text:
## The IBM BigFix Client setup could not find a valid masthead file.

Product:
Planning Analytics 2.0.8
Microsoft Windows 2016 Server

Problem:
How install the ILMT BIG FIX client to Planning Analytics servers?
IBM demand that virtual PA installations have the ILMT client installed and that the customer provide an ILMT report of the CPU usage for virtual installations.

How install the BIG FIX client:
You should have installed the ILMT and BIG FIX server in your organization on a separate server.
During the installation of the Big Fix server a license is created for your organization.
Above is a must for you to be able to install the client.

The simplest way is that you get a setup.exe and actionsite.afxm file from the server copied to your Planning Analytics (TM1) server – place all files in same folder (c:\install). If you do not have the afxm file you get a error like this:

Run the installation of the client program BigFix-BES-Client-9.5.14.73.exe

(The files to copy for the client, can be found on the BigFix server in folder C:\Program Files (x86)\BigFix Enterprise\BES Installers\Client)

Click OK to the language selection.

Click Next.

Click next to the selection of installation folder.

Click install.

Click finish.

The client should talk to the server and show up direct, inside the Big Fix server console.

Then the IT department need to activate the scans for IBM software.
https://help.hcltechsw.com/bigfix/9.5/inventory/Inventory/planinconf/t_scheduling_soft_scans.html

There is the possibility that IT department need to download other scan software, as part of the Big Fix server configuration to allow the Big Fix client to perform the software scan. This is done with the Big Fix system.

Install the installation generator first from here;

The server exe is found here;

After the installation generator is setup, you get instructions on how to install the clients.

The big fix program use port 52311 to talk to the server.

On the Big Fix server, launch console, create a lot of default values.

Please check the current license demands from IBM on the Cognos and TM1 products, as they may change.

## Exclude folders from anti-virus program

Product:
Planning Analytics Workspace
Microsoft Windows 2016 server

Problem:
PAW take long time to start when anti-virus software is running.
https://success.docker.com/article/endpoint-security-for-windows-containers

Suggested solution:
Depending on your anti virus program, the settings are different.

exclude this folders and sub-folders from scanning
c:\programdata\docker
c:\Program Files\ibm\cognos\tm1_64
c:\Program Files\ibm\cognos\analytics\temp\
c:\ProgramData\Microsoft\Windows\HNS
d:\ibm\paw
d:\docker
d:\tm1applications folder ( can be d:\tm1server\app1\data )

Folders name depend on the installation made of docker and PAW on your server.

For McAfee you can setup the exclude like this:
On-Access Default Processes Policies
**\Docker\*
**\ProgramData\Microsoft\Windows\HNS\*
**\Install\ipa_workspace_local_win*\*

Exclude also all “active scanning” and instead run a scan on the disk once a week, when you have stopped the PAW and Docker. Turn off Windows Defender Real-Time Protection.

Exclude scan of the processes used by docker and cognos, like java.exe.

There will likely be many containers depending on the same package layers. The same data stream of a given package file will provide the data for placeholders on multiple container system volumes. As a result, there is potential for redundant AV scans of the same data in every container. This has an unnecessary negative impact on the performance of containers. This is a signification cost given that containers are expected to start quickly and may be short-lived.

https://docs.microsoft.com/en-us/windows-hardware/drivers/ifs/anti-virus-optimization-for-windows-containers

https://success.docker.com/article/endpoint-security-for-windows-containers

https://support.symantec.com/us/en/article.tech183201.html

https://support.symantec.com/us/en/article.tech246815.html

https://kc.mcafee.com/corporate/index?page=content&id=KB90041

https://www.ibm.com/support/pages/which-cognos-files-need-be-excluded-virus-scan

## Limits exceeded

Product:
Planning Analytics 2.0.8
Microsoft Windows server 2016

Problem:
Open a tm1web view in the new PA version give error, it worked in the old TM1 server.

Error message:
Limits exceeded
The action has been terminated because it exceeds the configured maximum memory limit.

