text:
## Calculus (3rd Edition)

\begin{align*} \lim _{x \rightarrow 0-} f(x)&=\lim _{x \rightarrow 0+} f(x)=1\\ \lim _{x \rightarrow 2-} f(x)&=\lim _{x \rightarrow 2+} f(x)=\infty\\ \lim _{x \rightarrow 4-} f(x)&=-\infty\\ \lim _{x \rightarrow 4+} f(x)&=\infty\\ \end{align*} The function is both left- and right-continuous at $x = 0$ and neither left- nor right-continuous at $x = 2$ and $x = 4$.
To find the limits, we notice what values the function tends toward to the left and right of each point. \begin{align*} \lim _{x \rightarrow 0-} f(x)&=\lim _{x \rightarrow 0+} f(x)=1\\ \lim _{x \rightarrow 2-} f(x)&=\lim _{x \rightarrow 2+} f(x)=\infty\\ \lim _{x \rightarrow 4-} f(x)&=-\infty\\ \lim _{x \rightarrow 4+} f(x)&=\infty\\ \end{align*} We see that the function is smoothly connected at $x=0$; thus, the function is both left- and right-continuous at $x = 0$. Because we get an infinite limit at the other points, we know that the function is neither left- nor right-continuous at $x = 2$ and $x = 4$.


text:
# Does the limit of a convergent sequence depend on the norm?

Let $X$ be a vector space, and $\|\cdot\|_1$ and $\|\cdot\|_2$ two different (non-equivalent) Norms on $X.$ Let $(x_n)\subset X$ be a sequence and $x\in X$ such that $\lim_{n\to\infty}\|x_n-x\|_1=0.$ The question is: If $\lim_{n\to\infty}\|x_n-x\|_2>0$ can you say that the sequence $(x_n)$ does not converge (to any other limit) with respect to $\|\cdot\|_2$? Proof?, example? Thanks in advance!

-
Is $X$ a finite dimensional space or infinite dimensional space? – Mhenni Benghorbal Dec 5 '12 at 8:39
@Mhenni: As you rightly pointed out, the hypothesis implies that $X$ must be infinite dimensional. – Jonas Meyer Dec 5 '12 at 8:40
Albert: Note that $\lim\limits_{n\to\infty}\|x_n-x\|_2>0$ is not the negation of $\lim\limits_{n\to\infty}\|x_n-x\|_2=0$. You just want to say that $\|x_n-x\|_2$ does not converge to $0$, i.e., that $(x_n)$ does not converge to $x$ in $\|\cdot\|_2$, which does not imply that $\lim\limits_{n\to\infty}\|x_n-x\|_2$ exists. (However, it is equivalent to the limsup being positive.) – Jonas Meyer Dec 5 '12 at 8:56
You are right, thanks for pointing that out! – Albert Dec 6 '12 at 10:46
In finite dimensions space the norms are equivalent here . However, that is not true in infinite dimensions spaces. See here. – Mhenni Benghorbal Apr 30 '13 at 5:55

This was asked and answered in a MathOverflow question, Example of sequences with different limits for two norms. The answer is that there are examples where $(x_n)$ converges in both norms, to different limits.
@Albert: That is incorrect. The answers give distinct limits that exist within a single set, with the two different norms. Take the accepted answer by Denis Serre, for example. The limit in $\|\cdot\|_1$ is the zero function, and the limit in $\|\cdot\|_2$ is the constant function always equal to $1$. Both of these are trigonometric polynomials functions, hence in the space, and they are the (unique) limits for the respective norms of the same sequence. The fact that the sequence is constructed from a function not in the space is not a problem; that function is not either of the limits. – Jonas Meyer Dec 6 '12 at 15:13

format: plain


text:
# If A=[0 -tan(alpha)/2] tan(alpha)/2 0 ] and I is an identity matrix of order 2, then (I-A)[ cos(alpha) -sin(alpha) sin(alpha) cos(alpha)]is equal to?(A)A(B)I(C)I+A(D)none of the above

We prove that, $I+A=I-A\left[\begin{array}{cc}\mathrm{cos}\alpha & -\mathrm{sin}\alpha \\ \mathrm{sin}\alpha & \mathrm{cos}\alpha \end{array}\right]$

Hence, the correct option is C.

• 13
What are you looking for?


text:
Timelord64 - 11 months ago 61
Android Question

# Resources$NotFoundException trying to set up a RecyclerView in Android Studio I've been trying to set up a RecyclerView for the past two days, and most of it appears to work as intended. However, whenever I attempt to run my app, it crashes with a fatal exception: E/AndroidRuntime: FATAL EXCEPTION: main Process: com.mad.exercise5, PID: 16634 android.content.res.Resources$NotFoundException: String resource ID #0x1
at android.content.res.Resources.getText(Resources.java:342)
at android.support.v7.widget.ResourcesWrapper.getText(ResourcesWrapper.java:52)
at android.widget.TextView.setText(TextView.java:4244)
at (...)

I have been following the slideview tutorials from YouTube, parts 1-4. In the context of the tutorials, the RecyclerView is being set up in a fragment, which I have not gone in to yet. Just in case the error came from creating the RecyclerView in the main activity, I also tried following a tutorial at StackTips.

To the best of my ability, I have been able to stop the error by commenting out the line that sets the Layout Manager (
mRecyclerView.setLayoutManager(new LinearLayoutManager(this));
), but the error appears to originate from the
onBindViewHolder
. Specifically, setting text to my ViewHolder TextViews.

Googling the error, and doing some research, I found lots of cases where this error popped up in other questions and issues. However, the problem always seems to be that the user is calling for a resource, but directly referencing it as an id. That is, they would use '
String string = R.string.string
Why am I receiving a Resources$NotFoundException, in my RecyclerView example activity? The two relevant java classes are as follows: MainActivity.java public class MainActivity extends AppCompatActivity { private ArrayList<Train> mTrains; private TrainAdapter mTrainAdapter; private RecyclerView mRecyclerView; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.activity_main); // Create the Recycler View mRecyclerView = (RecyclerView) findViewById(R.id.recycler_view); mRecyclerView.setLayoutManager(new LinearLayoutManager(this)); mTrains = FillTrainData(); mTrainAdapter = new TrainAdapter(this, mTrains); mRecyclerView.setAdapter(mTrainAdapter); } private ArrayList<Train> FillTrainData() { ArrayList<Train> mTempTrainData = new ArrayList<>(); String[] mPlatform = new String[]{"platA", "platB", "platC", "platD"}; String[] mStatus = new String[]{"arrive", "notarrive"}; String[] mDestination = new String[]{"destA", "destB", "destC", "destD"}; String[] mDestinationTime = new String[]{"destA", "destB", "destC", "destD"}; int[] mArrivalTime = new int[]{1,2,3,4}; for(int i = 0; i < 4; i ++) { Train mTrain = new Train(mPlatform[i], mArrivalTime[i], mStatus[i%2], mDestination[i], mDestinationTime[i]); mTempTrainData.add(mTrain); } return mTempTrainData; } } TranAdapter.java public class TrainAdapter extends RecyclerView.Adapter<TrainAdapter.CustomViewHolder> { private Context mContext; private ArrayList<Train> mTrains; public TrainAdapter(Context context, ArrayList<Train> trains) { mContext = context; mTrains = trains; } @Override public CustomViewHolder onCreateViewHolder(ViewGroup parent, int index) { View view = LayoutInflater.from(parent.getContext()) .inflate(R.layout.train_item, null); CustomViewHolder viewHolder = new CustomViewHolder(view); return viewHolder; } @Override public void onBindViewHolder(CustomViewHolder viewHolder, int position) { Train currentTrain = mTrains.get(position); viewHolder.platformText.setText(currentTrain.GetPlatform()); viewHolder.arrivalTimeText.setText(currentTrain.GetArrivalTime()); viewHolder.statusText.setText(currentTrain.GetStatus()); viewHolder.destinationText.setText(currentTrain.GetDestination()); viewHolder.destinationTimeText.setText(currentTrain.GetDestinationTime()); } @Override public int getItemCount() { return (mTrains != null ? mTrains.size() : 0); } class CustomViewHolder extends RecyclerView.ViewHolder { TextView platformText; TextView arrivalTimeText; TextView statusText; TextView destinationText; TextView destinationTimeText; public CustomViewHolder(View itemView) { super(itemView); platformText = (TextView)itemView.findViewById(R.id.train_item_platform_text); arrivalTimeText = (TextView)itemView.findViewById(R.id.train_item_arrival_time_text); statusText = (TextView)itemView.findViewById(R.id.train_item_status_text); destinationText = (TextView)itemView.findViewById(R.id.train_item_destination_text); destinationTimeText = (TextView)itemView.findViewById(R.id.train_item_destination_time_text); } } } Answer Source First Read Document of TextView. your arrival time is int and TextView have two type of setText method first is TextView.setText(String text); and second is TextView.setText(int resource); when you write viewHolder.arrivalTimeText.setText(currentTrain.GetArrivalTime()); then you second type of method will call. int that case you have four value 1,2,3,4 means it's like viewHolder.arrivalTimeText.setText(1); at this time R.java may not have resource of value "1"; that's why you got android.content.res.Resources$NotFoundException: String resource ID #0x1


text:
Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01rr1721032
DC FieldValueLanguage
dc.contributor.authorTseng, Tammy-
dc.date.accessioned2019-07-12T14:22:59Z-
dc.date.available2019-07-12T14:22:59Z-
dc.date.created2019-04-08-
dc.date.issued2019-07-12-
dc.identifier.urihttp://arks.princeton.edu/ark:/88435/dsp01rr1721032-
dc.description.abstractStudies of White flight from non-White minorities rarely distinguish Asians as a unique minority group. However, Asians' heterogeneous economic and educational outcomes may imply that Asian immigrants induce a differential response in White residents than other minorities do. I propose a theoretical model of White school district residence decisions which assumes that Asian inflows generate two countervailing effects: a negative amenity arising from White racial distaste, and a positive amenity arising from positive peer effects of Asians in schools. Net flight is determined by the degree of White response to each of these effects. I employ a common shift-share immigration instrument to address endogeneity in Asian settlement and find that in a reasonable subset of California public school districts between 2001-2016, each new Asian student causes approximately 1.5 White peers to leave, on average. The magnitude of flight implies both that White racial distaste for Asians exists, beyond racially-agnostic housing price effects, and that the magnitude of Whites' racial distaste outweighs the positive preference they have for any improvements in school quality that higher Asian enrollment also generates. An additional contribution of this paper is to show that the shift-share immigration instrument, when decomposed into only Asian components, fails to work over the time period that a significant share of immigration literature has used in predicting aggregate US inflows.en_US
dc.format.mimetypeapplication/pdf-
dc.language.isoenen_US
dc.titleWhite Flight from Asian Immigration: Evidence from California Schoolsen_US
dc.typePrinceton University Senior Theses-
pu.date.classyear2019en_US
pu.departmentEconomicsen_US
pu.pdf.coverpageSeniorThesisCoverPage-
pu.contributor.authorid961035530-
pu.certificateCenter for Statistics and Machine Learningen_US
Appears in Collections:Economics, 1927-2020

Files in This Item:
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text:
Exponential Growth and Decay

• June 4th 2010, 02:01 PM
Hockey_Guy14
Exponential Growth and Decay
a) If a boat slows according to an exponential function, and its initial speed was 40km/h, what does the formula need to be if the boat is only going 10km/h and it has been 10 minutes since letting off the gas? Define the variables

b) How fast was the boat going 1 minute after letting off the gas?

c) How fast will it be going 15 minutes after letting off the gas?

d) How long does it take to cut the boats original 40km/h speed in half?

