text:
Dataplot Vol 1 Vol 2

# CCPR PLOT

Name:
CCPR PLOT
Type:
Graphics Command
Purpose:
Generate a CCPR (component and component-plus-residual) plot.
Description:
When performing a linear regression with a single independent variable, a scatter plot of the response variable against the independent variable provides a good indication of the nature of the relationship. If there is more than one independent variable, things become more complicated. Although it can still be useful to generate scatter plots of the response variable against each of the independent variables, this does not take into account the effect of the other independent variables in the model.

Partial residual plots attempt to show the relationship between a given independent variable and the response variable given that other independent variables are also in the model.

Partial residual plots are formed as:

Res + $$\hat{\beta}_{i} X_{i}$$ versus Xi

where

Res = residuals from the full model
$$\hat{\beta}_{i}$$ = regression coefficient from the ith independent variable in the full model
Xi = the ith independent variable

Partial residual plots are widely discussed in the regression diagnostics literature (e.g., see the References section below). Although they can often be useful, be aware that they can also fail to indicate the proper relationship. In particular, if Xi is highly correlated with any of the other independent variables, the variance indicated by the partial residual plot can be much less than the actual variance. These issues are discussed in more detail in the references given below.

The CCPR plot is a refinement of the partial residual plot. It generates a partial residual plot but also adds

$$\hat{\beta}_{i} X_{i}$$ versus Xi

This is the "component" part of the plot and is intended to show where the "fitted line" would lie.

Dataplot provides two forms for the CCPR plot. You can generate either a single CCPR plot or you can generate a matrix of CCPR plots (one plot for each independent variable in the model).

For the matrix form of the command, a number of SET FACTOR PLOT options can be used to control the appearance of the plot (not all of the SET FACTOR PLOT options apply). These are discussed in the Notes section below.

Syntax 1:
CCPR PLOT <y> <x1> ... <xk> <xi>               <SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<x1> ... <xk> are the independent variables;
<xi> is the independent variable for which the CCPR plot is being generated
(note that <xi> must be one of the variables listed in <x1> ... <xk>;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.

This is the syntax for generating a single CCPR plot.

Syntax 2:
MATRIX CCPR PLOT <y> <x1> ... <xk>               <SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<x1> ... <xk> are the independent variables;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.

This syntax is used to generate a matrix of CCPR plots.

Examples:
CCPR PLOT Y X1 X2 X3 X4 X2

MATRIX CCPR PLOT Y X1 X2 X3 X4

CCPR PLOT Y X1 X2 X3 X4 X2 SUBSET TAG > 2
MATRIX CCPR PLOT Y X1 X2 X3 X4 SUBSET TAG > 2

Note:
The following option controls which axis tic marks, tic mark labels, and axis labels are plotted.

SET FACTOR PLOT LABELS <ON/OFF/XON/YON/BOX>

OFF means that all axis labels are suppressed (this can be useful if a large number of variables are being plotted). ON means that both X and Y axis labels are printed. XON only plots the x axis labels and YON only plots the y axis labels.

BOX is a special option that creates an extra column on the left and an extra row on the bottom. The axis label is printed in this box. BOX is typically reserved for the plot types that plot the variable names in the axes labels.

The default is ON (both x and y axis labels are printed).

Note:
The following option controls where the x axis tic marks, tic mark labels, and axis label are printed.

SET FACTOR PLOT X AXIS <BOTTOM/TOP/ALTERNATE>

BOTTOM specifies that the x axis labels are printed on the bottom axis (on the last row only). TOP specifies that the x axis labels are printed on the top axis (first row only). ALTERNATE specifies that the x axis labels alternate between the top (first row) and bottom axis (last row). We recommend using the TIC OFFSET command to avoid overlap of axis labels and tic marks.

The default is ALTERNATE.

Note:
The following option controls where the y axis tic marks, tic mark labels, and axis label are printed.

SET FACTOR PLOT Y AXIS <LEFT/RIGHT/ALTERNATE>

LEFT specifies that the y axis labels are printed on the left axis (on the first column only). RIGHT specifies that the y axis labels are printed on the right axis (last column only). ALTERNATE specifies that the y axis labels alternate between the left (first column) and right axis (last column). We recommend using the TIC OFFSET command to avoid overlap of axis labels and tic marks.

The default is ALTERNATE.

Note:
Users have different preferences in terms of whether the plot frames for neighboring plots are connected or not. This is controlled with the following option.

SET FACTOR PLOT FRAME <DEFAULT/CONNECTED/USER>

DEFAULT connects neighboring frames (i.e., the FRAME CORNER COORDINATES are set to 0 0 100 100). USER uses whatever frame coordinates are currently set (15 20 85 90 by default) and makes no special provisions for axis labels and tic marks (i.e., you set them as you normally would, each plot uses whatever you have set). CONNECTED uses whatever frame coordinates have been set by the user, but it draws the axis labels and tic marks as if DEFAULT were being used (that is, as determined by the SET FACTOR PLOT commands described above). Typically, CONNECTED is used to put a small bit of space between plots. For example, you might use FRAME CORNER COORDINATES 3 3 97 97 before the PARTIAL RESIDUAL PLOT command.

Since the plots can often have different limits for the axes, the default is USER.

Note:
When the tic marks and tic mark labels are all plotted on the same side (i.e., SET FACTOR PLOT Y AXIS is set to LEFT or RIGHT or SET PARTIAL RESIDUAL PLOT X AXIS is set to BOTTOM or TOP), then overlap between plots is possible. The TIC OFFSET command can be used to avoid this. In addition, you can stagger the tic labels with the following command:

SET FACTOR PLOT LABEL DISPLACEMENT <NORMAL/STAGGERED/VALUE>

NORMAL means that all tic labels are plotted at a distance determined by the TIC LABEL DISPLACEMENT command. STAGGERED means that alternating plots will be staggered. That is, one will use the standard displacement while the next uses a staggered value. Entering this command with a numeric value specifies the amount of the displacement for the staggered tic labels. For example,

TIC MARK LABEL DISPLACEMENT 10
SET FACTOR PLOT LABEL DISPLACEMENT STAGGERED
SET FACTOR PLOT LABEL DISPLACEMENT 25

These commands specify that the default tic label displacement is 10 and the staggered tic mark label displacement is 25.

Note:
It is often helpful on scatter plot matrices to overlay a fitted line on the plots. The following command is used to specify the type of fit.

SET FACTOR PLOT FIT <NONE/LOWESS/LINE/QUAD/SMOOTH>

NONE means that no fitted line is plotted. LOWESS means that a locally weighted least squares line will be overlaid. LINE means that a linear fit (Y = A0 + A1*X) will be overlaid. QUAD means that a quadratic fit (Y = A0 + A1*X + A2*X**2) will be overlaid. SMOOTH means that a least squares smoothing will be overlaid.

For LOWESS, it is recommended that the lowess fraction be set fairly high (e.g., LOWESS FRACTION 0.6).

The fitted line is currently only generated if the factor plot type is PLOT.

The default is for no fitted line to be overlaid on the plot. If a overlaid fit is desired, the most common choice is to use LOWESS.

Note:
Dataplot allows you to set axis limits with the LIMITS command. For the factor plot, it is often desirable to set the axis limits for each plot. This can be done with the command

SET FACTOR PLOT YLIMITS <LOW1> <UPP1> <LOW2> <UPP2> ...
SET FACTOR PLOT XLIMITS <LOW1> <UPP1> <LOW2> <UPP2> ...

The default is to allow the axis limits to float with the data.

Note:
You can use standard plot control commands to control the appearance of the factor plot.

For example,

MULTIPLOT CORNER COORDINATES 5 5 95 95
MULTIPLOT SCALE FACTOR 3
TIC OFFSET UNITS SCREEN
TIC OFFSET 5 5
Default:
None
Synonyms:
None
Related Commands:
 FIT = Perform a multi-linear fit. PARTIAL RESIDUAL PLOT = Generates a partial residual plot. PARTIAL REGRESSION PLOT = Generates a partial regression plot. PARTIAL LEVERAGE PLOT = Generates a partial leverage plot. VIF = Compute variance inflation factors for a multi-linear fit. CONDITION INDICES = Compute condition indices for a design matrix. SCATTER PLOT MATIRX = Generate a scatter plot matrix plot. FACTOR PLOT = Generate a plot for a response against a number of different independent variables. CONDITIONAL PLOT = Generate a conditional (subset) plot.
References:
Tom Ryan (1997), "Modern Regression Methods," John Wiley,

Neter, Wasserman, and Kunter (1990), "Applied Linear Statistical Models", 3rd ed., Irwin.

Draper and Smith (1998), "Applied Regression Analysis," 3rd. ed., John Wiley.

Cook and Weisberg (1982), "Residuals and Influence in Regression," Chapman and Hall.

Belsley, Kuh, and Welsch (1980), "Regression Diagnostics," John Wiley,

Paul Velleman and Roy Welsch (1981), "Efficient Computing of Regression Diagnostiocs," The American Statistician, Vol. 35, No. 4, pp. 234-242.

Applications:
Multi-linear Regression
Implementation Date:
2002/6
Program:

SKIP 25
READ HALD647.DAT Y X1 X2 X3 X4
.
MULTIPLOT CORNER COORDINATES 5 5 95 95
MULTIPLOT SCALE FACTOR 2
LINE BLANK
CHARACTER X
X1LABEL DISPLACEMENT 12
Y1LABEL DISPLACEMENT 12
TIC OFFSET UNITS SCREEN
TIC OFFSET 5 5
.
MATRIX CCPR PLOT Y X1 X2 X3 X4

NIST is an agency of the U.S. Commerce Department.

Date created: 08/19/2002
Last updated: 10/13/2015

Please email comments on this WWW page to alan.heckert@nist.gov.


text:
# NIELIT DEC 2017 SET-C 62

1 vote
450 views

If a random coin is tossed 11 times then what is the probability that for 7th toss head appears exactly 4 times?

A) 5/32

B) 15/128

C) 35/128

D) None of the options

For $7^{th}$ toss head appears exactly $4$ times means in the previous $6$  tosses head can appear at any position but exactly thrice and the $7^{th}$ toss will result in a head.

Required probability  $= \binom{6}{3}\left ( \frac{1}{2} \right )^{3}\left ( \frac{1}{2} \right )^{3}\left ( \frac{1}{2} \right )$  $= \frac{5}{32}$

$A$ is correct.

selected
0
What about the tosses done from 8,9,10,11 do we need to ignore them ? if to ignore why so ?
0
yes it is redundant information, here we're more concerned about the 7th toss, it might even be possible that coin is tossed infinite times
1
Thanks got it .
0
Official answer is C for this question.
1 vote
→ To find probability that for the 7th toss head appears exactly 4 times. We have find that past 6 run of tosses.
→ Past 6 tosses, What happen we don’t know. But according to 7th toss, we will find that head appeared for 3 times.
→ Any coin probability of a head=½ and probability of a tail=½
→ If we treat tossing of head as success,then this leads to a case of binomial distribution.
→ According to binomial distribution, the probability that in 6 trials we get 3 success is
6C3*(½)3 *(½)3= 5/16 (3 success in 6 trials can happen in 6C3 ways). → 7th toss, The probability of obtaining a head=1/2
→ In given question is not mention that what happens after 7 tosses.
→ Probability for that for 7th toss head appears exactly 4 times is =(5/16)*(1/2)
=5/32.