Solution:

Open tm1web_config.xml file in Notepad++. It is in folder C:\Program Files\IBM\cognos\tm1_64\webapps\tm1web\WEB-INF\configuration\

Having WorkbookMaxCellCount blank or setting it to less than 0 indicates that an unlimited cell count for workbooks is allowed.
Before IBM Planning Analytics version 2.0.7, the default value is -1, which indicates an unlimited number of cells are allowed in a workbook.
After IBM Planning Analytics version 2.0.7, the default value is 500000.
Setting this parameter to 0 indicates that workbooks cannot have any cells.
You must set it to above 0.

You must do a restart of the application server for the change to take affect.

If WorkbookMaxCellCount parameter is in tm1web_config.xml and it is not the default, when the user opens a workbook, the server validates its cell count against WorkbookMaxCellCount. If the cell count of the workbook exceeds WorkbookMaxCellCount, an error message is logged and the workbook is not opened.

Can also be other values that affect how tm1 show cells.

## cannot connect to TM1 Admin server

Product:
Cognos TM1 10.2.2
Microsoft Windows 2008 server

Problem:
Can not connect to TM1 server from client TM1 Architect.
Error:  TM1 Perspectives cannot connect to TM1 Admin server

Solution:

TELNET  servername  5498

if you get a black screen, and no error message, then you have a open connection.

Then check that you have the same version on you client as on the TM1 server.

You can check the file tm1api.dll in folder C:\Program Files\ibm\cognos\tm1_64\bin

Above the 10.2.2 TM1 client in RTM version.

Above the TM1 10.2.2 fix pack 7 version.  Version number 10.2.20700

https://www.ibm.com/support/pages/how-determine-exact-version-tm1-installed

If you do not have correct version, uninstall the TM1 client from your computer.

Install the correct program and update, and try again.

Run tm1_64b_10.2.2_win_ml.tar.gz and then up_tm1client_win64_10.2.5270.109_ml.tar.gz to install the needed software.

https://www.ibm.com/support/knowledgecenter/SSD29G_2.0.0/com.ibm.swg.ba.cognos.tm1_inst.2.0.0.doc/c_tm1_inst_defaultvaluesfortm1installation.html

https://www.ibm.com/support/pages/ibm-cognos-tm1-ssl-expiration-manual-fix-approach-landing-page

https://www.ibm.com/support/pages/node/553505

## PAW administration is grey for memory and cpu

Product:
Planning Analytics 2.0.8
Planning Analytics Workspace 46
Microsoft Windows 2016 Server

Problem:
When going to Planning Analytics Workspace Administration dialog some of the TM1 instance are grey, in the area for memory and cpu usage.

Error:

On PA Agent version 2.0.47.1220, you can find this kind of error in the file c:\Program Files\ibm\cognos\tm1_64\paa_agent\wlp\usr\servers\kate-agent\logs\messages.log
00000048 SystemErr R ERROR:status.py:Error while getting windows registry details: ERROR: The system was unable to find the specified registry key or value
00000048 SystemErr R ERROR:status.py:error getting the registry details: local variable ‘return_object’ referenced before assignment
00000048 com.ibm.pa.kate.agent.scheduler.ScheduledTask E Error occurred while updating the server info json file
Traceback (most recent call last):
File “kateagent/scripts\status.py”, line 1053, in <module>
logger.exception(
NameError: name ‘data_directory’ is not defined
0000005d com.ibm.pa.kate.agent.service.PAKateAgentService E Exception
java.lang.NullPointerException

Solution:
A space have come into the TM1 application name in the TM1S.CFG file, and when registered with Microsoft Service, got a extra space in the name.
The PA Agent can not parse the space in the name, and therefor will not display all values.

Stop the TM1 instance.
Edit the tm1s.cfg file to correct value.
ServerName= proven_techniques
should be changed to
ServerName=proven_techniques
Delete the TM1 instance in Cognos Configuration.
Check that the instance is gone – if not, you may need to delete it from a DOS command line.

Enter
sc delete ” proven_techniques”
in the elevated cmd prompt.
Now the windows service should be gone.
Go into Cognos Configuration again,
Add the new TM1 instance and enter a correct name “proven_techniques”
and point out the folder for the configuration file.
Start the service from inside Cognos Configuration.