Im having trouble figuring out the formula needed to determine the answers. Any help would be great. Thanks
• June 4th 2010, 02:23 PM
e^(i*pi)
Quote:

Originally Posted by Hockey_Guy14
a) If a boat slows according to an exponential function, and its initial speed was 40km/h, what does the formula need to be if the boat is only going 10km/h and it has been 10 minutes since letting off the gas? Define the variables

b) How fast was the boat going 1 minute after letting off the gas?

c) How fast will it be going 15 minutes after letting off the gas?

d) How long does it take to cut the boats original 40km/h speed in half?

Im having trouble figuring out the formula needed to determine the answers. Any help would be great. Thanks

The standard exponential decay formula is $A(t) = A_0e^{-kt}$ where

• A(t) = Amount left at any given time
• A_0 = Amount at time 0
• e = the base of the natural logarithm
• k = a positive constant
• t = time

In part a you are given A(t), A_0 and t so use those to find k
• June 4th 2010, 02:52 PM
Hockey_Guy14
We dont use that formula, we use Y=A(1+r)^n when trying to solve these, so any help on how to arrange it using this formula?
• June 6th 2010, 05:40 AM
skeeter
Quote:

Originally Posted by Hockey_Guy14
We dont use that formula, we use Y=A(1+r)^n when trying to solve these ... in future, please make sure that such information is provided in your original post.

$v = 40(1+r)^t$ , where $v$ is in km/hr and $t$ is in hrs

at $t = \frac{1}{6}$ hr (10 minutes) , $v= 10$

$10 = 40(1+r)^{\frac{1}{6}}$

solve for $r$ to get the boat's speed as a function of time.
• June 6th 2010, 05:42 AM
Hockey_Guy14
Ahh I get it now..thanks alot eh..appreciate it :)


text:
# Finding mean and total probabilities

A population of widgets produced by a machine has weights with mean of 200mg and standard deviation of 10mg. If a random sample of 36 was taken,

a) What is the probability that the mean weight is at least 196mg?

b) What is the probability that the total weight is at most 3100mg?

I think I'm on the right trick with part a), but my approach to part b) must contain some kind of mistake...

a) Standardizing...

$E(M_{36}) = 200$

$var(M_{36}) = \frac{10^2}{36} = \frac{25}{9}$

$P(M_{36} \geq 196) = P(\frac{M_{36} - 200}{\sqrt\frac{25}{9}} \geq \frac{196 - 200}{\sqrt\frac{25}{9}}) = P(Z \geq \frac{196 - 200}{\frac{5}{3}}) = P(Z \geq \frac{196 - 200}{\frac{5}{3}}) = P(Z \geq -2.4) = 1 - P(Z < 2.4) = 1 - .9918 = .0082$

b) Standardizing...

$E(S_{36}) = 36 * 200 = 7200$

$var(S_{36}) = 36 * 10^2 = 3600$

$P(S_{36} \leq 3100) = P(\frac{S_{36} - 7200}{\sqrt{3600}} \leq \frac{3100 - 7200}{\sqrt{3600}}) = P(Z \leq \frac{3100 - 7200}{60}) = P(Z \leq \frac{-205}{3})$

This is where I get stuck.

• So far so good. The number in part b) is far away from mean, in terms of the standard score you computed, so you may answer something like approximately zero. – BGM Oct 10 '17 at 11:22
• Do we know that the distribution of weights for the population of widgets is normal, or do we just know that the distribution has the given mean and standard deviation? – MMASRP63 Oct 12 '17 at 3:14
• Your calculations are fine. It is simply extremely unlikely that the total weight does not exceed 3100 mg. – H. H. Rugh Oct 12 '17 at 17:15

This, definitely, is not a complete answer, but I believe it might be helpful. The key point, I would say, is to understand how your samples are drawn, i.e., to define the joint distribution over 36 random variables
\begin{align} p(x_{1},\cdots,x_{36}), \label{eq:joint} \end{align} then your problems can be formalized through expectations and solved using Chebychev's inequality. That is,

(a) \begin{align} E[g(x_{1},\cdots,x_{36})] = \int g(x_{1},\cdots,x_{36})p(x_{1},\cdots,x_{36}) \end{align} where \begin{align} g(x_{1},\cdots,x_{36}) = \frac{1}{36}\sum^{36}_{i=1} x_{i} \end{align} (b) is define as \begin{align} E[f(x_{1},\cdots,x_{36})] = \int f(x_{1},\cdots,x_{36})p(x_{1},\cdots,x_{36}) \end{align} where \begin{align} f(x_{1},\cdots,x_{36}) =\sum x_{i} \end{align}

Hint on solving both equations:

The probablity of the samples can be factored as $p(x_{1},\cdots,x_{36}) = p(x_{1}\mid x_{2},\cdots,x_{36}),\cdots,p(x_{35}|x_{36})$. Now, this amounts to knowing how the samples are drawn. e.g., are the samples drwan independnetly of each other. Using such kind of prior knowldege about your samples, you can try to compute the expectations given in (a) and (b) and use the Chebychev’s inequality to solve for you specific problem. Fo instance,

\begin{align} p(f(x_{1},\cdots,x_{36}) \geq 3100) \leq \frac{E[f(x_{1},\cdots,x_{36})]}{3100} \end{align}

which you can use to compute the probability of it being $\leq$ 3100 directly.

Good luck.

Your calculations in both parts are correct, assuming that the population is Normal; otherwise, these are Normal approximations relying on the Central Limit Theorem and the fact that the term "random sample" ordinarily refers to observations that are independent and identically distributed (i.i.d.) random variables.

Note that $P(Z \leq \frac{-205}{3})<<0.0001$: as you may recall, more than $99.99$% of any Normal distribution falls within $4$ standard deviations of the mean, whereas here we have $\frac{205}{3}= 68\frac{1}{3}$ standard deviations!

Also, note that $S_n = n\,M_n,$ so when you standardize $S_n$ you get the same random variable $Z$ as when you standardize $X_n$:

$$\frac{S_n - E[S_n]}{\sqrt{V[S_n]}}=\frac{n\,M_n - E[n\,M_n]}{\sqrt{V[n\,M_n]}}=\frac{n\,M_n - n\,E[M_n]}{\sqrt{n^2\,V[M_n]}}=\frac{M_n - E[M_n]}{\sqrt{V[M_n]}}=Z.$$

In other words, both of your probabilities pertain to the distribution of $Z$, which it appears you are expected to assume is at least approximately Standard Normal.

You are right in the answer to part (b). However, I think the answer to part (a) is incorrect.

Part a:

Part (a) asks the probability that the sample mean is in the range [true mean - 4mg, true mean + $\infty$mg] This amswer should be more than 50% since it fully covers the higher side of the true mean (which is 50%) plus some part on the lower side.

$P(Z≥−2.4)=1−P(Z<-2.4)$ you missed the minus sign above.

We should have $P(Z≥−2.4)=P(Z<2.4)$ (since the distribution is symmetric). Finally,we have $P(Z<2.4) =.9918$

So, answer to part a is $.9918$.

Part b Your derivation is right. The last step is to evaluate the probability $P(Z < -K)$ where $K$ is a large positive number.

Now, $P(Z<-K)=P(Z>K)$. This is very close to zero, so don't try plugging this in directly into some function call. A common approximation is the following: $P(Z>K)\leq e^{-K^2/2}/(K\sqrt{2\pi})$ Your class notes might give you a different approximation - you can use that.
For part b, we need to evaluate the above for $K=205/3=68.33$. $e^{-K^2/2}=e^{-2334}=10^{-1014.}$

Answer to part b is $10^{-1014}/171.2$ For all practical purposes, you can treat this as zero.


text:
# [OS X TeX] Automatic labeling for references

Peter Pagin peter.pagin at philosophy.su.se
Sat Apr 5 17:03:27 EDT 2008

Alain Schremmer wrote:
>
> On Apr 5, 2008, at 1:02 PM, Peter Pagin wrote:
>
>> I thought a little more about it.
>>
>> Why not define a new counter, \newcounter{labb},
>> and a new command, like
>> \newcommand[1]{\lab}{\label{#1}\stepcounter{labb}}
>
> 1) Without writing anything in the document, I am getting the error
> message
>
> Missing \endcsname inserted.
>        \@ne
> l.82             \newcommand[1]
>             {\lab}{\label{#1}\stepcounter{labb}}
>
> Companion2ed p905 says
>
>     Missing \endcsname inserted
>         This error can arise from using commands as part of the name
> of a counter or environment
>
> So, the error seems to occur as LaTeX is reading the \newcommand but I
> have no idea what the above means.
>
>> Then in the document run \lab{newlabel}\thelabb. Typeset and check
>> the output result of \thelabb. Replace "newlabel" with that result,
>> and delete "\thelabb". One has to make sure to get the value from the
>> end of the document, even the label is put into the middle. So it is
>> sort of quarter-automatic. But if 391 \lab commands have been given,
>> the next gets the label content "392".
>> Maybe of no interest.
>
> I have a feeling that this is what I want but I won't know for sure
> until I can fix the \newcommand,
>
> Hopeful regards
> --schremmer
\newcommand{\laban}[1]{\label{#1}\stepcounter{labb}}

Peter
> ------------------------------------------------------------------------
>
>
> TeX FAQ: http://www.tex.ac.uk/faq
> List Reminders and Etiquette: http://www.esm.psu.edu/mac-tex/list/
> List Info: http://email.esm.psu.edu/mailman/listinfo/macosx-tex
> List Archive: http://tug.org/pipermail/macostex-archives/
> Mac-TeX Website: http://www.esm.psu.edu/mac-tex/
>
>

--

-----------------------------------------------------
Professor
Department of Philosophy, Stockholm University
106 91 Stockholm, Sweden
tel: +46-8-162813, fax: +46-8-152226
email: peter.pagin at philosophy.su.se
http://people.su.se/~ppagin/pagineng.htm


text:
# Natural suppression of symmetry violation in gauge theories: Muon- and electron-lepton-number nonconservation

@article{Lee1977NaturalSO,
title={Natural suppression of symmetry violation in gauge theories: Muon- and electron-lepton-number nonconservation},
author={B. Lee and R. Shrock},
journal={Physical Review D},
year={1977},
volume={16},
pages={1444-1473}
}
• Published 1977
• Physics
• Physical Review D
• We analyze the circumstances under which the violations of an approximate symmetry in a unified gauge theory of weak interactions are naturally suppressed; in particular, we investigate approximate muon- and electron-type lepton-number conservation as an example of such a symmetry. Extending earlier work, we propose a unified treatment of this symmetry together with strangeness conservation by the weak neutral current and $\mathrm{CP}$ invariance. The rate for the decay \$\ensuremath{\mu… CONTINUE READING
280 Citations