## Related questions

1
192 views
When we use slow start algorithm ,the size of CW increases ...............until it reaches threshold. A)Additively B)Multiplicatively C)Exponentially D) NONE
The excess $3$ code is also called cyclic redundancy code weighted code self complimenting code algebraic code

format: plain


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# Setting up dataset for Cox regression with time-dependent variables

I posed a question about how to set up the code for this question here (Psychometrics: Survival analysis of help seeking behaviors) and @Fomite suggested that I pose a separate question about whether my data are set up correctly.

Background:
I'm studying people seeking help. Participants described contacts with between 1 and 3 "responders" (e.g., friends, the police) in order- for example, a participant could have contacted just responder 1, or responder 1, then responder 2, then responder 3. I'm trying to predict help-seeking dropout, meaning that, for example, a participant contacted responder 1 but did not go on to contact a second or third responder- that participant would have a dropout at responder 1. So unlike other survival analysis, the observations are responders rather than time points- but they're still ordered in time. The independent variables in my model include characteristics of the people seeking help (e.g., gender) and aspects of their interactions with the responders (e.g., whether they liked the interaction). The data are right-censored for those participants who said that they contacted more than three responders because they could not record more than three responders in the survey. There are two people who only reported on responder 3; those people are left-censored because data are missing for responders 1 and 2.

Current data setup:
The data are set up as a person-period dataset such that there is a line for each responder, which means that most participants have multiple lines. Responders are nested within participants. So a participant that contacted two responders would have two lines in the dataset; the participant-level data is the same in both lines and the responder-level data is different.

Here's what the data looks like now: https://flic.kr/p/s1fW6k

Variables:

Participant-level:

• id is the ID number for the participant.
• gender is the participant's gender; 1 = woman and 2 = man

Responder-level:
• responder represents the responder number in the order that the participant contacted them. Possible values are 1, 2, and 3.
• stoppedhelpseeking represents whether the participant stopped seeking help/dropped out after contacting that responder; 0 = no and 1 = yes. This is the event that I'm hoping to predict.
• likedresponder represents whether the participant liked their interaction with the responder; 0 = no and 1 = yes
• censor represents whether the participant did not report a dropout by responder 3. It is coded for their last recorded responder in the dataset.

My question:
@Fomite pointed out that, the way I've set up my data, each responder with a 0 value for stoppedhelpseeking will have a censored event time. This doesn't sound like what I want- what I'm hoping to predict is the point at which people drop out of help-seeking. Censoring should occur at the participant level, not the responder level. What's the right way to set this up to model things correctly?

Actually, the censoring of everything with stoppedhelpseeking=0 is what you want.
Having a subject with a time-dependent covariate is the same as having that subject leave the study when the covariate changes, and another subject with slightly different covariates enter the study at the same time. So you do want it to look like everything with stoppedhelpseeking=0 got censored. And then you want to claim that this person only entered that study when their previous version exited the study--so the person with responder 1 was in the study for the "time" interval (0, 1], the person with responder 2 was there for time interval (1, 2], etc. In this case your censor variable is actually irrelevant.
I don't know SAS, but in R, the way you would set up the response variable here is Surv(responder-1, responder, stoppedhelpseeking). For more info, see section 2 of this vignette. I imagine it's similar in other statistical software.
EDIT: based on @Fomite's answer to your other question, you can probably use something like entry=responder-1 to encode this?


text:
Free Version
Moderate

# Triangle Perimeter

ALGEBR-1X6Y1V

A triangle has sides lengths of $2 + x$, $3 + x$, and $5 + x$. What is the perimeter?

A

$10 + 3x$

B

$15 + x^{3}$

C

$13x$

D

$13 + x$


text:
# How can I use orientation to compute position in Windows Phone 7? [closed]

I'm using the motion API and I'm trying to figure out a control scheme for the game I'm currently developing.

What I'm trying to achive is for an orienation of the device to correlate directly to a position, such that tilting the phone forward and to the left represents the top left position and back to the right would be the bottom right position.

Here are some photos that should make my intent clearer. The red dot would be the calculated position.

Tilted forward and left, upper left selected.

Tilted back and right, bottom right selected.

I also have to make sure that the algorithm takes into account left landscape and right landscape device orientations (portrait is the default so no calculations would be needed for it).

I've tried using the yaw, pitch, roll and Quaternion readings. Here's a sample of my current attempt:

// Get device facing vector
public static Vector3 GetState()
{
lock (lockable)
{
var down = Vector3.Forward;
var direction = Vector3.Transform(down, state);

switch (Orientation)
{
case Orientation.LandscapeLeft:
return Vector3.TransformNormal(direction, Matrix.CreateRotationZ(-rightAngle));
case Orientation.LandscapeRight:
return Vector3.TransformNormal(direction, Matrix.CreateRotationZ(rightAngle));
}

return direction;
}
}

## closed as off-topic by Josh♦Dec 2 '13 at 16:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Programming questions that aren't specific to game development are off-topic here, but can be asked on Stack Overflow. A good rule of thumb is to ask yourself "would a professional game developer give me a better/different/more specific answer to this question than other programmers?"" – Josh
If this question can be reworded to fit the rules in the help center, please edit the question.

• I would recommend you ask this on SO instead. It isn't directly related to gaming, and you'll probably get better answers (esp. regarding the questions on the Windows Phone API). I'll vote to transfer if you think this is a good idea. – DMan Jan 4 '12 at 2:14
• @DMan yeah maybe – Daniel Little Jan 4 '12 at 2:31
• Try multiplying accelerometer values by 1 or -1 depending on screen orientation. it worked for me. – Vector3d Aug 13 '12 at 9:50
• Cross-posted, with an accepted answer (for posterity): stackoverflow.com/questions/8721561/… – Josh Dec 2 '13 at 16:45

Vector3 rotatedVector = Vector3.Transform(vectorToRotate, Matrix.CreateFromAxisAngle(rotationAxis, radiansToRotate));

Your approach seems to be similar, but I'm not sure what your state value is. I would recommend trying Matrix.CreateFromAxisAngle(PhoneFaceAxis, radiansToRotate);, where PhoneFaceAxis is the axis represented by the normal vector of the screen of your phone (which may be Z, judging by your code sample', and radiansToRotate would be some form of a 90 or 180 degree rotation depending on the phone's orientation.


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# Storing recordset data

I have managed to write some pretty terrible code in VBA and I know there has to be a better way. I have made all of these variables to store some data which I eventually reference in the application. I'm fairly certain I need to loop over the recordset or fields and store the variables that way but I can't figure out how to do it in a way that actually simplifies this code. It's torturing me because it seems like a painfully simple concept.

The columns of the table that it is gathering data from looks similar to this:

[LANGUAGE_ID]-[NEXT]-[PREVIOUS]-[RESUME] -- and so on

sqlstmt = "SELECT *" & _
" FROM cf_Language" & _
" WHERE LANGUAGE_ID = " & parm_language & ";"

Set rst_t = dbs.OpenRecordset(sqlstmt, dbOpenSnapshot)

If (parm_language <> 1) Then
parm_surveyname = UnicodeToAscii(parm_surveyname, True)
aNext = UnicodeToAscii(rst_t![Next], True)
aPrevious = UnicodeToAscii(rst_t![previous], True)
aResume = UnicodeToAscii(rst_t![resume], True)
aFinish = UnicodeToAscii(rst_t![finish], True)
aSave = UnicodeToAscii(rst_t![Save], True)
aComplete = UnicodeToAscii(rst_t![complete], True)
aThanks = UnicodeToAscii(rst_t![thanks], True)
aSavedMsg = UnicodeToAscii(rst_t![saved_msg], True)
aBlankCode = UnicodeToAscii(rst_t![blank_code], True)
aInvalidCode = UnicodeToAscii(rst_t![invalid_code], True)
aTimeOut = UnicodeToAscii(rst_t![timeout], True)
aClosed = UnicodeToAscii(rst_t![closed], True)
aAccessCode = UnicodeToAscii(rst_t![access_code], True)
aConFirstTime = UnicodeToAscii(rst_t![con_first_time], True)
aAnonFirstTime = UnicodeToAscii(rst_t![anon_first_time], True)
aAnonResume = UnicodeToAscii(rst_t![anon_resume], True)
Else
aNext = rst_t![Next]
aPrevious = rst_t![previous]
aResume = rst_t![resume]
aFinish = rst_t![finish]
aSave = rst_t![Save]
aComplete = rst_t![complete]
aThanks = rst_t![thanks]
aSavedMsg = rst_t![saved_msg]
aBlankCode = rst_t![blank_code]
aInvalidCode = rst_t![invalid_code]
aTimeOut = rst_t![timeout]
aClosed = rst_t![closed]
aAccessCode = rst_t![access_code]
aConFirstTime = rst_t![con_first_time]
aAnonFirstTime = rst_t![anon_first_time]
aAnonResume = rst_t![anon_resume]
End If

• You might be interested in this code, which pretty much mootinates my answer. Came up with it recently ;) – Mathieu Guindon May 28 '14 at 19:11

# Key Points

• I would not use local variables to hold recordset values, because you'd need one for every field of every returned row... which is just wrong.
• As already pointed out, good variable names are readable variable names (which can be pronounced!).
• You're not showing it, but make sure you close that connection once you're done.

## "Entities"

Here's an [perhaps less appealing] alternative to Dictionary. Since I learned about C# and started using Entity Framework, I like wrapping each record in an object - given a class called MyEntity (which does nothing more than exposing getters and setters for all properties):

Private Type ThisEntity
SomeProperty As String
AnotherProperty As String
End Type

Private this As ThisEntity
Option Explicit

Public Property Get SomeProperty() As String
SomeProperty = this.SomeProperty
End Property

Public Property Let SomeProperty(value As String)
this.SomeProperty = value
End Property

Public Property Get AnotherProperty() As String
AnotherProperty = this.AnotherProperty
End Property

Public Property Let AnotherProperty(value As String)
this.AnotherProperty = value
End Property

You define this object only once, and then you instantiate it as many times as you have records in your recordset. With a dictionary you get easier loading (each "entity" is a dictionary where you identify each field by their names to retrieve the corresponding value), but with this object you completely avoid referring to your properties using strings - the only hard-coded strings involved are the field names.

All you need to do is exactly as you said, loop through the recordset:

Dim entity As MyEntity
Dim entities As New Collection
While Not rs.EOF And Not rs.BOF 'BOF check if result has 0 rows

Set entity = New MyEntity
With entity
.SomeProperty = rs.Fields("SomeField").Value 'or rs!SomeField
.AnotherProperty = rs.Fields("AnotherField").Value 'or rs!AnotherField
End With

entities.Add entity ', entity.Id /PK property if you want to use a key here
rs.MoveNext 'don't forget this, or your loop will be endless!
Wend

rs.Close
Set rs = Nothing
'at this point you have your entire recordset's content in the "entities" collection

For what it's worth, I've recently fixed a bug in our codebase that was caused by the rs!FieldName notation in a specific circumstance: rs!FieldName was being added to a collection, and the recordset was left open. When I found that out, I added rs.Close and all of a sudden the collection, although non-empty, could no longer be read. This is because adding rs!FieldName to the collection (or, if you take the dictionary road) was actually adding a reference to the Field object, and the original programmer was left with the impression it worked fine because he could retrieve any entry in the collection and see the field's value - when he was actually retrieving a Field object and through the default property was fetching the value from the recordset: that stopped working when I added code to close the recordset after all records were read.