Surf to PAW and check that the administration page is correct.

## How copy files between windows servers

Product:
Planning Analytics 2.0.5
Microsoft Windows 2016 server

Problem:
How do i copy files from one server to another server and keep the NTFS security?

Solution:
Check you have a file share on source server.
Create a file share on destination server.
Ensure that the user running the script is local administrator on both servers
Enter below line (modify it for your server names)
robocopy \\serverSource\ShareA \\serverDestination\ShareA /MIR /COPYALL /DCOPY:T /R:1 /W:1 /NP /TEE /LOG:d:\temp\ShareA-RobocopyResults.log
Save file as d:\temp\robocopyfiles.cmd

Open a command prompt as administrator
move to the folder where the script is
run
robocopyfiles.cmd

Check the log files for errors when it is done.

(solution is from this site https://social.technet.microsoft.com/Forums/en-US/3c42061c-d739-4c44-90ec-014b0e08c45c/how-to-copy-file-security-permissions-from-one-server-to-another-one-by-using-robocopy?forum=winserverfiles)
======================
TITLE “Copying ShareA from TestSource to TestDestination”
robocopy \\TestSource\ShareA \\TestDestination\ShareA /MIR /COPYALL /DCOPY:T /R:1 /W:1 /NP /TEE /LOG:C:\Logs\ShareA-RobocopyResults.log
PAUSE
======================

The switch functions are as follows:

/MIR – Creates a mirror copy.

/COPYALL – Copies ALL file attributes

/DCOPY – Copies (ie retains original) Directory Creation/Modified Timestamps. If you don’t include this, all directories will be recreated with NEW timestamps

/R:1 – Retry copy once on failure

/W:1 – Wait 1 second between retries

/NP – No Percentage Completion written to output

/TEE – Output progress to Command Window (I guess TEE = Terminal Emulation)

/LOG – Write out and save to a specified Log File

## Missing colors in TM1 App Web

Product:
Planning Analytics 2.0.8
Microsoft Windows 2016 server

Problem:
When you click on a TM1 application Web link in Cognos Connection (from Cognos Analytics 11) you get a error. You also noted that the SSO did not work to your Cognos Analytics 11 webpage.
You can also get into pmpsvc page, but with wrong colors.

Error:
error, see logs

Solution:
To make a link from CA work to Planning Analytics Web (PMPSVC) you must login correct to Cognos Analytics (CA11) first.
Ensure you surf to http://servername/ibmcognos and not to http://servername:9300/bi/v1/disp

Please also try to, in Internet Explorer change the Compatibility view settings to be blank.

“Display intranet sites in Compatibility View” should be checked off.

https://docs.microsoft.com/en-us/internet-explorer/ie11-deploy-guide/missing-the-compatibility-view-button

## Blank page when surf to contributor page

Product:
Planning Analytics 2.0.9
Microsoft Windows 2016 server

Problem:
After upgrade to Planning Analytics 2.0.9, user of TM1 Application web (contributor) get a blank page or below error in web browser when they try to open a link.

Error:

Solution:
Clear the web browser cache before you surf to PMPSVC webpage.

You can also try a script to clean the browser cache, please test the script in your lab first.
Below script will clean Chrome browser cache:


text:
# Nested roots

Algebra Level 3

$\Large \sqrt[5x]{\sqrt[4x]{\sqrt[3x]{\sqrt[2x]{x}}}} = \sqrt[x]{\sqrt[x^2]{\sqrt[x^3]{\sqrt[x^4]{x^5}}}}$

Find the real value of $x$ satisfying the real equation above.

The answer is of the form $\sqrt[6]{a}$, then submot the value of $a$.

Note:- $x \neq 1$

×

format: plain


text:
# Exponent Definition

An exponent is the number of times that a certain number is multiplied by itself. It is written as a small-sized number at the top right of the number.

In the example above, 5^2 is read as "5 to the power of 2" or "5 raised to 2".

## Product form of numbers

Large numbers broken down into smaller forms is known as product form of numbers.