#### References

SHOWING 1-10 OF 50 REFERENCES
44Such a conjecture has recently been made by
• 44Such a conjecture has recently been made by
5 The literature on the production and decay of heavy leptons is quite considerable. A partial list of references is Y.-S~ Tsai
• 5 The literature on the production and decay of heavy leptons is quite considerable. A partial list of references is Y.-S~ Tsai
• 1560
6For a recent comparison of the pxedictions of several gauge models with the data of Refs. 23 and 24 and with bubble-chamber experiments, see C
• 6For a recent comparison of the pxedictions of several gauge models with the data of Refs. 23 and 24 and with bubble-chamber experiments, see C
A. Salam, in Elementary Particle Theory: Relativistic Groups and Analyticity
• Svartholm (Almqvist and Wiksell, NATURAL SUPPRESSION OF SYMMETRY VIOLATION IN. . ~ 1471
• 1967
A5) and (A6) prove our assertions at the beginning of this Appendix; note further that V, (q') — V, (0) is finite whether or not V " and Z' have been renormalized
• Equations
B109 Phys. Lett. 57B
• B109 Phys. Lett. 57B
• 1975
Bardeen, R. Gastmans, and B. Lautrup, Nucl. Phys. B46 Phys. Rev. D
• Bardeen, R. Gastmans, and B. Lautrup, Nucl. Phys. B46 Phys. Rev. D
• 1972
Feist, invited talk at the Chicago APS meeting
• Feist, invited talk at the Chicago APS meeting
• 1977
For a brief discussion of muon-and electron-number nonconservation, see B
• For a brief discussion of muon-and electron-number nonconservation, see B


text:
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Eigenspaces of a Matrix

Tool to calculate eigenspaces associated to eigenvalues of any size matrix (also called vectorial spaces Vect).

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# Eigenspaces of a Matrix

## Eigenspaces Calculator

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## Answers to Questions (FAQ)

### What is an eigenspace of an eignen value of a matrix? (Definition)

For a matrix $M$ having for eigenvalues $\lambda_i$, an eigenspace $E$ associated with an eigenvalue $\lambda_i$ is the set of eigenvectors $\vec{v_i}$ which have the same eigenvalue and the zero vector. That is to say the kernel (or nullspace) of $M - I \lambda_i$.

### How to calculate the eigenspaces associated with an eigenvalue?

For an eigenvalue $\lambda_i$, calculate the matrix $M - I \lambda_i$ (with I the identity matrix) (also works by calculating $I \lambda_i - M$) and calculate for which set of vector $\vec{v}$, the product of my matrix by the vector is equal to the null vector $\vec {0}$

Example: The 2x2 matrix $M = \begin {bmatrix} -1 & 2 \\ 2 & -1 \end {bmatrix}$ has eigenvalues $\lambda_1 = -3$ and $\lambda_2 = 1$, the computation of the proper set associated with $\lambda_1$ is $\begin{bmatrix} -1 + 3 & 2 \\ 2 & -1 + 3 \end {bmatrix}. \begin {bmatrix} v_1 \\ v_2 \end {bmatrix} = \begin {bmatrix} 0 \\ 0 \end {bmatrix}$ which has for solution $v_1 = -v_2$. The eigenspace $E_{\lambda_1}$ is therefore the set of vectors $\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ of the form $a \begin{bmatrix} -1 \ \\ 1 \end{bmatrix}, a \in \mathbb {R}$. The vector space is written $\text{Vect} \left\{ \begin{pmatrix} -1 \\ 1 \end{pmatrix} \right\}$

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text:
# ADF: EntityImpl.refresh(…) has no effect

Many times we use entity.refresh(REFRESH_FORGET_NEW_ROWS | REFRESH_UNDO_CHANGES) to prevent any changes we made to that entity from getting committed to the DB. What the refresh() method effectively does is change the post state of the entity. (See difference between entity-state and post-state.) The post-state is later used to decided which DML operation to use for the entity, i.e. DML_DELETE, DML_INSERT or DML_UPDATE (ref).

I recently came across a code which called refresh() from the entity’s prepareForDML() but that had no effect. ADF documentation says nothing about such a behaviour. I read the EntityImpl code and got my answer there. When we are saving data to DB the call chain is like – transaction.postChanges() -> entity.postChanges() -> entity.prepareForDML() -> entity.doDML().

prepareForDML() and doDML() have first argument operation; this is the DML operation to undertake. The code to decide which DML operation to perform based on entity’s post-state is in EntityImpl.postChanges(). That method invokes prepareForDML() method with the DML operation to undertake. Changing post-state from prepareForDML() or doDML() is useless. So, post-state needs to be set no later than postChanges(). However, if you do need to do this from prepareForDML() or doDML() then you could modify the operation param’s value appropriately.


text:
## Training a Pitman-Yor process tree with observed data at the leaves (part 1)

November 12, 2014

A simple problem with hierarchical Pitman-Yor processes (PYPs) or Dirichlet processes (DP) is where you have a hierarchy of probability vectors and data only at the leaf nodes.  The task is to estimate the probabilities at the leaf nodes.  The hierarchy is used to allow sharing across children without resorting to Bayesian model averaging as I did for decision trees (of the Ross Quinlan variety) in my 1992 PhD thesis (journal article here) and done for n-grams with the famous Context Tree Weighting compression method.

The technique used here is somewhat similar to Katz’s back-off smoothing for n-grams but the discounts and back-off weight are estimated differently.  Yeh Whye Teh created this model originally, and it is published in an ACL 2006 paper, though the real theory is buried in an NUS technical report (yes, one of those amazing unpublished manuscripts). However, his sampler is not very good, so the purpose of this note is to explain a better algorithm.  The error in his logic is exposed in the following quote (page 16 of the amazing NUS manuscript):

However each iteration is computationally intensive as each $c_{uw\cdot}$ and $t_{uw}$ can potentially take on many values, and it is expensive to compute the generalized Stirling numbers $s_d (c, t)$.

We use simple approximations and caching to overcome these problems. His technique, however, has generated a lot of interest and a lot of papers use it.  But the new sampler/estimate is a lot faster, gives substantially better predictive results, and requires no dynamic memory.  Note that when the data at the leaves is also latent, such as with clustering, topic models, or HMMs, then completely different methods are needed to enable faster mixing.  This note is only about the simplest case where the data at the leaves is observed.

The basic task is we want to estimate a probability of the $k$-th outcome at a node $\theta$, call it $\hat{p}^\theta_k$, by smoothing the observed data $n^\theta_k$ at the node with the estimated probability at the parent node $\phi$ denoted $\hat{p}^\phi_k$.  The formula used by Teh is as follows:

$\hat{p}^\theta_k ~=~ \frac{n^\theta_k - t^\theta_k d}{N^\theta+c} + \frac{c+T^\theta \,d}{N^\theta+c} \,\, \hat{p}^\phi_k$             (1)

In this, the pair $(d,c)$ are the parameters of the PYP called discount and concentration respectively, where $0 \le d < 1$ and $c \ge -d$.  The $N$s and $T$s are totals, so $N^\mu=\sum_k n^\mu_k$ and $T^\mu=\sum_k t^\mu_k$ for all nodes $\mu$ in the graph.  The DP is the case where $d=0$.

For discount of $d=0$, its easy to see that for large amounts of data the probability estimate converges to the observed frequency, so this behaves well.

If the data is at the leaf nodes, where do the counts $n^\mu_k$ for non-leaf nodes $\mu$ come from?  Easy, they are totalled from their children’s $t$ counts $n^\mu_k=\sum_{\phi \in \mu} t^\phi_k$ (using $\phi \in \mu$ to denote $\phi$ are the children).

But what then are the $t^\mu_k$?  In the Chinese Restaurant Process (CRP) view of a PYP, these are the “number of tables” at the node.  To understand this, you could go and spend a few hours studying CRPs.  You don’t need to if you don’t know it already.  The way to think of it is:

The $t^\mu_k$ are the subset of counts $n^\mu_k$ that you will pass from the node $\mu$ to its parent $\phi$ in the form of a multinomial message.  So the contribution from $\mu$ to $\phi$ is a likelihood with the form $\prod_k \phi_k^{t^\mu_k}$.  The more or less counts you pass up the stronger or weaker you are making the message.

The idea is that if we expect the probability vector at $\mu$ to be very similar to the vector at $\phi$, then most of the data should pass up, so $t^\mu_k$ is close to $n^\mu_k$. In this case, the first term in Equation (1) shrinks and the loss is transferred via $T^\mu$ so the parent probability contributes more to the estimate. Conversely, if the probability vector at $\mu$ should be quite different, then only a small amount of data should pass up and $t^\mu_k$ is closer to 1 when $n^\mu_k>0$.  Consequently $T^\mu$ is smaller so the parent probability contributes less to the estimate.   Similarly, increasing the concentration makes the first term smaller and allows the parent probability to contribute more to the estimate, and vice versa.  Finally, note for the DP the $t^\mu_k$ don’t seem to appear in Equation (1), but in reality thet do they are just hidden.  They have propagated up to the parent counts so are used in the parent probability estimate $\hat{p}^\phi_k$.  So the more of them there are, the more confident you are about your parent’s probability.

PYP theory and sampling has a property shared by many other species sampling distributions that this just all works.   But there is some effort in sampling the $\vec{t}^\mu$.

This Equation (1) is a recursive formula.  The task of estimating probabilities is now reduced to:

1. Build the tree of nodes and populate the data in counts $\vec{n}^\theta$ at leaf nodes $\theta$.
2. Run some algorithm to estimate the $\vec{t}^\mu$ for nodes $\mu$, and hence the $\vec{n}^\mu$ for non-leaf nodes $\mu$.
3. Estimate the probabilities from the root node down using Equation (1).

So the whole procedure for probability estimate is very simple, except for the two missing bits:

• How do you estimate the $\vec{t}^\mu$ (and thus $\vec{n}^\mu$ for non-leaf nodes)?
• How do you estimate the discount and concentration parameters?  By the way, we can vary these per node too!

We will look at the first of these two tasks next.

### One comment

1. […] Topical insights into components of prior research and aspects of latent academia « Training a Pitman-Yor process tree with observed data at the leaves (part 1) […]


text:
# Closed geodesic on a Riemannian manifold is homotopic to a closed curve of strictly shorter length

This is an exercise from chapter 9 of do Carmo.

Let $M^n$ be an orientable Riemannian manifold with positive sectional curvature and even dimension. Let $\gamma$ be a closed geodesic in $M$, that is, $\gamma$ is an immersion of $S^1$ into $M$ that is geodesic at all of its points. Prove that $\gamma$ is homotopic to a closed curve whose length is strictly less than that of $\gamma$.

What I am really confused about is the hint:

Hint: The parallel transport along the closed curve leaves a vector orthogonal to $\gamma$ invariant (this comes from the orientability of $M$ and the fact that the dimension is even). Therefore there exists a vector field $V(t)$ parallel along the closed curve $\gamma$.