That bug was fixed by explicitly adding rs.Fields("FieldName").Value to the collection instead.

• This was very informative, as was tomdemuyt's answer, I wish I could accept both answers. – Jake Zeitz Sep 12 '13 at 14:54

From the help : "The default collection of a Recordset object is the Fields collection, and the default property of a Field object is the Value property. Use these defaults to simplify your code."

I would either :

• Loop over the fields and create a dictionary with name/value pairs
• Make sure that rst_t is of type Snapshot and just keep using rst_t

Other than that:

• Hungarian notation, just don't do it
• Disemvoweling ( sqlstmt ? ), just don't do it either
• Seems like you could simply always call UnicodeToAscii and cut half of the code

On dictionaries in VBa: https://stackoverflow.com/questions/915317/does-vba-have-dictionary-structure On Hungarian notation ( and why it is bad ): http://en.wikipedia.org/wiki/Hungarian_notation

I could not find a good article on disemvoweling ( remove all vowels from the name and then some like 'statement' -> 'sttmnt' -> 'stmt' ), I was sure Coding Horror had one. Basically it just makes your code harder to read for other people, it is one more barrier to (mis)understand code.

• I'm not sure how to create a 'dictionary' like you suggest. Could you give an example? I can't keep using the UnicodeToAscii function because of other complications.. (I should probably put it on codereview as well). Lastly, could you explain what you mean by 'hungarian notation' and 'disemvoweling' and why those are bad things? – Jake Zeitz Sep 11 '13 at 14:20
• @tomdemuyt found this article: researchonlyclayton.blogspot.ca/2010/02/… – Mathieu Guindon Sep 12 '13 at 0:15
• @retailcoder Awesome, the guy responsible for Hungarian Notation is also an avid disemvoweler ;) – konijn Sep 12 '13 at 12:16
• Argh just noticed I linked to he wrong article - here's the one I meant to link to:folklore.org/… – Mathieu Guindon Sep 12 '13 at 14:06
• Thanks for the help, but I could only accept one answer, otherwise I would accept yours too. – Jake Zeitz Sep 12 '13 at 14:54


text:
# What key exchange do OpenSSL and CryptoAPI prefer by default?

Here is what I know.

When I connect to a brand-new Linux Web server with Chrome and click the "lock" icon, the key exchange algorithm is described as "ECDHE_RSA". (Yes I know what that is.)

When I connect with Internet Explorer 8 on Windows 7, the key exchange is described as "RSA with 2048 bit exchange". I know what that is, too.

With Firefox, it says I am secured by "AES-256, 256 bit keys". Which is (a) redundant and (b) tells me nothing about the key exchange algorithm.

When I run openssl ciphers on my brand-new Linux machine, the output starts with:

ECDHE-RSA-AES256-GCM-SHA384:ECDHE-ECDSA-AES256-GCM-SHA384:...

(I suspect these are listed in order of OpenSSL's preference, but I do not know where this is documented.)

ECDHE stands for "elliptic curve Diffie-Hellman, ephemeral". That's great, but I want to know the actual key exchange algorithm, including the specific elliptic curve.

Here is what I really want: A big table, containing all possible pairs of SSL client and server (all versions), telling me what specific key exchange algorithm (including the actual elliptic curve) would be negotiated between them. Or enough information to construct that table.

I would settle for simply knowing how the latest versions of OpenSSL and CryptoAPI behave by default, and where this is documented.

-

Given a SSL-enabled web site, the Qualys SSL tester will tell what ciphersuite would be negotiated by a bunch of different browsers if they connected to that web site. It will also tell you the list of ciphersuites supported by that server and the list of ciphersuites are supported by each of those major browsers.

For example, here is the output for one web site you mentioned elsewhere: output. The section labelled "Handshake Simulation" will tell you what cipher suite many different major browsers would negotiate with this server. For instance:

• Chrome 29 (Win 7) would negotiate TLS_ECDHE_RSA_WITH_AES_256_CBC_SHA over TLS 1.2;

• IE 8 (Win 7) would negotiate TLS_RSA_WITH_AES_128_CBC_SHA over TLS 1.0 (which does not provide forward secrecy);

• Firefox 22 (Win 7) would negotiate TLS_ECDHE_RSA_WITH_AES_256_CBC_SHA over TLS 1.0;

and so on. This doesn't tell you what elliptic curve they would use or the key size, but it gives you the other information you were looking for.

By testing one representative web site for each possible SSL server, you should be able to get close to the grid you want of the key exchange algorithm negotiated by all (browser, server) pairs.

-
Awesome resource; thank you. Of course, for obvious reasons, I am also interested in the specific elliptic curve(s) in use... As are we all, I think. But this is a great start. –  Nemo Sep 11 '13 at 17:56


text:
# non-trivial zeros of partial zeta functions

Let $N,a\in\mathbf{Z}_{\geq 1}$. Define a partial $\zeta$-function as $$\zeta(s;N,a):=\sum_{\substack{n\geq 1\newline n\equiv a\pmod{N}}} \frac{1}{n^s}$$ where $Re(s)>1$. Let $\omega$ be either the trivial character or the sign character i.e. $x\mapsto sign(x)$. Define a partial $\Psi$-function as $$\Psi(s,\omega;N,a):=\sum_{\substack{0\neq n\in\mathbf{Z}\newline n\equiv a\pmod{N}}}\frac{\omega(n)}{|n|^s},$$ for $Re(s)>1$. Then it is well known that $\zeta(s;N,a)$ and $\Psi(s,\omega;N,a)$ admit a meromorphic continuation to $\mathbf{C}$ with at most of pole of order $1$ at $s=1$. If $\omega$ is the sign character then $$\Psi(s,\omega;N,a)+\Psi(s,1;N,a)=2\zeta(s;N,a)$$ and $$\zeta(s;N,a)-\zeta(s;N,-a)=\Psi(s,\omega;N,a).$$ Note that when $N>1$, the functions $\Psi$ and $\zeta$ do not have an Euler product. So here are 2 natural questions:

Q1: For a fixed $N$ do we know if there exists a constant $C_N>1$ such that if $Re(s)>C_N$ then $\Psi$ and $\zeta$ do not vanish (if the answer is yes then how to prove it)?

Q2: What do we know in general about the nontrivial zeros of $\Psi$ and $\zeta$?

-

The answer to question 1 is classical: Any Dirichlet series which has a finite abscissa of absolute convergence has a zero-free half-plane.

Suppose the Dirichlet series $$A(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ has an abscissa of absolute convergence $\sigma_a$. If $a_m$ is the first non-zero coefficient, then for $\Re(s)>\sigma_a$ $$m^s A(s) = a_m + a_{m+1} \Big(\frac{m}{m+1}\Big)^s + a_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \to a_m \ \text{ as } \ \Re(s) \to +\infty.$$ Hence, there exists an absolute constant $C$ such that for $\Re(s)>C$ we have $$\left|a_{m+1} \Big(\frac{m}{m+1}\Big)^s + a_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \right| \leq \frac{|a_m|}{2}.$$ Consequently, $m^s A(s)$ has no zeros in the half-plane $\Re(s)>C$.

-
Thanks Micah, do you think it is possible to give a sharp approximation of $C_N$? So for example, do you think that $C_N=1$ should work? –  Hugo Chapdelaine May 17 '12 at 2:48
With a little work, one should be able to prove an explicit value of for $C_N$. Proving "sharpness" may not be so easy. It is unclear to me that $C_N=1$ will work. There are plenty of fairly basic Dirichlet series with zeros to the right of $\Re(s)=1$. –  Micah Milinovich May 17 '12 at 14:40

Micah's answer answers your Q1. My reply gives some information on your Q2. Let us assume that $1 \leq a\leq N$ for simplicity (this condition can be removed). Your zeta-functions $\zeta$ and $\Psi$ can be expressed by the Hurwitz zeta-function http://en.wikipedia.org/wiki/Hurwitz_zeta_function. For your first function $$\zeta(s;N,a)=N^{-s} \zeta \left(s,\frac a N \right).$$ Davenport and Heilbronn proved that for any $c>0$, $a/N \neq 1/2,1$ there exists $\gg T$ zeros of this function in the strip $1<\Re(s)<1+c, | \Im (s)| < T$. By Voronin universality for the Hurwitz zeta-function the same can be said in any strip $1/2<\sigma_0< \Re (s)<\sigma_1<1$ (this can be found for example in Jörn Steuding's SLN or Garunkstis-Laurincikas book on the Lerch zeta-function).

Your second function can also be expressed in terms of the Hurwitz zeta-function $$\Psi(s,\omega;N,a)=N^{-s} \left(\zeta\left(s,\frac a N\right)+\omega(-1)\zeta \left( s,1-\frac a N \right) \right).$$ In this case joint universality of the Hurwitz zeta-function (see same references as above) can be used to show that except when $\Psi$ has an Euler-product (which only occurs for some small special cases when $a / N$ equals 1/6, 1/4, 1/3, 1/2, 2/3, 3/4, 5/6, or 1.) there are $\gg T$ zeros in any strip $1/2<\sigma_0< \Re (s) < \sigma_1 < 1$ and $|\Im(s)| < T$. Davenport-Heilbronn's result for $\Re(s)>1$ can be obtained in this case as well.

For the special cases where your function has an Euler-product it will essentially be a Dirichlet L-function or the Riemann zeta-function, and its zeros will be the same.

-
Dear Johan, this is quite an interesting result! It seems to me that this result suggests that given $c>0$, if we look for zeros with $1<Re(s)<1+c$ up to some large imaginary part $T$ then "most of them" will have a real part close to $1$. Are analytic number theorists close to prove anything like that? –  Hugo Chapdelaine May 19 '12 at 16:23
No, I do not think so. In fact if $\sup_{\zeta(s,N,a)=0}\Re(s)=c_1$ then $c_1>1$ and given any $c_0>1$ there will be $\gg T$ zeros in $c_0<\Re(s)<c_1$. These results follows from results of Bohr and Kronecker's theorem. –  Johan Andersson May 19 '12 at 16:41
I see, thanks for setting me on the right path! –  Hugo Chapdelaine May 20 '12 at 2:26

EDIT Experimentally the zeros of $\zeta(s;5,1)$ and $\zeta(s;6,1)$ are those of zeta (the previous revision incorrectly included wrong zeros caused by insufficient precision).

