning a negative sample labelled as positive) . Similarly precision is lower for negative ,higher FP (incase of negative class,meaning a positive sample labelled as negative one) .This made it look like the classifier is predicting negative class more which is opposite to what i understood from average graph..Any help is appreciated. – Ayza Jul 29 '18 at 9:42
• Yes, the $Precision$ for the negative case is lower as more negative cases are being classified as (false) positives, leading to a lower proportion of true negatives in the data due to class imbalance, skewing the ratio of true negatives to total negatives towards 0, lower than the $Precision$ for the positive case. – Data Science Officer Jul 29 '18 at 10:48
• Does the meaning of true positives or false positive changes w.r.t class? e.g. false positive w.r.t positive class is something actually negative labeled as positive..right? would it be same w.r.t negative as well or the meaning changes that something actually positive labeled negative? – Ayza Jul 29 '18 at 11:03


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## Notes for "Transitive action on finite points of a full shift and a finitary Ryan’s theorem"

In this paper, as a corollary of my transitive action, I obtained a finite set of automorphisms whose centralizers intersect to the shift group, which led me to define the automorphism group invariant $$k(X)$$. Johan Kopra has recently proved that $$k(X) = 2$$ for all transitive SFTs containing a unary point $$0^{\mathbb{Z}}$$. A main part of his construction is to show that any such SFT admits an automorphism where finite-support points turn into a starfleet over a smaller subshift, i.e. spaceships that never collide again. (On the full shift where I worked with this part is trivial, and my paper rather spends its time on details of the transitive action, which does not need to be done for getting just the Ryan's theorem part.)

Of course, after his result, we had $$k(X) \in \{2, \infty, \bot\}$$ for all subshifts $$X$$ that have been considered, so my question after Lemma 7.7 in the paper (together with the conclusion of this lemma) became more interesting. Here's that question: Let $$G$$ be a (discrete) group. Define $$k^*(G)$$ as the minimal cardinality of a set $$S \subset G$$ such that $$C_G(S) = Z(G)$$. Define $$k(G) = k^*(G)$$ if $$G$$ has trivial center (i.e. $$|Z(G)|=1$$), and $$k(G)=\bot$$ otherwise. If $$k(G)=\bot$$, then the convention is that neither $$k(G)\leq n$$ nor $$k(G) \geq n$$ holds, for any cardinal $$n$$. (The subshift version of $$k^*(G)$$ would be the minimal cardinality $$k^*(X)$$ of automorphisms whose centralizers intersect to the center).

Does there exist a finite group $$G$$ with $$k(G) \geq 3$$, and more generally does every number (other than $$1$$) appear as $$k(G)$$ for some finite group $$G$$?

Johan reminded me of this question recently, so I tried my best to construct examples with prescribed $$k(G)$$ value, but after my first $$12$$ or so ideas did not work, and after noticing that the commutation graph provided some sort of qualitative difference between $$k(G) = 2$$ and $$k(G) \geq 3$$, I was no longer completely sure whether an example even exists, and asked this on MathOverflow.

This was solved quite quickly: Derek Holt shows that all finite cardinals (except $$1$$ of course) indeed appear as $$k(G)$$ for some finite group (for infinite cardinals the question is easy). YCor showed a smaller construction showing that for any $$k \in \mathbb{N}$$, either $$k$$ or $$k+1$$ appears as $$k(G)$$ for a finite group, and probably one could also show that any number appears in this construction, even if not with as simple a trick as in the construction of Derek Holt. Peter Mueller showed that $$k(\mathrm{SmallGroup}(486,176)) = 3$$ in the terminology of GAP or Magma (this group is $$3$$-generated, so $$3$$ is an upper bound, and at least an exhaustive search gives the lower bound).

Combining these with Lemma 7.7 we obtain the following:

Let $$K = \{k(X) \;|\; X \mbox{ is a minimal \(\mathbb{Z}$$-subshift} \} \). Then $$(\mathbb{N} \setminus \{1\}) \cup \{\bot\} \subset K \subset (\mathbb{N} \setminus \{1\}) \cup \{\aleph_0, \bot\}.$$

Note that all subshifts have $$k(X) \leq \aleph_0$$ because all automorphisms are cellular automata, i.e., they have local rule with a canonical finite representation with respect to any fixed expansive partition (this is the so-called Curtis–Hedlund–Lyndon theorem). In fact I am sure $$\aleph_0$$ is in $$K$$, but I do not want to try to explain the construction I have in mind (and maybe there is a simpler idea). In the countable category it's easy to characterize the corresponding $$K$$ in full:

Let $$K = \{k(X) \;|\; X \mbox{ is a countable \(\mathbb{Z}$$-subshift} \} \). Then $$K = (\mathbb{N} \setminus \{1\}) \cup \{\aleph_0, \bot\}.$$

Let us outline this construction: Here $$\aleph_0$$ and $$\bot$$ are easy: $$\aleph_0$$ is in my paper and $$k(\mathcal{L}^{-1}(0^*(12)^* + (12)^*0^*)) = \bot$$ where $$0^*(12)^* + (12)^*0^*$$ is a regular expression and $$\mathcal{L}^{-1}(L)$$ for a language $$L$$ denotes the smallest subshift whose language contains $$L$$, when this exists (it does here).