My question are:

1. Why do we need the statement about the parallel transport leaving a vector invariant at all? We know that given any smooth curve $\gamma$ and a vector $v_0$ tangent to $M$ at $\gamma(t_0)$, the parallel transport of $v_0$ exists (given by Proposition 2.6 p. 52 of do Carmo) and that the parallel transport map is an isometry so parallel transport is orthogonal to $\gamma$ at all points.

2. Assuming that I'm wrong and we do actually need the invariance statement, how does this actually follow from the orientability of $M$ and the even dimension? I thought this was from the Synge-Weinstein theorem (page 203) but I cannot see how to apply it.

• can you do this for $S^2$ and $\gamma$ the equator? – Will Jagy Aug 22 '18 at 1:39

I‘m first going to comment on your two questions and then, for completeness, I‘ll give a sketch of proof for the exercise.

1) You are right that if we look at the curve as $$t\to \gamma(e^{2\pi i t})$$ then there is a parallel vector field $$V(t)$$ along $$\gamma$$ with $$V(0)=v$$ for every $$v\in T_pM.$$ However it might be that $$V(1)\neq V(0).$$ But as now we see the curve as a mapping $$\gamma:S^1\to M$$ we also want a parallel VF $$V$$ as a map from $$S^1$$ sucht that $$V(t)\in T_{\gamma(t)}M$$ and therefore the parallel VF should have the same value when starting and when „finishing the whole round“.

If this is not quite clear, then don‘t worry: In my answer below you can still think of the curve as a map $$\gamma:[0,1]\to M,$$ but we will see that there is a parallel vector field with $$V(1)=V(0).$$

2) Let $$\omega$$ be an orientation form for $$M,$$ i.e. a non-vanishing $$n-$$form (where $$n$$ is the dimension of $$M$$). For each $$t\in [0,1]$$ we can define an orientation on the orthogonal complement of $$\gamma‘(t)$$ by declaring $$v_1,...,v_{n-1}$$ to be positively oriented if $$$$\omega_{\gamma (t)}(v_1,...,v_{n-1},\gamma'(t))>0.$$$$ Let $$A:\left(\mathbb{R}\gamma'(0)\right)^{\perp}\to \left(\mathbb{R}\gamma'(0)\right)^{\perp}$$ be the restriction of the parallel transport map along $$\gamma$$ to the orthogonal complement of $$\gamma'(0)=\gamma'(1).$$

$$A$$ is clearly an isometry and also it is orientation preserving. To see the latter just observe that when $$e_1,...,e_{n-1}$$ is a positively oriented ONB of the orthogonal complement of $$\gamma'(0)$$ and if $$E_1(t),...,E_{n-1}(t)$$ denote their parallel transports along $$\gamma,$$ then for all $$t$$ we have $$$$\omega_{\gamma (t)}(E_1(t),...,E_{n-1}(t),\gamma '(t))\neq 0$$$$ because $$E_1(t),...,E_{n-1}(t),\gamma'(t)$$ is a basis for $$T_{\gamma(t)}M$$ and therefore, as $$e_1,...,e_{n-1}$$ is positively oriented, $$$$\omega_{\gamma (t)}(E_1(t),...,E_{n-1}(t),\gamma '(t))>0.$$$$ So $$E_1(1),...,E_{n-1}(1)$$ has the same orientation as $$e_1,...,e_{n-1}.$$ Because of $$E_i(1)=A(e_i),$$ $$A$$ is orientation preserving.

Because $$n$$ is even, Lemma 3.8 in Do Carmo impliess that there is a $$v\in T_{\gamma(0)}M$$ such that $$$$V(1)=A(v)=v,$$$$where $$V(t)$$ denotes the parallel transport of $$v$$ along $$\gamma.$$

Now to the rest of the exercise: Let $$v$$ and $$V(t)$$ such as we obtained it in 2). Then $$$$h(s,t)=exp_{\gamma(t)}(sV(t))$$$$ is a variation of $$\gamma.$$ By the same calculation as in the proof of the Synge-Weinstein-Theorem we get $$$$\frac{1}{2}E''(0)=-\int_0^1K(V(t),\gamma'(t))dt<0$$$$ and so there is a $$s_0$$ such that the closed curve $$c(t):=h(s_0,t)$$ satisfies $$l(c) (details again at the end of the proof of the Synge-Weinstein-Theorem).


text:
size - Maple Help

MTM

 size
 size of an array in each dimension

 Calling Sequence size(M) size(M, n)

Parameters

 M - array n - positive integer dimension

Description

 • The size(M) function returns an array with the number of rows as the first entry, and the number of columns as the second entry.
 • When M is a row-vector, the result of size(M) will have a 1 in the entry specifying the number of rows.  Similarly when M is a column-vector, there will be a 1 in the second entry of the result.
 • When M has more than 2 dimensions, the number of entries in M will correspond to the number of dimensions of M.
 • When n is given, size(M,n) will return only the number corresponding to that dimension.  size(M,1) will give the number of rows and size(M,2) will give the number of columns.
 • When n is larger than the number of dimensions of M, 1 will be returned.

Examples

 > $\mathrm{with}\left(\mathrm{MTM}\right):$
 > $\mathrm{size}\left(\mathrm{Matrix}\left(2,3\right)\right)$
 $\left[\begin{array}{cc}{2}& {3}\end{array}\right]$ (1)
 > $\mathrm{size}\left(\mathrm{Matrix}\left(2,3\right),2\right)$
 ${3}$ (2)


text:
Math Fractions Domain of an algebraic fraction

The domain of an algebraic fraction

For fraction terms, certain numbers are sometimes not possible to set for a variable, if the denominator assumes the value zero. All permissible insertions are therefore specified as domain $D$.

!

Remember

The domain of a term is the set of numbers for which the term is defined.
It is given as follows:$D=\mathbb{Q}\backslash\{\text{number}\}$ (Domain is all rational numbers without "number")
i

Method

1. Write out the denominator of the fractional term and set it equal to zero
2. Dissolve after the variable
3. Specify the domain, exclude calculated numbers

Example

Specify the domain of the algebraic fraction: $\frac{15}{x+4}$

1. Write out the denominator of the algebraic fraction and set it equal to zero

$x+4=0$
2. Dissolve after the variable

$x+4=0\quad|-4$
$x=-4$
=> $x$ cannot be -4, otherwise it is divided by 0
3. Specify the domain

The domain is all rational numbers except -4, because if you use -4, the algebraic fraction would be invalid.
$\mathbb{D}=\mathbb{Q}\backslash\{-4\}$


text:
Are the eigenvectors of a real symmetric matrix always an orthonormal basis without change?

a real n×n matrix A is symmetric if and only if there is an orthonormal basis of Rn consisting of eigenvectors for A

Does this mean the eigenvectors of a symmetric matrix with real values always form an orthonormal basis, meaning that without changing them at all, they're always orthogonal and always have a norm of 1?

Or does it mean that based on the eigenvectors, we can manipulate them (e.g. divide them by their current norm) and turn them into vectors with a norm of 1?

There is no "the" eigenvectors for a matrix. That's why the statement in Wikipedia says "there is" an orthonormal basis...

What is uniquely determined are the eigenspaces. But you can make different choices of eigenvectors from the eigenspaces and make them orthogonal or not (and of course you can go in and out of "orthonormal" by multiplicating by scalars). In the special case where all the eigenvalues are different (i.e. all multiplicities are $1$) then any set of eigenvectors corresponding to different eigenvalues will be orthogonal.

As a side note, there is a small language issue that appears often. This is that matrices have eigenvalues, but to talk about eigenvectors you are seeing your matrix as a linear operator on a vector space (and that's of course where the notion of eigenvalue comes from)

To see a concrete example, consider the matrix $$\begin{bmatrix}1&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}$$ The orthonormal basis the Wikipedia article is talking about is $\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$.

But as the multiplicity of zero as eigenvalue is $2$, we can choose a different basis for its eigenspace, and then $\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$, $\begin{bmatrix}0\\2\\1\end{bmatrix}$ is another (not orthogonal) basis of eigenvectors.

Finally, if you don't want a basis, you can have an infinity of eigenvectors: for instance all vectors of the form $\begin{bmatrix}t\\0\\0\end{bmatrix}$, for any $t$, are eigenvectors. And all vectors $\begin{bmatrix}0\\t\\s\end{bmatrix}$, for any $t$ and $s$, are eigenvectors.

• Forgive me for not following, but just to be clear, for a 3x3 matrix the eigenvectors will always be orthogonal (and can be scaled to be orthonormal) if the matrix has 3 DISTINCT eigenvalues. Is that correct? – chessofnerd Jul 31 '15 at 2:19
• No. It will be like you say if the matrix is also selfadjoint (that is, symmetric in the real case). A $3/times 3$ matrix with three distinct eigenvalues will have a basis of eigenvectors, but not necessarily an orthogonal one. – Martin Argerami Jul 31 '15 at 2:23

There are two different things that might go wrong:

1) Eigenvectors can always be scaled. So if $v$ is an eigenvector then so is $av$ for $a\in k^\ast$. So the length one part is not automatic but can be forced easily - as you said yourself by dividing each eigenvector by it's length.

2) More importantly linear independent eigenvectors to the same eigenvalue do not need to be orthogonal. What is true however is that two eigenvectors to different eigenvalues of a symmetric matrix are orthogonal. So if each eigenvalue has multiplicity one a basis of eigenvectors is automatically orthogonal (and can be made orthonormal as above). In general we need to find an orthogonal basis of each eigenspace first, e.g. by Gram-Schmidt.

Edit: Part two is illustrated in @Martin's answer. The eigenvectors to the eigenvalue $1$ are always orthogonal to the eigenvectors to the eigenvalue $0$. However we can choose multifarious non-orthogonal bases of the eigenspace to $0$.

• So then what it should really say is that a matrix is symmetric iff it's eigenspaces are orthogonal? – Robert S. Barnes Oct 31 '12 at 8:05

No, if $v$ is an eigenvector then $\alpha v$ is also an eigenvector for any $\alpha\in F$.

In order for $v$ to be a part of an orthormal basis it must hold that $||v||=1$, but multiplying in $\alpha$ changes the norm of the vector

• which set of numbers is F? I'm not familiar with it – Neyo Jun 12 '12 at 15:01
• $F$ is your field, I belive that in your case $F=\mathbb{R}$ – Belgi Jun 12 '12 at 15:02

While we normally talk about a finite set of eigenvectors for a matrix, this finite set is not uniquely determined, and you could replace any element of it by a scalar multiple. In other words, if $v$ is an eigenvector for $\lambda$, then so is $\mu v$ for any $\mu\ne0$ in the base field. Hence in general an eigenvector need not have norm $1$, but we can divide by the norm to obtain a parallel eigenvector which does have norm $1$.

Probably you should consider $E_n$ which has only one eigenvalue - that is $1$ and only eigenvectors. Not every basis for $\mathbb R^n$ is orthonormal.