The zeros of $\zeta(s;6,2)$ (if computed correctly) don't appear related to those of zeta.

sage code:

import mpmath
mpmath.mp.pretty=True
mpmath.mp.dps=100
N=5
a=1
an=[0]*N
an[a-1]=1
def L(x):
return mpmath.dirichlet(x,an)

def search1():
cac={}
rr=list(range(-20,0))
P=[]
for k in xrange(1,40):
rr += [0.1 + mpmath.j*k]
for x in rr:
try:
r=mpmath.findroot(L,[x],maxsteps=1000)

except:  continue
print r
ks=str(r)
if ks in cac:  continue
cac[ks]=1
P += [(RR(r.real),RR(r.imag))]
return P

P=search1()
pt=points(P)
pt.show()

-
Thanks a lot joro for the data. So I find it quite interesting that all the zeroes you computed are located in the vertical strip $-1<Re(s)<1$. It would be interesting to find a zero $\rho$ with $Re(\rho)>1$. –  Hugo Chapdelaine May 17 '12 at 14:29
Also @Joro, note that $\zeta(s;6,2)$ is not primitive in the sense that $gcd(6,2)=2$ and therefore $\zeta(s;6,2)=2^{-s}\zeta(s;3,1)$ where $\zeta(s;3,1)$ is now primitive. –  Hugo Chapdelaine May 17 '12 at 14:38

This is only a comment regarding the functional equation of the $\Psi$-function. So define $$\Psi^*(s,\omega;N,a):=\sum_{0\neq n\in\mathbf{Z}}\frac{\omega(n)}{|n|^s}e^{-2\pi i\frac{a}{N}n}$$ Then one can show using Riemann's classical idea (see for example 1) that $$(-1)^p N^s\cdot\Gamma_{\infty}(s)\Psi(s,\omega;N,a)=\Gamma_{\infty}(1-s)\Psi^*(1-s,\omega;N,a),$$ where $p=0$ (resp. p=$1$) if $\omega=1$ (resp. $\omega=sign$) and $\Gamma_{\infty}(s):=\pi^{-s/2}\Gamma(\frac{s+p}{2})$.

So it seems that many of the non-trivial zeros of the $\Psi$-functions are located along the symmetry axis of the functional equation. Is there some heuristic argument that could explain this phenomenon?

-
That zeroes cluster around the cricital line is well-known phenomena, see for example my answer mathoverflow.net/questions/45912/function-zeros-in-strip-0-re-1/…, where I cite a paper of Steuding that proves this result in the case of the periodic zeta-function (Your zeta-function $\Psi^*$ is such a function). –  Johan Andersson May 19 '12 at 16:19
Thanks a lot for this reference of Steuding! –  Hugo Chapdelaine May 19 '12 at 16:31


text:
# How Much Over-parameterization Is Sufficient to Learn Deep ReLU Networks?

A recent line of research on deep learning focuses on the extremely over-parameterized setting, and shows that when the network width is larger than a high degree polynomial of the training sample size n and the inverse of the target accuracy ϵ^-1, deep neural networks learned by (stochastic) gradient descent enjoy nice optimization and generalization guarantees. Very recently, it is shown that under certain margin assumption on the training data, a polylogarithmic width condition suffices for two-layer ReLU networks to converge and generalize (Ji and Telgarsky, 2019). However, how much over-parameterization is sufficient to guarantee optimization and generalization for deep neural networks still remains an open question. In this work, we establish sharp optimization and generalization guarantees for deep ReLU networks. Under various assumptions made in previous work, our optimization and generalization guarantees hold with network width polylogarithmic in n and ϵ^-1. Our results push the study of over-parameterized deep neural networks towards more practical settings.

## Authors

• 2 publications
• 37 publications
• 8 publications
• 59 publications
• ### An Improved Analysis of Training Over-parameterized Deep Neural Networks

A recent line of research has shown that gradient-based algorithms with ...
06/11/2019 ∙ by Difan Zou, et al. ∙ 0

• ### Over-parameterization Improves Generalization in the XOR Detection Problem

Empirical evidence suggests that neural networks with ReLU activations g...
10/06/2018 ∙ by Alon Brutzkus, et al. ∙ 0

• ### Benefits of Jointly Training Autoencoders: An Improved Neural Tangent Kernel Analysis

A remarkable recent discovery in machine learning has been that deep neu...
11/27/2019 ∙ by Thanh V. Nguyen, et al. ∙ 0

• ### A Study of Neural Training with Non-Gradient and Noise Assisted Gradient Methods

In this work we demonstrate provable guarantees on the training of depth...
05/08/2020 ∙ by Anirbit Mukherjee, et al. ∙ 0

• ### Refined Generalization Analysis of Gradient Descent for Over-parameterized Two-layer Neural Networks with Smooth Activations on Classification Problems

Recently, several studies have proven the global convergence and general...
05/23/2019 ∙ by Atsushi Nitanda, et al. ∙ 2

• ### Trainability and Data-dependent Initialization of Over-parameterized ReLU Neural Networks

A neural network is said to be over-specified if its representational po...
07/23/2019 ∙ by Yeonjong Shin, et al. ∙ 3

• ### Optimal Lottery Tickets via SubsetSum: Logarithmic Over-Parameterization is Sufficient

The strong lottery ticket hypothesis (LTH) postulates that one can appr...
06/14/2020 ∙ by Ankit Pensia, et al. ∙ 0

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## 1 Introduction

Deep neural networks have become one of the most important and prevalent machine learning models due to their remarkable power in various real-world applications. However, the success of deep learning has not been well-explained in theory. It remains mysterious why standard training algorithms tend to find a globally optimal solution, despite the highly non-convex landscape of the training loss function. Moreover, despite the extremely large amount of parameters, deep neural networks rarely over-fit, and can often generalize well to unseen data and achieve good test accuracy. Understanding these mysterious phenomena on the optimization and generalization of deep neural networks is one of the most fundamental goals in deep learning theory.

Recent breakthroughs have shed light on the optimization and generalization of deep neural networks under the over-parameterized setting, where the hidden layer width is extremely large. In terms of optimization, a line of work (Du et al., 2019b; Allen-Zhu et al., 2019b; Zou et al., 2018; Oymak and Soltanolkotabi, 2019b; Arora et al., 2019b; Zou and Gu, 2019) proved that (stochastic) gradient descent with random initialization can successfully find a global optimum of the training loss function regardless of the labeling of the data, as long as the width of the network is larger than , where is the training sample size. For generalization, Allen-Zhu et al. (2019a); Arora et al. (2019a); Cao and Gu (2019b, a); Nitanda and Suzuki (2019) established generalization bounds of neural networks trained with (stochastic) gradient descent under certain data distribution assumptions, when the network width is at least . Although these results have provided important insights into the learning of extremely over-parameterized neural networks, the requirement on the network width is still far from the practical settings. Very recently, Ji and Telgarsky (2019b) showed that for two-layer ReLU networks, when the training data are well separated, polylogarithmic width is sufficient to guarantee good optimization and generalization performance of neural networks trained by GD/SGD. However, it remains unclear whether similar results can be developed for deep neural networks.

In fact, most of the aforementioned results can be categorized in the so called neural tangent kernel (NTK) (Jacot et al., 2018; Du et al., 2019b) regime or lazy training regime (Chizat et al., 2019), where along the whole training process, the neural network function behaves similarly as its first-order Taylor expansion at initialization (Jacot et al., 2018; Lee et al., 2019; Arora et al., 2019b; Cao and Gu, 2019a). It is recognized that in order to make the learning of neural networks stay in the NTK regime, a proper scaling with respect to the network width is essential. For example, Cao and Gu (2019a) introduced a scaling factor in their definition of the neural network function, where is the network width. Same scaling factor has also been applied to the initialization of the output weights in Allen-Zhu et al. (2019b); Zou et al. (2019); Cao and Gu (2019b); Zou and Gu (2019). Many other results in the NTK regime used a different type of parameterization, but essentially have the same scaling factor (Jacot et al., 2018; Du et al., 2019b, a; Arora et al., 2019a, b). In fact, without such a scaling factor, it has been shown that the training of two-layer networks falls in a different regime, namely the “mean-field” regime (Mei et al., 2018; Chizat and Bach, 2018; Chizat et al., 2019; Sirignano and Spiliopoulos, 2019; Rotskoff and Vanden-Eijnden, 2018; Wei et al., 2019; Mei et al., 2019; Fang et al., 2019a, b).

In this paper, we study the optimization and generalization of deep ReLU networks for a wider range of scaling. Specifically, for a ReLU network with hidden nodes per layer, we generalize the scaling factor introduced in Cao and Gu (2019a) to , where is a constant. Note that similar scaling has been studied in Nitanda and Suzuki (2019). We show that for all such , as long as there exists a good neural network weight configuration within certain distance to the initialization, the global convergence property as well as good generalization performance can be provably established under mild condition on the neural network width, which is polylogarithmic in sample size and inverse target accuracy . At the core of our analysis is a milder requirement on the first-order approximation of neural network function, which allows the algorithm to travel longer to find the global minima. Our contributions are highlighted as follows:

• [leftmargin = *]

• We establish the global convergence guarantee of GD for training deep ReLU networks for binary classification. Specifically, we prove that for any positive constant , if there exists a good neural network weight configuration within distance to the initialization, and the neural network width satisfies , GD can achieve -training loss within iterations, where is the neural network width, is the scaling factor of the neural network and is the neural network depth.

• We also establish the generalization guarantees for both GD and SGD in the same setting. Specifically, for GD, we establish a sample complexity for a wide range of network width. For SGD, we prove a sample complexity. For both algorithms, our results provide tighter sample complexities based on milder network width conditions compared with existing results.

• Our theoretical results can be generalized to the scenarios with different data separability assumptions studied in the literature, and therefore can cover and improve many existing results in the NTK regime. Specifically, under the data separability assumptions studied in Cao and Gu (2019a); Ji and Telgarsky (2019b), our results hold with , where

is the failure probability parameter. This suggests that a neural network with width

can be learned by GD/SGD with good optimization and generalization guarantees. Moreover, we also show that under a very mild data nondegeneration assumption in Zou et al. (2019), our theoretical result can lead to a sharper over-parameterization condition, which improves the existing results in Zou et al. (2019) if the neural network depth satisfies .

In terms of optimization, a line of work focuses on the optimization landscape of neural networks (Haeffele and Vidal, 2015; Kawaguchi, 2016; Freeman and Bruna, 2017; Hardt and Ma, 2017; Safran and Shamir, 2018; Xie et al., 2017; Nguyen and Hein, 2017; Soltanolkotabi et al., 2018; Zhou and Liang, 2017; Yun et al., 2018; Du and Lee, 2018; Venturi et al., 2018; Nguyen, 2019). They study the properties of landscape of the optimization problem in deep learning, and demonstrate that under certain situations the local minima are also globally optimal. However, most of the positive results along this line of work only hold for simplified cases like linear networks or two-layer networks under certain assumptions on the input/output dimensions and sample size.

For the generalization of neural networks, a vast amount of work has established uniform convergence based generalization error bounds (Neyshabur et al., 2015; Bartlett et al., 2017; Neyshabur et al., 2018; Golowich et al., 2018; Arora et al., 2018; Li et al., 2018a). While such results can be applied to the mean-field regime to establish certain generalization bounds (Wei et al., 2019), the bounds are loose when applied to the NTK regime due to the larger scaling of network parameters. For example, some case studies in Cao and Gu (2019b) showed that the resulting uniform convergence based generalization bounds are increasing in the network width .

Another important topic on neural networks is the implicit bias of training algorithms such as GD and SGD. Overall, the study of implicit bias aims to figure out the specific properties of the solutions given by a certain training algorithm, as the solutions to the optimization problem may not be unique. Along this line of research, many prior work (Gunasekar et al., 2017; Soudry et al., 2018; Ji and Telgarsky, 2019a; Gunasekar et al., 2018a, b; Nacson et al., 2019b; Li et al., 2018b)

has studied implicit regularization/bias of gradient flow, GD, SGD or mirror descent for matrix factorization, logistic regression, and deep linear networks. However, generalizing these results to deep non-linear neural networks turns out to be much more challenging.