One construction for finite values is the following: let $$\Gamma$$ be a connected undirected graph with $$0 \notin V(\Gamma)$$. For any $$a \in V(\Gamma)^{\mathbb{N}}$$ let $$x_a = ...0000 a_{0}^{2^0} a_{1}^{2^1} a_{2}^{2^2}...$$ and let $$X = \overline{\bigcup_{a \in P} O(x_a) }$$ where $$P \subset V(\Gamma)^{\mathbb{N}}$$ is the set of all (cyclic) paths of the form $$u^\omega$$ in the graph where $$|u| \leq m$$, where $$m$$ is picked large enough so that there is a cycle of length at most $$m$$ which traverses every edge at least once.

Here's a proof sketch that $$\mathrm{Aut}(X) = \mathrm{Aut}(G) \times \langle \sigma \rangle$$, which then proves the claim similarly as in Lemma 7.7 of the paper by Frucht's theorem (every group is the automorphism group of an undirected graph).

If $$g : \Gamma \to \Gamma$$ is an automorphism, then clearly there is a corresponding automorphism $$f$$ of $$X$$ which simply applies $$g$$ in every cell (and preserves $$0$$), no matter what the choice of $$m$$ is.

Suppose then that $$f : X \to X$$ is an automorphism, then observe that the unary points (fixed points of the dynamics) can be defined by a first-order sentence over the language of dynamics, so $$f$$ preserves them. The unary point $$0^{\mathbb{Z}}$$ can be defined by a second-order sentence: points left-asymptotic to it but not right-asymptotic to it are not right-asymptotic to any unary point. The points which are of the form $$x_a$$ for some $$a \in P$$ are then also definable, so $$f$$ must preserve them. We claim that if $$m$$ is large enough, then $$f$$ must come from a graph automorphism as in the previous paragraph, up to a shift. More precisely, let $$g$$ be the automorphism of $$\Gamma$$ defined by how $$f$$ permutes the unary points, and associate $$f'$$ to $$g$$ as in the previous paragraph. We claim that for some $$n$$, $$f = \sigma^n \circ f'$$. To see this, pick $$x_a$$ such that $$a = u^\omega$$ and $$u$$ traverses every edge. Pick $$n$$ so that $$f \circ \sigma^{-n}$$ does not move the origin of $$x_a$$, so that $$(f \circ \sigma^{-n})(x_a) = x_{a'}$$ for some $$a' \in \Gamma$$. Since $$f \circ \sigma^{-n}$$ has a finite radius, we see that it must act as $$f'$$ on $$x_{a'}$$, and the only point bi-asymptotic to $$x_{a'}$$ is $$x_{a'}$$ itself, so $$(f \circ \sigma^{-n})(x_a) = x_{a'}$$.

For other $$x_b$$, every edge in $$b$$ appears in $$a$$, so we see that $$(f \circ \sigma^{-n})$$ also maps $$x_{b}$$ to some $$x_{b'}$$. Again by a radius consideration, $$(f \circ \sigma^{-n})$$ must act precisely as $$f'$$.

For all but $$\aleph_0$$, the previous construction even gives a countable quasiminimal subshift, that is, one that has only finitely many subsystems. I do not immediately see whether countable quasiminimal subshifts can have $$k(X) = \aleph_0$$.

Last update: 26.4.2018


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# Selling a mixture of rice

Algebra Level 2

A trader takes 26 kg of jasmine rice, which sells at $20 per kg, and mixes it with 30 kg of long grained rice, which sells at$36 per kg.

If he sells the mixture at \$30 per kg, what is his percentage profit?

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## The z-Transform

The z transform is a discrete time counterpart of Laplace transform. The Laplace transform is used on continuous signal while z transform is used for the discrete signal.

The z- transform of a general discrete signal x[n] is defined as

$X(Z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n}$

## Region of Convergence(ROC)

property 1:- The ROC of X(z) consists of a ring in the z-plane centered about the origin.

Property 2:- The ROC does not contain any poles.

Property 3:- If x[n] is of finite duration, then ROC is the entire z-plane, except possibly z=0 and/or $z=\infty$

Property 4:- If x[n] is a right sided sequence and if the circle |z|=$r_0$ is in the ROC then all finite values of z for which |z|>$r_0$ will also be in the ROC.