• Of course not every basis is orthonormal, I'm talking about basis of symmetric matrices here, not matrices in general. – Neyo Jun 12 '12 at 14:59
• $E_n$ is symmetric – Blah Jun 12 '12 at 15:07
• I presume that by $E_n$ you mean the identity matrix in $\mathbb{R}^{n\times n}$. – talmid Jun 12 '12 at 21:25


text:
# Citi Tops $1 Billion After Q3, Leading US Banks In FX Results ## FRONT PAGE NEW YORK--Citibank is on course to break its record earnings of 1997 following strong third quarter FX trading results, which have pushed the bank's year-to-date revenues over the$1 billion mark.

Overall, however, results were mixed among the US banks, with the biggest slide recorded by BankAmerica, which showed a 52 per cent decline from last year.

"It has certainly been an eventful quarter," says one forex manager. "I think the bread and butter FX business is doing very well at the moment--


text:
# Given a Message and Decryption Key, is it possible to find the Cipher in RSA?

This is a slight reversal from the norm but...

Suppose Alice wants to ensure that everyone knows that a message will be from her. Instead of giving out the encryption key, she broadcasts the decryption key, and encrypts her messages privately. She doesn't care that anybody can read her messages, only that someone can't imitate her.

So if I want to imitate Alice, and all I have is a decryption key and the message, is it possible to fake a message from Alice?

Edit: so to clarify, we'll assume a literate message is valid. We'll also add the constraint that the attacker wants to mimic a specific Message (M), not just any message, or previous messages will do.

• This seems to be trying to reinvent RSA signature, but in a way that I do not quite get. Suggestion: precisely state what receivers do to decide if something received is accepted as coming from Alice; then try to fool that, possibly assuming availability of cryptograms genuinely produced by Alice. For a start, do receivers get one or two numbers? Assuming one, on what grounds exactly will a receiver accept or refuse $0$? $1$? $42$? $N-1$, where $N$ is Alice's public modulus? $C+N$, where $C$ is genuinely from Alice? Assuming two, what about $(0,0)$? $(1,1)$? etc..
– fgrieu
Oct 3, 2017 at 20:23
• I'll edit the question, but for the purposes of the question, lets assume a literate message is valid. I'll also add that+ the attacker has a specific message M which they want to use, and can't just use any message they like. Oct 3, 2017 at 20:34


text:
# Chapter 8 Regularization

library("glmnet")
## Loading required package: Matrix
##
## Attaching package: 'Matrix'
## The following object is masked from 'package:tidyr':
##
##     expand
## Loading required package: foreach
##
## Attaching package: 'foreach'
## The following objects are masked from 'package:purrr':
##
##     accumulate, when
## Loaded glmnet 2.0-16
library("tidyverse")
library("broom")
UScrime <- MASS::UScrime %>%
mutate_at(vars(y, M, Ed, Po1, Po2, LF, M.F, Pop,
NW, U1, U2, GDP, Ineq, Prob, Time),
funs(log))

varlist <- c("M", "Ed", "Po1", "Po2", "LF", "M.F", "Pop", "NW",
"U1", "U2", "GDP", "Ineq", "Prob", "Time")

By default, glmnet will return and entire range of coefficients.

mod_lasso <- glmnet(as.matrix(UScrime[, varlist]), UScrime[["y"]])
mod_ridge <- glmnet(as.matrix(UScrime[, varlist]), UScrime[["y"]], alpha = 0)
bind_rows(
mutate(tidy(mod_lasso), model = "Lasso"),
mutate(tidy(mod_ridge), model = "Ridge")
) %>%
filter(term != "(Intercept)") %>%
ggplot(aes(x = dev.ratio, y = estimate, colour = term)) +
geom_line() +
facet_wrap(~ model, ncol = 1)

Alternatively, the lasso and ridge regression models are the solutions to the problems $\hat{\beta}_{\text{lasso}} = \arg \min_\beta \left\{ \sum_{i =1}^n \left(y_i - \beta_0 - \sum_{j = 1}^p \beta_j x_{ij} \right)^{2} \right\} \text{s.t.} \sum_{j = 1}^p \beta_j^2 \leq c,$ and \begin{aligned}[t] \hat{\beta}_{\text{lasso}} &= \arg \min_\beta \left\{ \sum_{i =1}^n \left(y_i - \beta_0 - \sum_{j = 1}^p \beta_j x_{ij} \right)^{2} \right\} \\ \text{s.t.}& \sum_{j = 1}^p |\beta_j| \leq c \end{aligned}

In other words, these methods try to find the $$\Vec{\beta}$$ with the smallest sum of squared errors that has a $$\Vec{\beta}$$ with a norm less than $$c$$. The value of $$c$$ corresponds to some value of $$\lambda$$ in the previous methods.

Think of $$c$$ as a fixed budget. The lasso and ridge regressions try to find the variables that explain $$y$$ the best without going over the budget (James et al. 2013, 221):

Consider the case with only coefficients: $$\beta_1$$ and $$\beta_2$$. In the lasso, we want to find the values of $$\beta_1$$ and $$\beta_2$$ $|\beta_1| + |\beta_2| \leq c$

never trust OLS with more than five regressors — Zvi Grilliches

Regularization theory was one of the first signs of the existence of intelligent inference — Zapnik

Rather than choose the best fit, there is some penalty to avoid over-fitting. This is to choose the optimal optimal point on the expected predicted value.

There are two questions

1. method of regularization
2. amount of regularization

There are several choices of the former, chosen for different reasons.

The latter is almost always chosen by cross-validation.

While OLS is okay for estimating $$\beta$$ (best linear unbiased property). However, with $$K \geq 3$$ regressors, OLS is poor.

The approaches to regularization in regression are

1. Shrink estimates to zero (Ridge)
2. Sparsity, limit number of non-zero estimates (Lasso)
3. Combination of the two (Bridge)

## 8.1 Ridge Regression

$\hat{\beta}_{\text{OLS}} = \arg \min_{\beta} \sum_{i=1}^{n} (y_i - \Vec{x}_{i} \Vec{\beta})^{2}$

Regularized regression adds a penalty that is a function of $$\beta$$. This encourages $$\beta$$ to be close to zero. $\hat{\beta}_{\text{regularized}} = \arg \min_{\beta} \sum_{i=1}^{n} (y_i - \Vec{x}_{i} \Vec{\beta})^{2} + \lambda f(\beta)$

Where $$\lambda$$ is a penalty parameter, and $$f(\beta)$$ is a function that increases in the total magnitudes of the coefficients.

• $$\lambda \to \infty$$: all coefficients are zero
• $$\lambda \to 0$$: same as OLS

How do we choose the value of $$\lambda$$?

• Currently: cross-validation
• Historically: there were some default plug-in estimators, especially for ridge regression.

Ridge regression penalizes the $$\Vec{\beta}$$ vector by the $\hat{\beta}_{\text{ridge}} = \arg \min_{\beta} \sum_{i=1}^{n} (y_i - \Vec{x}_{i} \Vec{\beta})^{2} + \sum_{k = 1}^{p} \beta_k^2$

Lasso penalizes the coefficients by an the $$L1$$ norm. Suppose we want to find the best subset of $$\leq k$$ covariates . $\hat{\beta}_{\text{lasso}} = \arg \min_{\beta} \sum_{i=1}^{n} (y_i - \Vec{x}_{i} \Vec{\beta})^{2} + \lambda \sum_{k = 1}^p |\beta_k|$

• If true distribution of coefficients is a few big ones and many small ones, LASSO will do better. If many small/modest sized effects, ridge may do better.

• LASSO does not work well with highly correlated coefficients.

• Ridge: $$\hat{\beta}_{1} + \hat{\beta}_{2} \approx (\beta_1 + \beta_2)/ 2$$.
• LASSO: Indifferent between $$\hat{\beta}_1 = 0$$, $$\hat{\beta}_2 = \beta_1 + \beta_2$$, $$\hat{\beta}_1 = \beta_1 + \beta_2$$, and $$\hat{\beta}_2 = 0$$.
• Approximate best-subset selection. Suppose that we would really like to select the best subset of $$q < k$$ coefficients and set the rest to zero (this is the variable selection problem). That is a hard problem since there are $$\binom{k}{q}$$. Lasso can be viewed as an approximation of the problem.

• Oracle property. If the true model is sufficiently sparse, we can ignore the selection stage and use OLS standard errors of the non-zero variables for inference.

Bridge regression penalizes the $$\Vec{\beta}$$ vector by the $\hat{\beta}_{\text{bridge}} = \arg \min_{\beta} \sum_{i=1}^{n} (y_i - \Vec{x}_{i} \Vec{\beta})^{2} + \lambda_1 \sum_{k = 1}^{p} |\beta_k| + \lambda_2 \sum_{k = 1}^{p} \beta_k^2$

Bridge regression has some of the properties of both ridge and Lasso. It will select correlated regressors, yet also shrink coefficients to zero for a sparse solution.

The R package glmnet is the most commonly used package to estimate Lasso, ridge, and bridge regression for linear and generalized linear models. However, these methods are common enough that all machine learning frameworks will have some implementation of them. See other packages for variations on the lasso that take into account other dependencies in the data.

How to find the value of $$\lambda$$? Cross validation. The function cv.glmet() uses cross-validation to select the penalty parameter.

## 8.2 Regularization for Causal Inference

Belloni, Chernozhukov, and Hansen (2014) propose a simple method for using Lasso for causal effects.

What’s the problem with regularized regression for causal inference? Suppose we estimate a model with the aim to recover $$\beta_1$$. $\Vec{y} = \alpha + \beta x + \gamma_1 z_1 + \cdots + \gamma_k z_{k-1} + \epsilon$ If we estimate it with a regularized model, like lasso, then $$\beta_1$$ will be shrunk in addition to the controls. If we instead do not shrink $$\beta_1$$ but we shrink the controls enough. It will be closer to not controlling for the other variables since any part of of the treatment prediction of the outcome explained by the controls will be shrunk since those coefficients are penalized, but the treatment coefficient is not.

1. Run Lasso with the outcome $$y$$ on all controls, $$z_1, \dots, \z_k$$. Keep all non-zero coefficients.

2. Run Lasso with the treatment $$z$$ on all controls, $$z_1, \dots, z_k$$. Keep all non-zero coefficients.

3. Run OLS with the outcome $$y$$ on the treatment, $$x$$, and all variables with a non-zero coefficient in either step 1 or 2.

If the true model is sparse (and asymptotics), then by the Oracle property, we can treat the standard errors of the OLS coefficients in the last step as if the selection stage did not occur.

See https://arxiv.org/pdf/1603.01700.pdf and the hdm which implements this method, and extensions to work with high dimensional data in R.

## 8.3 References

It is a few years old, but the 2015 NBER Summer course has a good introduction to machine learning that is targeted at social scientists.

### References

James, Gareth, Daniela Witten, Trevor Hastie, and Robert Tibshirani. 2013. An Introduction to Statistical Learning with Applications in R. 6th ed. Springer.


text:
Generate a cluster assignment from a CTS, SRS, or ASRS similarity matrix.

LCE(E, k, dc = 0.8, R = 10, sim.mat = c("cts", "srs", "asrs"))

Arguments

E

is an array of clustering results. An error is thrown if there are missing values. impute_missing() can be used beforehand.

k

requested number of clusters

dc

decay constant for CTS, SRS, or ASRS matrix

R

number of repetitions for SRS matrix

sim.mat

similarity matrix; choices are "cts", "srs", "asrs".