Nacson et al. (2019a); Lyu and Li (2019) studied the implicit bias of deep homogeneous model trained by gradient flow, and proved that the convergent direction of parameters is a KKT point of the max-margin problem. Nevertheless, they cannot handle practical optimization algorithms such as GD and SGD, and did not characterize how large the resulting margin is.

Several recent results have proved that neural networks can outperform kernel methods or behave differently than NTK-based kernel regression under certain conditions. Wei et al. (2019) studied the convergence of noisy Wasserstein flow in the mean-field regime, while Allen-Zhu and Li (2019) studied three layer ResNets with a scaling similar to the mean-field regime. Moreover, Allen-Zhu et al. (2019a); Bai and Lee (2019) studied fully-connected three-layer or two-layer networks with a scaling similar to the NTK regime, but utilized certain randomization tricks to make the network “almost quadratic” instead of “almost linear” in its parameters, making the network behave differently from the standard NTK regime.

Notation. For two scalars and , we use and to denote and respective. Given two scalars , we denote by if and if

. For a vector

we use to denote its Euclidean norm. For a matrix , we use and to denote its spectral norm and Frobenius norm respectively, and denote by the entry of at the -th row and -th column. Given two matrices and with the same dimension, we denote . Given a collection of matrices and a function mapping , we define by the partial gradient of with respect to and . Given two collections of matrices and , we denote and . Given two sequences and , we denote if for some absolute positive constant , if for some absolute positive constant , and if for some absolute constants and . We also use notations and to hide logarithmic factors in and respectively. Moreover, given a collection of matrices and a positive scalar , we denote .

## 2 Preliminaries on Learning Neural Networks

In this section we introduce the problem setting studied in this paper, including definitions of the network function and loss function, and the detailed training algorithms, i.e., GD and SGD with random initialization.

Neural network function. Given an input , the output of deep fully-connected ReLU network is defined as follows,

 f\Wb(\xb)=mα\WbLσ(\WbL−1⋯σ(\Wb1\xb)⋯),

where is a scaling parameter, , and . We denote the collection of all weight matrices as .

Loss function. Given training dataset with input and output , we define the training loss function as

 LS(\Wb)=1nn∑i=1Li(\Wb),

where is defined as the cross-entropy loss.

Algorithms. We consider both gradient descent and stochastic gradient descent with Gaussian random initialization. These two algorithms are displayed in Algorithms 1 and 2 respectively. Specifically, the entries in

are generated independently from univariate Gaussian distribution

and the entries in are generated independently from . For GD, we consider using the full gradient to update the model parameters. For SGD, we consider only using one training data in each iteration.

## 3 Main Theory

In this section, we present the main theoretical results about the optimization and generalization guarantees of GD and SGD for learning deep ReLU networks. We first make the following assumption on the training data points. All training data points satisfy , . This assumption has been widely made in many previous work (Allen-Zhu et al., 2019b, c; Du et al., 2019b, a; Zou et al., 2019) in order to simplify the theoretical analysis. We also make the following assumption regarding the loss function . There exists a positive constant and such that for all . Considering a sufficiently small , Assumption 3 spells out that there exists a neural network model with parameters

such that all training data points can be correctly classified, i.e., achieving zero training error. We claim that this is a common empirical observation, thus Assumption

3 can be easily satisfied in practice. Moreover, note that we consider cross-entropy loss, therefore, Assumption 3 is equivalent to . In Section 4, we will show that Assumption 3 can be implied by a variety of assumptions made in prior work.

In what follows, we are going to deliver our main theoretical results regarding the optimization and generalization guarantees of learning deep ReLU networks. Specifically, we consider two training algorithms: GD and SGD with random initialization in Algorithms 1 and 2. We will thoroughly analyze these two algorithms separately.

The following theorem establishes the global convergence of GD for training deep ReLU networks for binary classification. For any , there exists that satisfies

 m∗(δ,R,L,α)=~\cO([poly(R,L)]1/α⋅log2/(2−α)(n/δ)),

such that if , with probability at least over the initialization, GD with step size can train a neural network to achieve at most training loss within iterations.

Theorem 3.1 suggests that the minimum required neural network width, i.e., , is polynomially large in and and has a logarithmic dependency on the training sample size and the failure probability parameter . As will be discussed in Section 4, if the training data can be separated by neural tangent random feature model or shallow neural tangent kernel, is in the order of . This further implies that is sufficient to guarantee the global convergence of GD. We would also like to remark that Theorem 3.1 will not hold for larger given that , which implies that one needs to apply early stopping when running Algorithm 1.

Then we characterize the generalization performance of the neural network trained by GD in the following theorem. Under the same assumptions as Theorem 3.1, with probability at least , the iterate of Algorithm 1 satisfies that

for all .

Theorem 3.1 provides an algorithm independent generalization bound. Note that the second term in the bound distinguishes our result from most of the previous work on the algorithm-dependent generalization bounds of over-parameterized neural networks (Allen-Zhu et al., 2019a; Arora et al., 2019a; Cao and Gu, 2019b; Yehudai and Shamir, 2019; Cao and Gu, 2019a; Nitanda and Suzuki, 2019). Specifically, while these previous results mainly focus on establishing a bound that does not explode when the network width goes to infinity, our result covers a wider range of , and therefore implements different bounds for small or large ’s. As will be shown in Section 4, under various assumptions made in previous work, Assumption 3 holds for , and therefore Theorem 3.1 guarantees a sample complexity of order for , which has not been covered by previous results.

A trend can be observed in Theorem 3.1: the generalization error bound first increases with the network width and then starts to decrease when becomes even larger. This to certain extent bears a similarity to the “double descent” phenomenon studied in a recent line of work (Belkin et al., 2019a, b; Hastie et al., 2019; Mei and Montanari, 2019). However, since Theorem 3.1 only demonstrates a double descent curve for an upper bound of the generalization error, it is not sufficient to give any conclusive result on the double descent phenomenon. In fact, for two-layer networks, Ji and Telgarsky (2019b) has proved a generalization error bound that does not depend on over the range , under certain data separability assumptions. Therefore, it is possible that the double descent curve in our bound is an artifact of our analysis. We believe a further analysis on the generalization error and its relation to the double descent curve is an important future direction.

In this part, we are going to characterize the performance of SGD for training deep ReLU networks. Specifically, the following Theorem establishes an generalization error bound in terms of the output of SGD, under certain condition on the neural network width.

For any , there exists that satisfies

 m∗(δ,R,L,α)=~\cO([poly(R,L)]1/α⋅log2/(2−α)(n/δ)),

such that if , with probability at least , SGD with step size achieves

 \EE[L0−1\cD(^\Wb)]≤8L2R2n+8log(1/δ)n+24ϵ,

where the expectation is taken over the uniform draw of from . Theorem 3.2 gives a sample complexity for deep ReLU networks trained with SGD. Treating as a constant, then as long as (which we will verify in Section 4 under various conditions), this is a sample complexity of order . Our result improves the results given by Allen-Zhu et al. (2019a); Cao and Gu (2019a) in two aspects. First, the sample complexity is improved from (Allen-Zhu et al., 2019a; Cao and Gu, 2019a) to . Moreover, while Allen-Zhu et al. (2019a); Cao and Gu (2019a) requires , our result works for .

## 4 Discussions on Data Separability

In this section, we will discuss different data separability assumptions made in existing work. Specifically, we will show that the assumptions on training data made in Cao and Gu (2019a), Ji and Telgarsky (2019b) and Zou et al. (2019) can imply Assumption 3, and thus our theoretical results can be directly applied to these settings.

### 4.1 Data Separability by Neural Tangent Random Feature Model

We formally restate the definition of Neural Tangent Random Feature (NTRF) introduced in Cao and Gu (2019a) as follows. Let be the initialization weights, the NTRF function class is defined as follows

 \cF(\Wb(0),R,α)={f(⋅)=f\Wb(0)(⋅)+⟨∇f\Wb(0)(⋅),\Wb⟩:\Wb∈\cB(0,R⋅m−α)}.

The NTRF function class is closely related to the neural tangent kernel. For wide enough neural networks, it has been shown that the functions NTRF model can learn are in the NTK-induced reproducing kernel Hilbert space (RKHS) (Cao and Gu, 2019a). The following proposition states that if there is a good function in NTRF function class that achieves small training loss, Assumption 3 can also be satisfied. Suppose there is a function such that for all , then Assumption 3 can be satisfied with .

Proposition 4.1 states that if the training data can be well classified by a function in the NTRF function class, they can also be well learned by deep ReLU networks. However, one may ask in which case there exists such a good function in the NTRF function class, and what is the corresponding value of ? We further provide such an example by introducing the following assumption on the neural tangent random features, i.e., .

There exists a collection of matrices satisfying , such that for all

 yi⟨∇f\Wb(0)(\xbi),\Ub∗⟩≥mαγ,

where is an absolute positive constant111The factor is introduced here since the Frobenius norm is typically in the order of . . Then based on Proposition 4.1, the following corollary shows that under Assumption 4.1, Assumption 3 can be satisfied with a certain choice of .

Under Assumption 4.1, Assumption 3 can be satisfied with for some absolute constant . Corollary 4.1 shows that if NTRF of all training data are linear separable with constant margin, Assumption 3 can be satisfied with the radius parameter logarithmic in , and . Substituting this result into Theorems 3.1, 3.1 and 3.2, it can be shown that a neural network with width suffices to guarantee good optimization and generalization performances for both GD and SGD.

### 4.2 Data Separability by Shallow Neural Tangent Model

In this subsection we study the data separation assumption made in Ji and Telgarsky (2019b) and show that our resutls covers this particular setting. We first restate the assumption made in Ji and Telgarsky (2019b) as follows. There exist and such that for any , and for all ,

 yi∫\RRdσ′(⟨\zb,\xbi⟩)⋅⟨¯\ub(\zb),\xbi⟩dμN(\zb)≥γ.

Assumption 4.2 is related to the linear separability of the gradients of the first layer parameters at random initialization, where the randomness is replaced with an integral by taking the infinite width limit. Note that similar assumptions have also been studied in Cao and Gu (2019b); Frei et al. (2019), where the gradients with respect to the second layer weights instead of the first layer weights are considered. In the following, we mainly focus on Assumption 4.2. However we remark that our result also covers the setting studied in Cao and Gu (2019b); Frei et al. (2019).

In order to make a fair comparison, we reduce our results for multilayer networks to the one-hidden-layer setting:

 f\Wb(\xb)=mα\Wb2σ(\Wb1).

Then we provide the following proposition, which states that Assumption 4.2 can also imply Assumption 3 with a certain choice of .

Suppose the training data satisfy Assumption 4.2, then if the neural network width satisfies , Assumption 3 can be satisfied with for some absolute constant . Proposition 4.2 suggests that for two-layer ReLU networks, under Assumption 4.2, Assumption 3 can be satisfied with . Plugging this into Theorem 3.1, and setting , the condition on the neural network width becomes 222Similar to Ji and Telgarsky (2019b), the margin parameter is considered as a constant and thus will not appear in the condition on ., which matches the condition proved in Ji and Telgarsky (2019b) if choosing .