Property 5:- If x[n] is a left sided sequence and if the circle |z|=$r_0$ is in the ROC, then all vaues of z for which 0<|z|<$r_0$ will also be in the ROC.

Property 6:- If x[n] is two sided and if the circle |z|=$r_0$ is in the ROC, then the ROC will consist of a ring in the z-plane that includes the circle |z|=$r_0$.

Property 7:- If the z-transform X(z) of x[n] is rational, then its ROC is bounded by pole or extends to infinity.

Property 8:- If the z-transform X(z) of x[n] is rational and if x[n] is right sided then the ROC is the region in the z-plane outside the outermost pole.

Property 9:- If the z-transform X(z) of x[n] is rational and if x[n] is left sided then the ROC is the region in the z-plane outside the innermost pole.

ECE462 Survivor

Seraj Dosenbach


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# When 16 cubic meters of water are poured into an empty cubic container

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When 16 cubic meters of water are poured into an empty cubic container  [#permalink]

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08 Jan 2019, 04:13
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When 16 cubic meters of water are poured into an empty cubic container, it fills the container to 25 percent of its capacity. What is the length of one edge of the container, in meters?

A. $$2\sqrt{2}$$

B. 4

C. $$4\sqrt{2}$$

D. 8

E. 16

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Re: When 16 cubic meters of water are poured into an empty cubic container  [#permalink]

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08 Jan 2019, 04:17
16 cubic meters = 25% of total capacity.

Total capacity = 64 cubic meters.
Volume = side^3 = 64.
Side = 4.

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When 16 cubic meters of water are poured into an empty cubic container  [#permalink]

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Updated on: 08 Jan 2019, 08:21
Bunuel wrote:
When 16 cubic meters of water are poured into an empty cubic container, it fills the container to 25 percent of its capacity. What is the length of one edge of the container, in meters?

A. $$2\sqrt{2}$$

B. 4

C. $$4\sqrt{2}$$

D. 8

E. 16

volume 16 m3 fills 1/4 of cubic container
or say
x/4 = 16
x^3=64
x=4
IMO B
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Originally posted by Archit3110 on 08 Jan 2019, 04:24.
Last edited by Archit3110 on 08 Jan 2019, 08:21, edited 1 time in total.
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Re: When 16 cubic meters of water are poured into an empty cubic container  [#permalink]

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08 Jan 2019, 05:13
Archit3110 wrote:
Bunuel wrote:
When 16 cubic meters of water are poured into an empty cubic container, it fills the container to 25 percent of its capacity. What is the length of one edge of the container, in meters?

A. $$2\sqrt{2}$$

B. 4

C. $$4\sqrt{2}$$

D. 8

E. 16

volume 16 m3 fills 1/4 of cubic container
or say
x/4 = 16
x=64
x=4
IMO B

Hey Archit3110, The highlighted portion should be $$x^3$$
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Re: When 16 cubic meters of water are poured into an empty cubic container &nbs [#permalink] 08 Jan 2019, 05:13
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## Thinking Mathematically (6th Edition)

Consider the conditional statement that is written below: “If Ram is ill, then he is vomiting”. Take $p$statement as“Ram is ill” and $q$ statement as “Ram is vomiting”. Then, write symbol notation of the conditional statement: $p\to q$ It signifies that it is a conditional statement if Ram is ill then he must be vomiting. Then write, the provided statement into a converse statement as the converse statement is the reverse of the statement. So, the converse statement is as follows: If Ram is vomiting, then he is ill. It’s symbolic statement, $q\to p$ It signifies that it is basically a conditional statement, if Ram is vomiting, then he is ill. The truth value of a conditional statement may or may not be changed if first part and the second part of a conditional statement are reversed then that statement $q\to p$ is known as the converse of a conditional statement, which is$p\to q$. Further, write the provided statement into an inverse statement as the inverse statement is the negation of the statement. So, the inverse statement is as follows: If Ram is not ill, then he is not vomiting. It’s symbolic statement, $\sim p\to \sim q$ It signifies that it is a basically conditional statement if Ram is not ill then he is not vomiting. The truth value of a conditional statement may or may not be changed if first part and the second part of a conditional statement are negated them individually, then that statement $\sim p\to \sim q$ is known as the converse of a conditional statement, which is$p\to q$. Here, a truth table is listed below to show that their truth values need not be necessarily true: $p$ $q$ $p\to q$ $q\to p$ $\sim p\to \sim q$ T T T F F Example: Take a conditional statement that is written below: “If Ram is ill then he is vomiting”. Then write its converse statement, If Ram is vomiting, then he is ill. Further, write its inverse statement, If Ram is not ill, then he is not vomiting.