Value

a vector containing the cluster assignment from either the CTS, SRS, or ASRS similarity matrices

Other consensus functions: CSPA(), LCA(), k_modes(), majority_voting()

Johnson Liu

Examples

data(hgsc)
dat <- hgsc[1:100, 1:50]
x <- consensus_cluster(dat, nk = 4, reps = 4, algorithms = c("km", "hc"),
progress = FALSE)
if (FALSE) {
LCE(E = x, k = 4, sim.mat = "asrs")
}

x <- apply(x, 2:4, impute_knn, data = dat, seed = 1)
x_imputed <- impute_missing(x, dat, nk = 4)
LCE(E = x_imputed, k = 4, sim.mat = "cts")
#>   [1] 1 1 1 1 1 1 1 1 1 2 1 1 2 1 3 1 1 2 1 1 2 1 2 1 1 1 2 1 2 1 1 1 1 1 2 2 1
#>  [38] 1 4 1 2 1 1 1 1 1 1 1 1 2 1 1 2 2 2 1 1 1 1 1 2 2 1 1 2 2 2 1 1 2 1 2 1 1
#>  [75] 1 1 1 1 4 2 1 1 1 1 1 1 1 1 3 1 1 1 2 1 1 2 2 1 2 1


text:
# Microstation DGN v8¶

New in version 2.2.

Driver short name

DGNv8

Build dependencies

Open Design Alliance Teigha library

Microstation DGN files from Microstation version 8.0 are supported for reading and writing. Each model of the file is represented by a OGR layer.

This driver requires to be built against the (non open source) Open Design Alliance Teiga library.

DGN files are considered to have no georeferencing information through OGR. Features will all have the following generic attributes:

• Type: The integer type code as listed below in supported elements.

• Level: The DGN level number.

• GraphicGroup: The graphic group number.

• ColorIndex: The color index from the dgn palette.

• Weight: The drawing weight (thickness) for the element.

• Style: The style value for the element.

## Driver capabilities¶

Supports Create()

This driver supports the GDALDriver::Create() operation

Supports Georeferencing

This driver supports georeferencing

Supports VirtualIO

This driver supports virtual I/O operations (/vsimem/, etc.)

## Supported Elements¶

The following element types are supported in reading:

• Cell Header (2): used for polygons with holes

• Line (3): Line (2 points) geometry.

• Line String (4): Multi segment line geometry.

• Shape (6): Polygon geometry.

• TextNode (7): Container of Text elements.

• Curve (11): Approximated as a line geometry.

• ComplexString (12): Treated as line string or compound curve.

• ComplexShape (14): Treated as polygon or curve polygon.

• Ellipse (15): Approximated as a line geometry or a circular string.

• Arc (16): Approximated as a line geometry or a circular string.

• Text (17): Treated as a point geometry.

• B-Spline (21): Treated as a line geometry.

• PointString (22): Treatead as multi point.

• Shared cell reference (35): Treated as point.

Generally speaking any concept of complex objects, and cells as associated components is lost. Each component of a complex object or cell is treated as a independent feature.

## Styling Information¶

Some drawing information about features can be extracted from the ColorIndex, Weight and Style generic attributes; however, for all features an OGR style string has been prepared with the values encoded in ready-to-use form for applications supporting OGR style strings.

The various kinds of linear geometries will carry style information indicating the color, thickness and line style (i.e. dotted, solid, etc).

Polygons (Shape elements) will carry styling information for the edge as well as a fill color if provided. Fill patterns are not supported.

Text elements will contain the text, angle, color and size information (expressed in ground units) in the style string.

The various metadata items that can be set in the DGN header with the dataset creation options (see below) can be retrieved in the “DGN” metadata domain.

## Creation Issues¶

DGN files may be written with OGR with limitations:

• Output features have the usual fixed DGN attributes. Attempts to create any other fields will fail.

• Translation from OGR feature style strings back into DGN representation information is limited to a few properties of LABEL (text, font name, size, angle, color), PEN (color) and BRUSH (fill color) tools.

• POINT geometries that are not text (Text is NULL, and the feature style string is not a LABEL) will be translated as a degenerate (0 length) line element.

• Geometries which fall outside the “design plane” of the seed file will be discarded, or corrupted in unpredictable ways.

The dataset creation supports the following options:

• SEED=filename: Specify the seed file to use.

• COPY_SEED_FILE_COLOR_TABLE=YES/NO: Indicates whether the color table should be copied from the seed file. Only taken into account if SEED is specified. By default this is NO.

• COPY_SEED_FILE_MODEL=YES/NO: Indicates whether the existing models (without their graphic contents) should be copied from the seed file. This holds as well for the view groups and named views to which they are linked to. Only taken into account if SEED is specified. By default this is YES.

• COPY_SEED_FILE_MODEL_CONTROL_ELEMENTS=YES/NO: Indicates whether the existing control elements of models should be copied from the seed file. Only taken into account if COPY_SEED_FILE_MODEL=YES. By default this is YES.

• APPLICATION=string: Set Application field in header. If not specified, derived from seed file when set. Otherwise mentions the version of GDAL and the Teigha library used.

• TITLE=string: Set Title field in header. If not specified, from the seed file.

• SUBJECT=string: Set Subject field in header. If not specified, from the seed file.

• AUTHOR=string: Set Author field in header. If not specified, from the seed file.

• KEYWORDS=string: Set Keywords field in header. If not specified, from the seed file.

• TEMPLATE=string: Set Template field in header. If not specified, from the seed file.

• LAST_SAVED_BY=string: Set LastSavedBy field in header. If not specified, from the seed file.

• REVISION_NUMBER=string: Set RevisionNumber field in header. If not specified, from the seed file.

• CATEGORY=string: Set Category field in header. If not specified, from the seed file.

• MANAGER=string: Set Manager field in header. If not specified, from the seed file.

• COMPANY=string: Set Company field in header. If not specified, from the seed file.

The layer creation supports the following options:

• DESCRIPTION=string: Description associated with the layer. If not specified, from the seed file.

• DIM=2/3: Dimension (ie 2D vs 3D) of the layer. By default, 3, unless the model is reused from the seed file.


text:
# Resonances and close approaches. I. The Titan-Hyperion case

@article{Bevilacqua1980ResonancesAC,
title={Resonances and close approaches. I. The Titan-Hyperion case},
author={R. Bevilacqua and O. Menchi and A. Milani and A. Nobili and P. Farinella},
journal={The moon and the planets},
year={1980},
volume={22},
pages={141-152}
}
• R. Bevilacqua, +2 authors P. Farinella
• Published 1980
• Physics
• The moon and the planets
• AbstractThe orbits of Titan and Hyperion represent an interesting case of orbital resonance of order one (ratio of periods 3/4), which can be studied within a reasonable accuracy by means of the planar restricted three-body problem. The behaviour of this resonance has been investigated by numerical integrations, of which we show the results in terms of the Poincaré mapping in the plane of the coordinates η = ✓[(2L − 2G)] cos ( $$\tilde \omega$$ H − t)and ξ = −,✓[(2L − 2G)] sin ( \tilde… CONTINUE READING
13 Citations

#### References

SHOWING 1-10 OF 15 REFERENCES


text:
# Net force between two identical dipoles

## Homework Statement

I need to find force between two dipoles. I was thinking of finding F between all the four charges separately and then using superposition principle to get the net charge. but is there an easier way. Say, if I use the formula of E (dipole) for one dipole and multiply it with the charges in another dipole and use superposition principle , will it work?


text:
Homework Help: Help on calculating Mk

1. Nov 19, 2004

(* ^_^ *)

Help on calculating Coefficent of kinetic friction

Here is the problem: A box of books weighing 325 N moves with a constant
velocity across the floor when it is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Find the Coefficent of kinetic friction between the box and the floor.

Thanks for the help!

Last edited: Nov 19, 2004
2. Nov 19, 2004

James R

1. Draw a vector free body diagram showing all forces acting on the box.
2. Since it is moving at constant velocity the net force must be zero.
3. Resolve forces horizontally and vertically to get two equations. Remember that the friction force is $\mu N$, where $\mu$ is the coefficient of friction and $N$ is the normal reaction force.
4. Solve your two equations simultaneously to find $\mu$.

3. Nov 20, 2004

Mk

Hey! I'm Mk!

4. Nov 20, 2004

UrbanXrisis

hahaha :rofl:


text:
# How to quickly and robustly convert between mmCIF and PDB?

There is already a question on PDB/CIF to MMTF, however what is a robust way to programmatically go between PDB and CIF files?

For example I can use a python script from this gist that relies on Biopython.

id="4ckh"
wget http://www.ebi.ac.uk/pdbe/static/entry/download/$$id-assembly-1.cif.gz gunzip$$id-assembly-1.cif.gz
python cif2pdb.py $$id-assembly-1.cif$$id.pdb

This process yields a pdb structure as shown below.

The issues are:

• Requires python/Biopython to be installed.
• The output isn't perfectly clean.
• The gist is relatively undeveloped and hasn't got much evidence of testing.
• I can't really address your issues certainly so I thought I would comment rather than answer, but I've had good results using Maxit (sw-tools.rcsb.org) which is produced by the RCSB. At least for submitting to the PDB Maxit is endorsed so I would hope it's robust.
– TW93
Nov 29 '19 at 10:44

Converting a PDB file to some version of an mmCIF file is always possible provided the PDB file doesn't have any "problems". However grouping identical molecules into the same entity is non-trivial so the Biopython and BioJulia mmCIF writers (which I wrote) do not try and do this, simply assigning new entites for each molecule. These files will also be missing other things like the metadata parts of the mmCIF dictionary not involving the atoms.

In Biopython:

from Bio.PDB import PDBParser, MMCIFIO
p = PDBParser()
struc = p.get_structure("", "file.pdb")
io = MMCIFIO()
io.set_structure(struc)
io.save("file.cif")

In BioJulia:

using BioStructures
struc = read("file.pdb", PDB)
writemmcif("file.cif", struc)

Going the other way is, however, not always possible. You can run out of atom serials or chain IDs to use in the output PDB file (this is one reason for the switch to mmCIF files in general). The gist you have linked is a way to get round this.

If this is not a problem for your structure you can invert the above snippets to get a decent PDB file - see the relevant docs. Note that in both cases it will use the auth_seq_id and auth_asym_id fields to get the residue number and chain ID rather than the relabelled label versions. The test suites of Biopython and BioJulia do test these conversions pretty thoroughly.

There is also pdb-tools if you do not want to install Biopython, though both involve a simple pip install so it's not much difference.

For some speed benchmarks of various software see here and here.