### 4.3 Class-dependent Data Nondegeneration

In Zou et al. (2019), an assumption on the minimum distance between inputs from different classes is made to guarantee the convergence of gradient descent to a global minimum. We restate this training data assumption as follows. For all if , then for some absolute constant . In contrast to the data nondegeneration assumption (i.e., no duplicate data points) made in Allen-Zhu et al. (2019b); Du et al. (2019b, a); Oymak and Soltanolkotabi (2019a); Zou and Gu (2019)333Specifically, Allen-Zhu et al. (2019b); Oymak and Soltanolkotabi (2019a); Zou and Gu (2019) require that any two data points are separated by a positive distance. Zou and Gu (2019) shows that this assumption is equivalent to those made in Du et al. (2019b, a), which require that the composite kernel matrix is strictly positive definite., Assumption 4.3 only requires that the data points from different classes are nondegenerate, thus we call it class-dependent data nondegeneration assumption. It is clear that Assumption 4.3 is milder since it can allow the data points to be arbitrary close as long as they are from the same class, while the data nondegeneration assumption requires that any two data points should be separated by a constant distance.

Then we provide the following proposition which shows that Assumption 4.3 also implies Assumption 3 for certain choices of and . Suppose the training data points satisfy Assumption 4.3, then if

 m=Ω([L2n9/2ϕ−2log(n/(δϵ))]1/α),

Assumption 3 can be satisfied with for some absolute constant . Proposition 4.3 suggests that when the neural network is sufficiently wide, as long as there exists no duplicate training data from different classes, Assumption 3 can still be satisfied with . We can also plug this result into Theorem 3.1, which gives the over-parameterization condition 444The detailed dependency on and of defined in Theorem 3.1 can be found in (4). Here hides logarithmic dependency on , and . if choosing . Compared with the counterpart proved in Zou et al. (2019), i.e., , our result is strictly sharper if the network depth satisfies .

## 5 Proof of the Main Theory

In this section we present the proofs of our main results in Section 3.

### 5.1 Proof of Theorem 3.1

We first present the following lemma which states that the neural network function is almost linear in terms of its weights. [Lemma 4.1 in Cao and Gu (2019a)] With probability at least over the randomness of initialization, for all and with , it holds that

 ∣∣f\Wb′(\xbi)−f\Wb(\xbi)−⟨∇f\Wb(\xbi),\Wb′−\Wb⟩∣∣=O(τ1/3L2mα√logm)L−1∑l=1∥\Wb′l−\Wbl∥2. (1)

We make a slight modification of its original version in Cao and Gu (2019a) as our neural network function encloses an additional scaling parameter . Then assuming that all iterates are close to the initialization, we establish a convergence guarantee of GD in the following lemma. Set the step size and . Then given and suppose for all , with probability at least over the randomness of initialization it holds that

 L∑l=1∥\Wb(0)l−\Wb∗l∥2F−L∑l=1∥\Wb(t′)l−\Wb∗l∥2F≥η⋅[t′−1∑t=0LS(\Wb(t))−2t′ϵ],

Then the remaining part is to characterize that under which condition on , we can guarantee all iterates are staying inside the required region until convergence. Based on Lemmas 5.1 and 5.1, we complete the remaining proof as follows.

###### Proof of Theorem 3.1.

In the following proof we choose and . Note that Lemmas 5.1 and 5.1 hold with probability at least over the randomness of initialization and . Therefore, if the neural network width satisfies

 m=Ω[L1/(2−α)log(m)2/(2−α)log(nL2/δ)2/(2−α)], (2)

we have with probability at least all results in Lemmas 5.1 and 5.1 hold.

Then we prove the theorem by two parts: 1) we show that all iterates will stay inside the region ; and 2) we show that gradient descent can find a neural network with at most training loss within iterations.

All iterates stay inside . We prove this part by induction. Specifically, given , we assume for all and prove that . First, it is clear that . Then by Lemma 5.1 and apply the fact that , we have

 L∑l=1∥\Wb(t′)l−\Wb∗l∥2F≤L∑l=1∥\Wb(0)l−\Wb∗l∥2F+2ηt′ϵ

Note that and , we have

 L∑l=1∥\Wb(t′)l−\Wb∗l∥2F ≤CLR2m−2α,

where is an absolute constant. Therefore, by triangle inequality, we further have the following for all ,

 ∥\Wb(t′)l−\Wb(0)l∥F≤∥\Wb(t′)l−\Wb∗l∥F+∥\Wb(0)l−\Wb∗l∥F≤√CLRm−α+Rm−α≤2√CLRm−α. (3)

Therefore, in order to guarantee that , by our choice of , it suffices to ensure that . Combining with the condition on provided in (2), we have if

 m≥m∗(δ,R,L,α)=~\cO([R4L11]1/α⋅log2/(2−α)(n/δ)), (4)

the iterate will be staying inside the region , which completes the proof of the first part.

Convergence of gradient descent. By Lemma 5.1, we have

 L∑l=1∥\Wb(0)l−\Wb∗l∥2F−L∑l=1∥\Wb(T)l−\Wb∗l∥2F≥η(T−1∑t=0LS(\Wb(t))−2Tϵ).

Dividing by on the both sides, we get

 1TT−1∑t=0LS(\Wb(t)) ≤∑Ll=1∥\Wb(0)l−\Wb∗l∥2F−∑Ll=1∥\Wb(T)l−\Wb∗l∥2FηT+2ϵ ≤LR2m−2αηT+2ϵ ≤3ϵ,

where the second inequality is by the fact that and the last inequality is by our choices of and which ensure that . Notice that . This completes the proof of the second part, and we are able to complete the proof. ∎

### 5.2 Proof of Theorem 3.1

Following Cao and Gu (2019b), we first introduce the definition of surrogate loss of the network, which is defined by the derivative of the loss function. We define the empirical surrogate error and population surrogate error as follows:

 \cES(\Wb):=−1nn∑i=1ℓ′[yi⋅f\Wb(\bxi)], \cE\cD(\Wb):=\EE(\xb,y)∼\cD{−ℓ′[y⋅f\Wb(\xb)]}.

The following lemma gives uniform-convergence type of results for utilizing the fact that is bounded and Lipschitz continuous. For any , suppose that . Then with probability at least , it holds that

 |\cE\cD(\Wb)−\cES(\Wb)|≤~\cO(min{4LL3/2~R√mn,L~R√n+L3~R4/3mα/3})+\cO(√log(1/δ)n)

for all

We are now ready to prove Theorem 3.1, which combines the trajectory distance analysis in the proof of Theorem 3.1 with Lemma 5.2.

###### Proof of Theorem 3.1.

With exactly the same proof as Theorem 3.1, by (3) and induction we have with . Therefore by Lemma 5.2, we have

 |\cE\cD(\Wb(t))−\cES(\Wb(t))|≤~\cO(min{4LL2R√mn,L3/2R√n+L11/3R4/3mα/3})+\cO(√log(1/δ)n)

for all . Note that we have . Therefore,

 \EE[L0−1\cD(^\Wb)] ≤2\cE\cD(\Wb(t)) ≤2\EE[LS(^\Wb)]+~\cO(min{4LL2R√mn,L3/2R√n+L11/3R4/3mα/3})+\cO(√log(1/δ)n).

This finishes the proof. ∎

### 5.3 Proof of Theorem 3.2

In this section we provide the proof of Theorem 3.2. The following result is the counterpart of Lemma 5.1 for SGD. Set the step size and . Then given a positive integer and suppose for all , with probability at least over the randomness of initialization it holds that

 L∑l=1∥\Wb(0)l−\Wb∗l∥2F−L∑l=1∥\Wb(n′)l−\Wb∗l∥2F≥η(n′∑i=1Li(\Wb(i−1))−2n′ϵ),

Our proof is based on the application of Lemma 5.3 and an online-to-batch conversion argument (Cesa-Bianchi et al., 2004), which is inspired by Cao and Gu (2019a); Ji and Telgarsky (2019b). We denote . The following lemma is provided in Ji and Telgarsky (2019b), whose proof only relies on the boundedness of and therefore is applicable in our setting.

For any , with probability at least , the iterates of Algorithm 2 satisfies that

 1nn∑i=1\cE\cD(\Wb(i−1))≤4nn∑i=1\cEi(\Wb(i−1))+4log(1/δ)n.
###### Proof of Theorem 3.2.

Similar to the proof of Theorem 3.1, we prove this theorem in two parts: 1) all iterates stay inside ; and 2) convergence of SGD.

All iterates stay inside . Similar to the proof of Theorem 3.1, we prove this part by induction. Assuming satisfies for all , by Lemma 5.3, we have

 L∑l=1∥\Wb(n′)l−\Wb∗l∥2F≤L∑l=1∥\Wb(0)l−\Wb∗l∥2F+2n′ηϵ≤LR2⋅m−2α+2nηϵ,

where the second inequality is by and . Then by triangle inequality, we further get

 ∥\Wb(n′)l−\Wb(0)l∥F≤∥\Wb(n′)l−\Wb∗l∥F+∥\Wb∗l−\Wb(0)l∥F≤\cO(√LRm−α+√nηϵ).

Then by our choices of , and , it can be easily verified that if , we have . This completes the proof of the first part.

Convergence of online SGD By Lemma 5.3, we have

 L∑l=1∥\Wb(0)l−\Wb∗l∥2F−L∑l=1∥\Wb(n)l−\Wb∗l∥2F≥η(n∑i=


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The radical symbol (√) represents the square root of a number. You can encounter the radical symbol in algebra or even in carpentry or another trade that involves geometry or calculating relative sizes or distances. You can multiply any two radicals that have the same indices (degrees of a root) together. If the radicals do not have the same indices, you can manipulate the equation until they do. If you want to know how to multiply radicals with or without coefficients, just follow these steps.

### Method 1 Multiply Radicals Without Coefficients

1. 1
Make sure that the radicals have the same index. To multiply radicals using the basic method, they have to have the same index. The "index" is the very small number written just to the left of the uppermost line in the radical symbol. If there is no index number, the radical is understood to be a square root (index 2) and can be multiplied with other square roots. You can multiply radicals with different indexes, but that is a more advanced method and will be explained later. Here are two examples of multiplication using radicals with the same indexes:
• Ex. 1: √(18) x √(2) = ?
• Ex. 2: √(10) x √(5) = ?
• Ex. 3: 3√(3) x 3√(9) = ?
2. 2
Multiply the numbers under the radical signs. Next, simply multiply the numbers under the radical or square root signs and keep them there. Here's how you do it:
• Ex. 1: √(18) x √(2) = √(36)
• Ex. 2: √(10) x √(5) = √(50)
• Ex. 3: 3√(3) x 3√(9) = 3√(27)
3. 3
Simplify the radical expressions. If you've multiplied radicals, there's a good chance that they can be simplified to perfect squares or perfect cubes, or that they can be simplified by finding a perfect square as a factor of the final product. Here's how you do it:
• Ex. 1: √(36) = 6. 36 is a perfect square because it is the product of 6 x 6. The square root of 36 is simply 6.
• Ex. 2: √(50) = √(25 x 2) = √([5 x 5] x 2) = 5√(2). Though 50 is not a perfect square, 25 is a factor of 50 (because it divides evenly into the number) and is a perfect square. You can break 25 down into its factors, 5 x 5, and move one 5 out of the square root sign to simplify the expression.
• You can think of it like this: If you throw the 5 back under the radical, it is multiplied by itself and becomes 25 again.
• Ex. 3:3√(27) = 3. 27 is a perfect cube because it's the product of 3 x 3 x 3. The cube root of 27 is therefore 3.