• hey @jgreener! Thanks for the clarification! I'm wondering how come MMCIFIO drops metadata and such? Isn't that pretty inconvenient? let's say i need to rename (change the ids of ) certain chains in a structure and nothing else... I'll be losing all the authors' annotation, seq data etc. just by doing io.set_structure(struct), right? Are there any alternatives that you know of? Jul 16 '21 at 12:42
• Yes, you would lose that info. The readers and writers focus on parsing the atomic structural data into a unified representation that can be modified and written back out to any file type. Developer focus has gone there rather than on writing a unified representation for the metadata. My recommendation would either be to concatenate the newly generated file onto the old metadata (beware some may be wrong now) or to use the MMCIF2Dict constructor and modify list items (writing that out preserves metadata, again some metadata may now be wrong). Jul 17 '21 at 15:27
• that's super useful. I have played with MMCIF2Dict before and that'd actually be ideal to preserve and edit the metadata. Any tips for how to write that representation (the dict resultant from MMCIF2Dict) to a .cif file if not through MMCIFIO? Jul 17 '21 at 23:09
• You can write a MMCIF2Dict with MMCIFIO, just use io.set_dict(d) rather than set_structure. Jul 18 '21 at 14:17


text:
# manim.utils.hashing.CustomEncoder¶

class CustomEncoder(*, skipkeys=False, ensure_ascii=True, check_circular=True, allow_nan=True, sort_keys=False, indent=None, separators=None, default=None)[source]

Bases: json.encoder.JSONEncoder

Constructor for JSONEncoder, with sensible defaults.

If skipkeys is false, then it is a TypeError to attempt encoding of keys that are not str, int, float or None. If skipkeys is True, such items are simply skipped.

If ensure_ascii is true, the output is guaranteed to be str objects with all incoming non-ASCII characters escaped. If ensure_ascii is false, the output can contain non-ASCII characters.

If check_circular is true, then lists, dicts, and custom encoded objects will be checked for circular references during encoding to prevent an infinite recursion (which would cause an OverflowError). Otherwise, no such check takes place.

If allow_nan is true, then NaN, Infinity, and -Infinity will be encoded as such. This behavior is not JSON specification compliant, but is consistent with most JavaScript based encoders and decoders. Otherwise, it will be a ValueError to encode such floats.

If sort_keys is true, then the output of dictionaries will be sorted by key; this is useful for regression tests to ensure that JSON serializations can be compared on a day-to-day basis.

If indent is a non-negative integer, then JSON array elements and object members will be pretty-printed with that indent level. An indent level of 0 will only insert newlines. None is the most compact representation.

If specified, separators should be an (item_separator, key_separator) tuple. The default is (’, ‘, ‘: ‘) if indent is None and (‘,’, ‘: ‘) otherwise. To get the most compact JSON representation, you should specify (‘,’, ‘:’) to eliminate whitespace.

If specified, default is a function that gets called for objects that can’t otherwise be serialized. It should return a JSON encodable version of the object or raise a TypeError.

Methods

 default This method is used to serialize objects to JSON format. encode Overriding of JSONEncoder.encode(), to make our own process.

Attributes

 item_separator key_separator
default(obj)[source]

This method is used to serialize objects to JSON format.

If obj is a function, then it will return a dict with two keys : ‘code’, for the code source, and ‘nonlocals’ for all nonlocalsvalues. (including nonlocals functions, that will be serialized as this is recursive.) if obj is a np.darray, it converts it into a list. if obj is an object with __dict__ attribute, it returns its __dict__. Else, will let the JSONEncoder do the stuff, and throw an error if the type is not suitable for JSONEncoder.

Parameters

obj (Any) – Arbitrary object to convert

Returns

Python object that JSON encoder will recognize

Return type

Any

encode(obj)[source]

Overriding of JSONEncoder.encode(), to make our own process.

Parameters

obj (Any) – The object to encode in JSON.

Returns

The object encoder with the standard json process.

Return type

str


text:
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# Due to a manufacturing defect, each copy of a 300-page book is missing

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Due to a manufacturing defect, each copy of a 300-page book is missing  [#permalink]

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26 Jul 2018, 00:29
1
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Difficulty:

25% (medium)

Question Stats:

77% (01:31) correct 23% (01:12) wrong based on 53 sessions

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Due to a manufacturing defect, each copy of a 300-page book is missing exactly one page,though each copy is not missing the same page. If the text of the book is only 288 pages (the other 12 are blank), what is the probability that any single copy is missing a page of text?

A. 1/30
B. 1/25
C. 11/12
D. 14/15
E. 24/25

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Re: Due to a manufacturing defect, each copy of a 300-page book is missing  [#permalink]

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26 Jul 2018, 01:19

as we can pick any of the 288 pages out of the 300
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Re: Due to a manufacturing defect, each copy of a 300-page book is missing  [#permalink]

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26 Jul 2018, 01:50
1
each copy of a 300-page book is missing exactly one page

each copy is not missing the same page => any page in the book can be missed => all pages are equally likely to be missed

text of the book is only 288 pages, the other 12 are blank

probability that any single copy is missing a page of text = $$\frac{288}{300}$$ = $$\frac{24}{25}$$

Hence option E
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Re: Due to a manufacturing defect, each copy of a 300-page book is missing  [#permalink]

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26 Jul 2018, 10:16
Bunuel wrote:
Due to a manufacturing defect, each copy of a 300-page book is missing exactly one page,though each copy is not missing the same page. If the text of the book is only 288 pages (the other 12 are blank), what is the probability that any single copy is missing a page of text?

A. 1/30
B. 1/25
C. 11/12
D. 14/15
E. 24/25

$$\frac{288}{30}0$$

= $$\frac{144}{150}$$

= $$\frac{24}{25}$$, Answer must be (E)
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Re: Due to a manufacturing defect, each copy of a 300-page book is missing  [#permalink]

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28 Jul 2018, 07:31
workout Abhishek009

How did you knew that the book that the question is asking about has 288 and not 300 pages?
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Re: Due to a manufacturing defect, each copy of a 300-page book is missing  [#permalink]

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28 Jul 2018, 08:19
1
workout Abhishek009

How did you knew that the book that the question is asking about has 288 and not 300 pages?

Hello,

As per the question statement, each copy of a 300-page book is missing one page.

The book has 300 pages of which 288 pages contain text and 12 pages are blank. Both of them add up to 300. Since the question asks for the probability of text page missing from any single copy, we have $$\frac{288}{300}$$
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Re: Due to a manufacturing defect, each copy of a 300-page book is missing &nbs [#permalink] 28 Jul 2018, 08:19
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text:
i⁴=i²⋅i²=(-1)²=1. More generally, you can find zn as the complex number (1) whose absolute value is |z|n, the nth power of the absolute value of z, and (2) whose argument is n times the argument of z. Get the value of input, textarea and radio button of different HTML form elements using our jQuery tutorial. value meaning: 1. the amount of money that can be received for something: 2. the importance or worth of something…. The number 1 is a square root of unity, (1 ± i√3)/2 are cube roots of unity, and 1 itself counts as a cube root, a square root, and a “first” root (anything is a first root of itself). It’s fun to find roots of unity, but we’ve found most of the easy ones already. In some cases (notably among diploid organisms), the terms C-value and genome size are used interchangeably; however, in polyploids the C-value may represent two or more genomes contained within the same nucleus. Ancienne définition du mot « plus-value » par l’Académie Française (édition de 1835). i − An int for which Integer representation would be returned. value verb. Get the contents of a text area: var x = document.getElementById("myTextarea").value; Try it Yourself » Textarea Object. Since |z| is less than one, it’s square is at 60° and closer to 0. The valueOf method returns the relevant Number Object holding the value of the argument passed. Value: The monetary, material or assessed worth of an asset, good or service. Aggregated value need to be arrived where future benefits need to be discounted with time value of money (usually done by dividing the profit in future by (1+ interest rate)^ years in … respect verb. 2. The remaining conductors looped together (with the help of crocodile clips) i. e. Conductor 2 and onwards, are connected to Earth terminal of megger. Computers can only understand numbers, so an ASCII code is the numerical representation of a character such as 'a' or '@' or an action of some sort. These six sixth roots of -1/2 are displayed in the figure as blue dots. Définition de « plus-value » issue … He can't value you more than you value yourself.” ― Toni Morrison tags: emotion, love, self-respect, value. For example, the following code produces a row vector 'M' that contains the maximum value of each column of 'A', which is 3 for the first column and 4 for the second column. But what guides the process behind making the choices that you do? Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f () = (2 – 1)^2 + 3f()=(2−1)^2+3 Hence, Minimum value of (2−1)^2 = 0 Minimum value of (2−1^2 )+3 = 0 + 3 = 3Also, there is no maximum value of ∴ There is no maximum For example on my homework, (2+4i) + (4-1) which i cant figure out. In nature these phenomena are all around us. Algebra. 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Calculate any Power of i (the Square Root of -1) - powered by WebMath Synonyms. The argument can be a primitive data type, String, etc. Star's way will set all values of 35 in column "L" to zero. Following are all the variants of this method −. By Administrator Relationships No Comments. The R-Value is calculated as R = l/λ where l = the thickness of the material in metres and λ (lambda) is the thermal conductivity of the material in W/m.K. Just click the start here page, read the information, and follow the prompts. Definition: Book value of equity, also known as shareholder’s equity, is a firm’s common equity that represents the amount available for distribution to shareholders. Value: the amount of money for which something will find a buyer. valueOf(String s, int radix) − This returns an Integer object holding the integer value of the specified string representation, parsed with the value of radix. For example, the brand value equation method calculates an overall brand value based on the values of a core trademark and other intangible assets associated with that mark. & Calculus. Trig. Over time, the train will move at lots of different speeds. Example: Shared storage. View options. Value definition, relative worth, merit, or importance: the value of a college education; the value of a queen in chess. If you stop to think about why you have chosen your career path or the city in which you live, you will realize that your core values start to come into play. Synonyms: valuation, worth, cardinal virtue… Antonyms: deficiency, demerit, … This page will calculate the approximate value of your Steam account by looking up your games on your Steam community profile, using the current prices for each game on the Steam store. Our free valuation tool gives a car's price instantly. In accounting, value describes what something is worth in terms of something else. This method is a static method. What is the definition of book value of equity? 2. What influences your decisions? 254; September 2, 2014 Good friends are vitally important to your mental health and to the quality of your life. If we take 1/6 of each of these angles, then we’ll have the possible arguments for z: Since each of the angles for z differs by 360°, therefore each of the possible angles for z will differ by 60°. What Does Book Value of Equity Mean? In general, the VLOOKUP function searches values from left to right in the array table, and it requires the lookup value must stay in the left side of target value. The value of teamwork is recognized by all players as they work together toward something bigger — the common team goal of winning the game. Conductors at the other end are left free. 1.]] (To calculate a value, you don't need to enter a serial number. Subtract any debts or liabilities. And just as a reminder, absolute value literally means-- whether we're talking about a complex number or a real number, it literally just means distance away from 0. (That's why you couldn't take the square root of a negative number before: you only had "real" numbers; that is, numbers without the "i" in them. 3. The remaining sixth roots are reflections of w in the real and imaginary axes. The argument can be a primitive data type, String, etc. Discover the trick to get the highest return on your savings bond investments. 26 synonyms of value from the Merriam-Webster Thesaurus, plus 79 related words, definitions, and antonyms. What's i⁵? Sid Sid. value definition: 1. the amount of money that can be received for something: 2. the importance or worth of something…. Like “I have always believed, and I still believe, that whatever good or bad fortune may come our way we can always give it meaning and transform it into something of value… valueOf(int i) − This returns an Integer object holding the value of the specified primitive. In the figure you see a complex number z whose absolute value is about the sixth root of 1/2, that is, |z| = 0.89, and whose argument is 30°. For example, if you value someone’s opinion, you will ask that person's advice before making a big decision. Learn more. A List of Over 200 Personal Values. [[10. Following are all … This pattern repeats itself all the time, so we can quickly tell what iⁿ for any positive integer n. appreciate verb. So we need to determine what value (if any) of the constant C 3 makes g(x) = f(x). Alternatively, a liquidation value approach may be used to determine the low-end of an IP asset’s current value in a situation of distress. The workplace also requires recognizing and appreciating the value of teamwork, but getting the entire staff to come together and work toward a common, big-picture goal is sometimes easier said than done. Instead of picking from a list, in Step 2 of the personal values process, I offer exercises to discover your personal values. IR Value Measurement (Conductors to conductor (Cross Insulation)) The first conductor for which cross insulation is being measured shall be connected to Line terminal of the megger. The value of i is √-1. An amount, as of goods, services, or money, considered to be a fair and suitable equivalent for something else; a fair price or return. Here, the unit circle is shaded black while outside the unit circle is gray, so z is in the black region. Learn more. To live and to love are inseparable from each other. This spiral is called a geometric or exponential sprial. For example, we have this set of numbers {4, 5, 6, 5}. Find out a car's value, whether it's being sold on the forecourt, as a private sale, or if it's being traded in. The value read from a capacitor changes depending on charge time and voltage. Value definition is - the monetary worth of something : market price. HOW TO. How to use value in a sentence. Improve this answer. Explore the Science of Everyday Life . i³=i²⋅i=-i. Friendship is an opportunity to love, to learn about yourself, to mature as a human being, and to open up to the full experience of life. Such roots are called primitive, so (1 ± i√3)/2 are the two primitive sixth roots of unity. The $i$ that is used here represents — 'Iota’ that is a type of imaginary number and is written as $\sqrt{-1}$. 8values is, in essence, a political quiz that attempts to assign percentages for eight different political values. The first six powers are displayed, as you can see, as points on a spiral. To get a value from non single element tensor we have to be careful: The next example will show that PyTorch tensor residing on CPU shares the same storage as numpy array na. Calculate the value of a paper bond based on the series, denomination, and issue date entered. We have taken three form elements in our example The method can take two arguments, where one is a String and the other is a radix. a gamete) or one half the amount in a diploid somatic cell of a eukaryotic organism. An amount, as of goods, services, or money, considered to be a fair and suitable equivalent for something else; a fair price or return. You will select personal values that “sound good” based on your desired self-image. Value of i. They will be placed around the circle at 60° intervals. Two of them, of course, are ±1. Anyway, this new number was called "i", standing for "imaginary", because "everybody knew" that i wasn't "real". The logic is the estimate the value of total benefit / profit owner is going to have after all expenses and taxes. Place Value of a Number - powered by WebMath. 90° = 360°, a full circle. To find the mean, we do the following: Finding the Blue Book value of your new and used firearms, including pistols, rifles, shotguns, airguns, and blackpowder guns is easy with the number one source of gun pricing. The book value of equity is equal to total assetsminus total liabilities, preferred stocks, and intangible assets. In summary, the six sixth roots of unity are ±1, and (±1 ± i√3)/2 (where + and  can be taken in any order). Let z be a sixth root of w. Then (1) |z| is |w|1/6 which is about 0.89. Value definition: The value of something such as a quality, attitude , or method is its importance or... | Meaning, pronunciation, translations and examples For example, the value of … Find out how to check and calculate the value of a savings bond online. So, for example, to work out the R-Value of 60mm of Kingspan Kooltherm K8 Cavity Board, which has a thermal conductivity or λ value of 0.021W/m.K: R-Value = l/λ = 0.06/0.021 = 2.86 m 2.K/W Let w be a complex number, and z any of its sixth roots. Plots & Geometry. Value - 7. something (as a principle or quality) intrinsically valuable or desirable example: Sought material values instead of human values
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ABM

### Program Information

Name: ABM
Domain: Algorithm
Functionality: Introduce the type of metamorphic properties for ABM.(To demonstrate the process of discovering metamorphic properties for validation, and identifying MR based on those properties, we use a agent-based model of angiogenesis to illustrate it.You can find it by searching Angiogenesis model")
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#### Reference

Metamorphic Validation for Agent-based Simulation Models https://dl.acm.org/citation.cfm?id=3015607

### MR Information

#### MR1------Environmental Parameters

Description:
Property: Changes to the environment with a predictable change in system behavior.
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#### MR2------Agent Parameters

Description:
Property: Changes to agent parameters that cause predictable change in the system behavior.
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#### MR3------Agent Topology

Description:
Property: If agent connections or situation are defined by a specific topology, there may be some simulation output metric that should not be impacted by a topology change, and other metrics that should undergo a predictable change when the interaction network topology changes. Although in some simulations this may be implemented as a parameter, in simulations in which a single topology is used this property will require adding new functionality into the system to test.
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#### MR4------Interactions Between Agents

Description:
Property: In many ABMs each agent may be able to interact with another agent. This interaction may be via direct and purposeful communication, or from specific interactions that occur when two collide, or any number of related events. Changes in what occurs when agents interact or under what circumstances they interact may lead to predictable changes.
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#### MR5------Interactions between Agent and Environment

Description:
Property: Agents may directly interact with artifacts in the environment, either through sensing or other communication activities. Changes to either the way interactions occur, the effect of interaction on the agent or environment, or the level of interaction, could be a metamorphic property.
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#### MR6------Individual Agent Behavior

Description:
Property: Agent behavior refers to any behavior that should be validated and does not occur due to interactions. For instance, the general decision process an agent may use before moving in a 2D world could be the basis of a property if a change to that decision process could lead to a predictable outcome.
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# Is the sum of two independent non-overlapping uniforms uniform?

Suppose $X_1\sim U[a,b]$ and $X_2\sim U[c,d]$ with $a<b<c<d$ and suppose they are independent. I guess that the sum must be a uniform but I don't know how to show it [EDIT: I was wrong].

I would like to show it using convolution (although I'm curious for other solutions as well). I have studied the proof of https://math.stackexchange.com/questions/220201/sum-of-two-uniform-random-variables but I could not adapt it to my case.

• You should make it explicit whether the variables are independent. From your mention of convolution I will assume they are. The shape of any convolution is unaffected by shifts in the arguments (so "non-overlapping" makes no difference). – Glen_b Nov 2 '16 at 23:03
• When you say "I know that the sum must be a uniform (just from guessing)" ... are all (or essentially all) of your guesses true? In several decades of statistics, I've found guessing to be a very unreliable means of making progress (and yet from observation of the guesses people make I expect my intuition is at least slightly better than average; this may be partly why I ended up being a statistician). It makes me wonder what you think the word "know" means; it doesn't mean the same thing as "hope" for example. – Glen_b Nov 2 '16 at 23:03
• There's an informal solution outlined here that should help you to set up the relevant integrals (in particular, the limits for the various pieces). Any diagonal line in the first image indicates the interval for the integral for a given value of the sum. – Glen_b Nov 2 '16 at 23:07
• Perhaps it might help your intuition to think about simpler discrete distributions. Say we have a die taking the values $\{1, 2, \ldots , 6 \}$ and another "negative die" taking the values $\{-1, -2, \ldots , -6 \}$. Now think about about how we can obtain $0$ as the sum compared to say $5$. Do you believe the sum is uniformly distributed? Would this change if we were working with continuous distributions? – dsaxton Nov 3 '16 at 2:26
• @Xu I won't deny that informed guessing can be useful, but you must be able to apply some critical thinking to it for it to have much value in general. Practice at the formalities will help improve your intuition (not least because you'll come to realize how dangerous untutored intuition can be in statistics and learn to be wary of it without some additional effort). A good second step to an initial intuition is attempting to show why it must be wrong. – Glen_b Nov 3 '16 at 4:16

I guess that the sum must be a uniform

It isn't. The middling values of the sum occur in "more" ways than the sums near the ends of the range.

I would like to show it using convolution

Yes, you just write the convolution integral.

$h(z)=(f*g)(z)=\int _{-\infty }^{\infty }f(z-t)g(t)dt=\int _{-\infty }^{\infty }f(t)g(z-t)dt$

Note that the $f$ and $g$ in the integral are both constant where they're non-zero, and so $f(z-t)g(t)$ is constant on an interval .... what drives the value of the integral at a specific value of the sum is the limits of where it's non-zero.

However, the integral is best split into three ranges because of the corners in the joint distribution (see the left most diagram here for the case where the lower limits are both 0; your case works similarly to that and you should use a similar diagram).

(although I'm curious for other solutions as well).

IN the case of problems very similar to this one, the approach at the link I gave earlier uses simple arguments to derive the distribution for a simpler case than yours, but then suggests how to generalize it to a case very like yours.

More generally there's a number of approaches related to using generating functions or Laplace transforms or Fourier transforms. They are all closely related. Sometimes they can make a problem easy.

• Moment generating function: $M_X(t) = E(e^{tX})$
• Laplace transform: $\mathcal{L}_X(t) = E(e^{-tX})$
• Characteristic function: $\phi_X(t) = E(e^{itX})$
• Fourier transform: $\mathcal{F}_X(t) = E(e^{-itX})$

There are several possible definitions (see toward the end of the linked section) of the Fourier transform in use. The above one is one of the forms listed at the linked section (chosen for the fact that it fits into this scheme of writing them all as very simple expectations); all the other forms of the Fourier transform work equally well, it's just a matter of consistently using whichever you like.

There are some other generating functions/transforms we could include in this list. In the ones above, aside from a change of sign or the inclusion of the constant $i$, they're all closely related; an argument involving one will have an explicit argument involving another. Note, however that the last two have a slight advantage, in that they always exist in a neighborhood of 0, which the first two might not (making them not useful on some problems).

They all have the property that the transformed product of a convolution is the product of the transforms. e.g. for $X$, $Y$ independent (so we have a convolution) $M_{X+Y}(t)=M_X(t)\,M_Y(t)$.

There are two main ways to use them.

1. Consider moment generating functions, $M_X(t)=E(e^{tX})$. One way to proceed is to proceed to calculate $M_X(t)\,M_Y(t)$ and simply recognize the result (perhaps after some simplification) as the MGF of a density, in much the same way that an applied mathematician, physicist, or engineer might use a table of Laplace transforms to solve a differential equation. In many simple problems (and a few not so simple ones) this will lead to a solution.

2. Explicit inversion. Consider say a Fourier transform. One computes $\mathcal{F}_X(t)\,\mathcal{F}_Y(t)$ and then uses the corresponding inverse Fourier transform (which depends on your units) to transform the result back to a density.


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# How do you use the exponent to determine the simplified form of any power of i?

${i}^{2 n} = {\left(- 1\right)}^{n} = \pm 1$, according as n is even or odd.
$i \left(2 n + 1\right) = {\left(- 1\right)}^{n} i = \pm i$, according as n is even or odd.