### Method 2 Multiply Radicals with Coefficients

1. 1
Multiply the coefficients. The coefficients are the numbers outside of a radical. If there is no given coefficient, then the coefficient can be understood to be 1. Multiply the coefficients together. Here's how you do it:
• Ex. 1: 3√(2) x √(10) = 3√( ? )
• 3 x 1 = 3
• Ex. 2: 4√(3) x 3√(6) = 12√( ? )
• 4 x 3 = 12
2. 2
Multiply the numbers inside the radicals. After you've multiplied the coefficients, you can multiply the numbers inside the radicals. Here's how you do it:
• Ex. 1: 3√(2) x √(10) = 3√(2 x 10) = 3√(20)
• Ex. 2: 4√(3) x 3√(6) = 12√(3 x 6) = 12√(18)
3. 3
Simplify the product. Next, simplify the numbers under the radicals by looking for perfect squares or multiples of the numbers under the radicals that are perfect squares. Once you've simplified those terms, just multiply them by their corresponding coefficients. Here's how you do it:
• 3√(20) = 3√(4 x 5) = 3√([2 x 2] x 5) = (3 x 2)√(5) = 6√(5)
• 12√(18) = 12√(9 x 2) = 12√(3 x 3 x 2) = (12 x 3)√(2) = 36√(2)

### Method 3 Multiply Radicals with Different Indices

1. 1
Find the LCM (lowest common multiple) of the indices. To find the LCM of the indexes, find the smallest number that is evenly divisible by both indices. Find the LCM of the indices for the following equation:3√(5) x 2√(2) = ?
• The indices are 3 and 2. 6 is the LCM of these two numbers because it is the smallest number that is evenly divisible by both 3 and 2. 6/3 = 2 and 6/2 = 3. To multiply the radicals, both of the indices will have to be 6.
2. 2
Write each expression with the new LCM as the index. Here's what the expressions would look like in the equation with their new indexes:
• 6√(5) x 6√(2) = ?
3. 3
Find the number that you would need to multiply each original index by to find the LCM. For the expression 3√(5), you'd need to multiply the index of 3 by 2 to get 6. For the expression 2√(2), you'd need to multiply the index of 2 by 3 to get 6.
4. 4
Make this number the exponent of the number inside the radical. For the first equation, make the number 2 the exponent over the number 5. For the second equation, make the number 3 the exponent over the number 2. Here's what it would look like:
• 2 --> 6√(5) = 6√(5)2
• 3 --> 6√(2) = 6√(2)3
5. 5
Multiply the numbers inside the radicals by their exponents. Here's how you do it:
• 6√(5)2 = 6√(5 x 5) = 6√25
• 6√(2)3 = 6√(2 x 2 x 2) = 6√8
6. 6
Place these numbers under one radical. Place them under a radical and connect them with a multiplication sign. Here's what the result would look like: 6√(8 x 25)
7. 7
Multiply them. 6√(8 x 25) = 6√(200). This is the final answer. In some cases, you may be able to simplify these expressions -- for example, you could simplify this expression if you found a number that can be multiplied by itself six times that is a factor of 200. But in this case, the expression cannot be simplified any further.

## Community Q&A

Search
• Can I multiply a number inside the radical with a number outside the radical?
wikiHow Contributor
Only if you are reversing the simplification process. For example, 3 with a radical of 8. 3 squared is 9, so you multiply 9 under the radical with the eight for the original. It would be 72 under the radical.
• Can you multiply radicals with the same bases but indexes?
Yes, though it's best to convert to exponential form first. To multiple squareroot2 by cuberoot2, write it as 2^(1/2)*2^(1/3) . Then the rules of exponents make the next step easy as adding fractions: = 2^((1/2)+(1/3)) = 2^(5/6).
• How do you multiply a coefficient and a radical by a radical?
wikiHow Contributor
Multipy the radicals together, then place the coeffcient in front of the result. For example, to multiply 2√2 and √3, first multiply √2 and √3 to get √6, then put the coeffcient of 2 in front to get 2√6.

## Tips

• If a "coefficient" is separated from the radical sign by a plus or minus sign, it's not a coefficient at all--it's a separate term and must be handled separately from the radical. If a radical and another term are both enclosed in the same set of parentheses--for example, (2 + (square root)5), you must handle both 2 and (square root)5 separately when performing operations inside the parentheses, but when performing operations outside the parentheses you must handle (2 + (square root)5) as a single whole.
• Radical signs are another way of expressing fractional exponents. In other words, the square root of any number is the same as that number raised to the 1/2 power, the cube root of any number is the same as that number raised to the 1/3 power, and so on.
• A "coefficient" is the number, if any, placed directly in front of a radical sign. So for example, in the expression 2(square root)5, 5 is beneath the radical sign and the number 2, outside the radical, is the coefficient. When a radical and a coefficient are placed together, it's understood to mean the same thing as multiplying the radical by the coefficient, or to continue the example, 2 * (square root)5.

## Article Info

Categories: Multiplication and Division

In other languages:

Italiano: Moltiplicare i Radicali, Español: multiplicar radicales, Português: Multiplicar Radicais, Deutsch: Wurzelterme multiplizieren, Français: multiplier des racines, Русский: умножать корни, 中文: 求根式的乘积, Bahasa Indonesia: Mengalikan Akar

Thanks to all authors for creating a page that has been read 207,961 times.


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# How can igneouse and sedimentary rocks change into metamorphic rocks

###### Question:

How can igneouse and sedimentary rocks change into metamorphic rocks

### Plz help me well mark brainliest if you are correct!..

Plz help me well mark brainliest if you are correct!.....

### How to write an equation and solve it for the followingOne fifth of the employees of Delta Industries, Inc., work in the Sotheastern region. If the company employs 252 workers in that region, what is the total number of employees working for the company.Moreover, what type of calculator would I need to do this type of math on.

how to write an equation and solve it for the followingOne fifth of the employees of Delta Industries, Inc., work in the Sotheastern region. If the company employs 252 workers in that region, what is the total number of employees working for the company.Moreover, what type of calculator would I need...

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### 1. Explain in words how you would solve the following equation: 4x + 1 = 6x - 11

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### A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.

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### 60 POINTS !!! I NEED HELPPP What are supporting details in an informational text? changes made to the text to make the writing more clear, precise, and effective a person, place, thing, or event that stands for something more than itself the most important thought in the text information that clarifies, proves, or explains the main idea of the text

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### 1.One factor is a two digit number between 1 and 10.2. One factor is a two digit number between 1 and 10(different from the first). 3. The product is greater then 8,but less than 10.

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### I WILL MARK BRAINLLIEST IF U ANSWER PLS FASTTTT

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### Imagine that a researcher has dissected out a single muscle bundle from a triceps muscle biopsy. She sections this bundle and stains it for a variety of muscle-associated proteins. After staining, when the researcher zooms in on a sarcomere, she notices that a certain protein always dots the periphery along the length of the thin filaments. This protein is most likely:

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# Measuring Relative Position

## Key Questions

• A z-score indicates how many standard deviations a data point is from the mean. It is calculated with the following formula:

z = (X - μ) / σ, where $X$ is the value of tha data point, μ is the mean, and  σ is the standard deviation.
`
Besides telling us where a data point lies compared with the rest of the data set in relation to the mean, a z-score also allows comparisons of data points across different normal distributions. (For example, we can compare the scores obtained by a student in two exams whose scores are normally distributed).

• This is a somewhat complicated and nuanced question.

First, one must know what hypothesis test they are performing. In addition, if one knows the true distribution, then it is simply a matter of comparing the means of these parameters, because the true distribution is a constant.

Where z-scores become most helpful is in comparing two samples to see if they are from the same distribution or not. The z-score will be most helpful in comparing samples from normally distributed distributions, but the Central Limit Theorem also states that for large enough samples, comparing the mean approaches a normal distribution.

The calculations are different if the two samples are matched or unmatched. For both, you can compare the differences between Sample 1 and Sample 2 to a normal distribution with mean 0 and standard error based on the sample standard deviation(s) and size(s). The major difference is how you calculate the standard error.

Once you have the mean difference between the two distributions ($\overline{X}$) and the standard error SE, then your z-statistic is $z = \frac{\overline{X}}{S E}$. You can use this to calculate a p-value. For example, if $| z |$ > 1.96, then the p-value is <0.05.


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# An element with density 10 g cm−3 forms a cubic unit cell with edge length of 3 × 10−8 cm. What is the nature of the cubic unit cell if the atomic mass of the element is 81 g mol−1? - Chemistry

An element with density 10 g cm−3 forms a cubic unit cell with edge length of 3 × 10−8 cm. What is the nature of the cubic unit cell if the atomic mass of the element is 81 g mol−1?

#### Solution

Given:

Density (D)               = 10 g/cm3

Edge length (a)        = 3 x 10−8 cm

Atomic mass            = 81 g/mol

Formula:

10=(Zxx81)/((3xx10)^3xx6.022xx10^23)

Z=(10xx(3xx10)^3xx6.022xx10^23)/81=2.0073

∴ Z = 2

The unit cell contains 2 atoms, so it is a body-centred cubic unit cell.

Concept: Calculations Involving Unit Cell Dimensions
Is there an error in this question or solution?


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Can you arrange the digits $1,2,3,4,5,6,7,8,9$ into three $3$ - digit numbers such that their total is close to $1500$?
You might like to think about how many different ways there are of getting as close as you can to $1500$.


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Volume 358 - 36th International Cosmic Ray Conference (ICRC2019) - GRI - Gamma Ray Indirect
Cherenkov Telescope Array potential in the search for Galactic PeVatrons
E.O. Angüner,* F. Cassol, H. Costantini, C. Trichard, G. Verna
*corresponding author
Full text: pdf
Pre-published on: 2019 July 22
Published on:
Abstract
One of the major scientific objectives of the future Cherenkov Telescope Array (CTA) Observatory is the search for PeVatrons. PeVatrons are cosmic-ray factories able to accelerate nuclei at least up to the knee feature seen in the spectrum of cosmic rays measured near the Earth. CTA will perform a survey of the full Galactic plane at TeV energies and beyond with unprecedented sensitivity. The determination of efficient criteria to identify PeVatron candidates during the survey is essential in order to trigger further dedicated observations. Here, we present results from a study based on simulations to determine these criteria. The outcome of the study is a PeVatron figure of merit, defined as a metric that provides relations between spectral parameters and spectral cutoff energy lower limits. In addition, simulations of the PeVatron candidate HESS\,J1641$-$463 and its parental particle spectrum are presented and discussed. Eventually, our work is applied to simulated population of Galactic PeVatrons, with the aim to determine the sensitivity of CTA.
Open Access
Copyright owned by the author(s) under the term of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.


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Algebra Level 3

Find the number of solution(s) of $$x$$ satisfying $$\log_{4}(x - 1) = \log_{2}(x - 3)$$.

×


text: Which of the following sets of parametric equations represent the ellipse given by $\displaystyle \frac{x^2}{4}+\frac{y^2}{64}=1$? To receive credit, you must select all correct choices.


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# why is the magnetic field circular

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference, why is the magnetic field around the wire is circular?

• I don't know enough to provide you with why the magnetic field is circular, but I don't think your terminology is quite correct. The magnetic field is certainly not the electric field viewed from another Gaililean frame. The magnetic field arises when the electric field is transformed under relativity. The exact transformation in your example above will have to come from someone else. – Reid Erdwien Mar 21 '15 at 5:16
• why do you think it is circular? – Skaperen Mar 23 '15 at 9:23
• what if the wires are charged but no current is flowing? – Skaperen Mar 23 '15 at 9:25
• If I assume the essence of the question is why the magnetic field force lines do not originate from anywhere ? That’s a question that just cannot be explained from “credible and/or official sources”. In other words: there is no answer yet. – Leon Sprenger Mar 24 '15 at 19:04
• @Skaperen: If the wires are charged with nu current flowing there is no magnetic field. – Leon Sprenger Mar 25 '15 at 16:07

Your statement is not really true, since if you only have a magnetic field in one frame of reference, then it can never be viewed as just an electric field in another frame of reference. And vice-versa.

As described here, the magnetic field can be defined as (e.g. in Jackson's Classical Electrodynamics) the field that is responsible for the Lorentz force $q\vec{v} \times \vec{B}$. Since in the example you show, the force would always be directed radially for charges moving parallel to a current carrying wire, then the field must circulate around the wire.

The reason that the force associated with the magnetic field is radial in such circumstances is one of the set-piece arguments in most textbooks that deal with these things, but arises from the requirement that a charge that is radially stationary with respect to the wire in one frame of reference is also stationary in any other frame of reference moving parallel to the wire. It goes something like this:

Consider the electric/magnetic fields due to a current carrying wire in the stationary frame and a frame moving uniformly, but parallel to the wire.

In the stationary frame, the wire is overall neutral, so there can only be a magnetic field. In the moving frame there is some transformed magnetic field and an electric field radial to the wire, caused by a difference in length contraction for the positive and negative charges in the wire, which due to the current flow, must be moving in opposite directions in the stationary frame.

This electric field in the moving frame clearly exerts a radial force on any test charge originally at rest with respect to the wire in the stationary frame. But, given that there is no radial force or acceleration in the stationary frame, there also cannot be a net radial force on the charge when it is in the moving frame either. The force that counteracts the radial electric field in the moving frame is the Lorentz force due to a mystery (B-)field. As the Lorentz force due to the mysterious (B-)field is observed to be both proportional to and perpendicular to the velocity, then it is natural to define it in terms of a vector product. And in that case, in order to act radially for a charge at any point around the wire, the B-field must circulate around the wire.

• Thanks. Why does B,according to Biot-Savart Law, have a cross product? Magnetic has a different direction from the magnetic force. Electric field and force have same direction, but this not true for magnetic field and force. I am trying to understand the logic behind the cross product in magnetism – user50322 Mar 24 '15 at 16:41
• @user50322 The Lorentz force is defined in that way. I suppose what you need to think about is if you wanted to define an intrinsic field that produced a radial force on a particle moving parallel to the wire, what else could it be but circulating around the wire. It must be a vector and it must produce a force perpendicular to the velocity for all positions around the wire. – Rob Jeffries Mar 24 '15 at 19:59
• Same direction with magnetic force. Like electric field and electric force. – user50322 Mar 24 '15 at 22:16
• @user50322 I understand (I think) your comment; but then how do you incorporate the fact that the magnitude of the force is proportional to the velocity? The force due to the electric field doesn't depend on the particle velocity. That is how they differ. In addition there are no sources or sinks of B-field, which precludes a radial field. – Rob Jeffries Mar 24 '15 at 22:19
• So can we say, magnetic field doesn't have physical meaning and we don't have any other way to define it besides defined in that way. So we define it that because we can not make any definition any other way. Right? Can we say that? – user50322 Mar 24 '15 at 22:47

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference

It is true that a pure electrostatic field in an inertial reference frame (IRF) will be observed as a mix of electric and magnetic fields in some relatively moving IRFs.

However, in the general (time varying) case, it is not possible to find an IRF in which the magnetic field vanishes.

why is the magnetic field around the wire is circular?

Consider the field of an isolated point charge at rest; a purely radial, static electric field.

From a relatively moving IRF, there is a magnetic field component in addition to the electric field. This magnetic field is perpendicular to the velocity vector and electric field in the rest frame and is given by

$$\mathbf {{B}_{\bot}}'= \gamma \left(-\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right)$$

A little reflection on the above should convince you that, looking at the charge along the direction of motion, the magnetic field lines form circles centered on the charge.

The extension to a line of charge is straightforward.

• Why does B have to be equal to v x E? Why does B have to be circular? Don't have the same direction with the magnetic(in fact electric) force? Thanks... – user50322 Mar 23 '15 at 23:07
• @user50322, what do you mean by "why?". If you are asking why nature is the way it is and not some other way, I don't have and answer (and there may not be one). – Alfred Centauri Mar 23 '15 at 23:15
• I mean "how". How does B equal to v x E? – user50322 Mar 23 '15 at 23:17
• @user50322, this is how the fields transform between relatively moving inertial reference frames according to special relativity. This is not something that can be explained in a comment or an answer for that matter. Simply put, Maxwell's equations, the equations that correctly describe classical electromagnetism, are relativistically covariant. – Alfred Centauri Mar 23 '15 at 23:33

Your question consists actually of two parts, I will answer them one-by-one:

Why is the magnetic field circular?

Any vector field $\vec F$ can be decomposed into a rotational part and a divergent part, according to the Helmholtz decomposition theorem
$$\vec F = - \vec \nabla \Phi + \vec \nabla \times \vec A$$
This is a purely mathematical statement and has nothing yet to do with physics. Physics comes into play when considering Maxwell's second equation
$$\vec \nabla \cdot \vec B = 0$$ Which means that $\vec B$ is divergence-free or source-free. Because we can use the decomposition theorem to calculate
$$\vec \nabla \cdot \vec B = \vec \nabla \cdot (-\vec \nabla \Phi + \nabla \times A) = \nabla \cdot \nabla \times A$$ We see that $B$ must be a purely rotational field and this results because the divergence of a gradient ($\vec \nabla \cdot \vec \nabla \Phi$) always vanishes. The latter statement also follows from math and is not a physical modell.

the second thing is:
How are magnetic and electric field related?

This now follows from Maxwells equations 3+4 other equations, that we didn't use until now. It's them that imply relativity and the transformations that my fellow posters noticed.

Concluding

So really the circularity of the B-field has nothing to do with relativity. But its relativity that allows us to transform between both, no matter how their geometries are.

• Yes but B is related with v x r (according to biot-savart). I understand why F is perpendicular to v(Lorentz Force). It is because energy conservation. But I don't understand why B and current must be perpendicular(Biot-Savart force) – user50322 Mar 25 '15 at 10:10
• The current is provided by the movement of the electrons. $\vec v \sim \vec j$ in metals. – AtmosphericPrisonEscape Mar 25 '15 at 12:36

Given a certain four-current $J^\mu = (c \varrho, \vec{j})$, that is a charge density $\varrho$ and current density $\vec{j}$. the four-potential $A^\mu = (\Phi / c, \vec{A})$ is given by: $$A^\mu(\vec{r},t) \propto \int \frac{j^\mu(\vec{r}\ ', t_r)}{|\vec{r}-\vec{r}\ '|} d^3r'$$ with $t_r = t - \frac{|\vec{r}-\vec{r}\ '|}{c}$ if one takes the retarded solution. Therefore the B-Field is: \begin{align} \vec{B} = \nabla \times \vec{A} & \propto \int \frac{\vec{j}(\vec{r}\ ', t_r) \times (\vec{r}-\vec{r}\ ')}{|\vec{r}-\vec{r}\ '|^3} d^3r' + \int \frac{\nabla \times \vec{j}(\vec{r}\ ', t_r)}{|\vec{r}-\vec{r}\ '|} d^3r' \\ & = \int \frac{\vec{j}(\vec{r}\ ', t_r) \times (\vec{r}-\vec{r}\ ')}{|\vec{r}-\vec{r}\ '|^3} d^3r' + \int \frac{\frac{\partial \vec{j}(\vec{r}\ ', t_r)}{\partial t_r} \times (\vec{r}-\vec{r}\ ')}{c|\vec{r}-\vec{r}\ '|^2} d^3r' \end{align}

The first term is which you would call "circular", since $\vec{j}(\vec{r}\ ', t_r) \times (\vec{r}-\vec{r}\ ')$ points always in a direction perpendicular to the current density and the point of interest (relative to the current density). The second term is zero if the current is stationary, that is if it is not time dependant. For example, this is the case if one looks at magnetic fields induced by a wire of constant charge flow.

So in general, the magnetic field is not circular around a wire.

• Yes but B is related with v x r (according to biot-savart). I understand why F is perpendicular to v(Lorentz Force). It is because energy conservation. But I don't understand why B and current must be perpendicular(Biot-Savart force) – user50322 Mar 25 '15 at 10:11
• Do you accept the maxwell equations? If so, my answer above shows you why. – image Mar 25 '15 at 14:47

since magnetic field in a wire carrying current is due to the movement of electrons, I assume that the magnetic field of a single isolated moving electron, or any other charged particle, can also be deduced by the right hand thumb rule i.e. the field will be much like Saturn's ring around it, where we assume saturn to be a charged particle, and its ring as its field. Now my question is: How is this field deduced. Why is it circular? Is it the result of superposition, if any? Secondly, I know that magnetic force exerted on a moving charge consists of a cross product of B and v. But I think that this force is the result of interaction between the charge's own magnetic field and the applied field, B. My question is: How does this interaction exactly happen? How does it result in this force? Most importantly, what causes this force to be perp to both B and v (I need an explanation other than that it is just the result of cross product)? I raised a similar question elsewhere but didn't get much help. I'd be grateful if somebody explains.

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference...

Relativity doesn't quite say this. Take a look at Minkowski's Space and Time: "In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy to the force-screw in mechanics; the analogy is, however, imperfect". Also see Jackson’s Classical Electrodynamics section 11.10 where he says "one should properly speak of the electromagnetic field Fμv rather than E or B separately". The field of the electron is the electromagnetic field, and it has a "screw" nature, which you can trace back to Maxwell. When you have two charged particles with no initial motion, you see linear electric force only. When you throw one past the other, you also see rotational magnetic force, as per positronium. IMHO the electron's electromagnetic field isn't totally unlike the frame-dragged gravitomagnetic field, and you can depict it by thinking "spinor" and combining radial electric field lines with concentric magnetic field lines, a bit like Maxwell's convergence + curl sketch on page 7 of this paper:

Why is the magnetic field around the wire circular?

Because it's rotationally symmetrical. Think of the wire as a column of electrons interleaved with a column of protons. Their electromagnetic fields cancel. But when you turn on the current and move the electrons, the fields don't quite cancel any more. The residual field has a cylindrical disposition, such that a charged particle thrown past the wire will loop around the magnetic field lines. We call it a magnetic field, but it's just one aspect of the greater whole that is the electromagnetic field.


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# Math Help - Integration of ( a gamma pdf)^-1

1. ## Integration of ( a gamma pdf)^-1

Hello,

could you please help me to calculate the integral of a pdf gama to the power of minus 1?

2. ## Re: Integration of ( a gamma pdf)^-1

Hey Kiki.

Check your other thread that you created on this topic.