To find the product form of numbers let's say 47
47 is written in the product form as 47= 4 × 4 × 4 × 4 × 4 × 4 × 4

## Expressing the product form of numbers by exponents

To express product from of the numbers by exponents we count that how many times the number multiplied by itself. That is the exponent of that number.

To find the exponent form of numbers let's say 32.
Write the product form of 32. 32= 2 x 2 x 2 x 2 x 2= 25.
Here 2 is multiplied by 5 times with itself. So that exponent is 5 and base is given 2.
We get the exponent form is 25.

## 1. Express 128 as a power 2.

#### Explanation:

Let us find the prime factors of 128.
128= 2 x 2 x 2 x 2 x 2 x 2 x 2 = 27
Here 2 is multiplied by 7 times with itself. So that exponent is 7 and base is given 2.
We get the exponent form is 27.

## 2. Express 32 as a power 2.

#### Explanation:

Let us find the prime factors of 32.
32= 2 x 2 x 2 x 2= 24
Here 2 is multiplied by 5 times with itself. So that exponent is 4 and base is given 2.
We get the exponent form is 24.

## Square of a Number

A square of number is the product of some integer with itself.
For example, 25 is a square number, since it can be written as 5 × 5.

#### Note:

•  To find a square of number we multiply the number two times by itself.
•  For example square of 4 can be represented as "4×4=16". We can express it in exponent form as 4^2.
•  Square of negative number is positive number.

## 1. Find the square of 35.

#### Explanation:

Square of 35.
= (-35)2
= (-35) × (-35)
= 1225

##### Examples square of variables:

• 1. ab2 = a × b × b
• 2. (ab)2 = ab × ab
• 3. (ab)-2 = -(ab) × (-ab)= (ab)2

## Cube of a Number

A cube of number is just the number multiplied by 3 times with itself.

For example, to find cube of 3.
Cube of 3 is 3 times 3.
Cube of 3 = 3 × 3 × 3= 27.
We can express it in exponent form as 33

#### Note:

•  Cube of negative number is negative number.

## 1. Find the cube of 25.

#### Explanation:

Find the cube of 25.
= (-25)3
= (-25) × (-25) × (-25)
= -15625

##### Examples cube of variables:
• 1. ab3= a × b × b x b
• 2. (ab)3= ab×ab×ab

## What is the value of numbers in exponent form?

We can find the value of numbers given in exponent form by multiplying the base with itself as many times the exponent is given.

To find the value of 35.
35 mean multiplying the 3 with itself 5 times.
35= 3 × 3 × 3 × 3 × 3
3 × 3 × 3 × 3 × 3= 243.

## Box Arrangement Examples

Please see the picture below. Boxes are in group of 4. Count the group number of boxes and write them in exponent & product form. How could you read this number? Find the value for this number in exponent form.

#### Explanation :

The group numbers of boxes are 3. In every group there are 4 boxes. So that base is 4 and exponent is 3. We can write this in exponent form as 43.
The product form of 43= 4×4×4
We can read 43 as 4 raised to 3.
The value of 43= 4×4×4= 64

## Star Eaxmple

Please, see the picture below. Stars are in group of 12. Count the group number of stars and write them in exponent & product form. How could you read this number? Find the value for this number in exponent form.

#### Explanation :

The group numbers of stars are 4. In every group there are 12 stars. So that base is 12 and exponent is 4. We can write this in exponent form as 124.
The product form of 124= 12 × 12 × 12 × 12
We can read 124 as 12 raised to 4.
The value of 124= 12 × 12 × 12 × 12= 20736

## Exponent rules

Any number with an exponent of 0 (except 0) is equals to 1. So that x0 = 1.
Examples: 50= 1, 150= 1, 1230= 1, 1540= 1, 1200= 1

When base raised to the positive power value; x1= x
Examples: (10)1= 10, (15)1= 15, (124)1= 124 and (-10)1= -10, (-15)1= -15, (-124)1= -124

## Simple Example

#### Explanation:

(2/3)6
Here 2/3 is the base and 6 is exponent.
For (2/3)6 mean 2/3 is 6 times. So that,
(2/3)6= 2/3× 2/3 × 2/3× 2/3 × 2/3 ×2/3= 2 × 2 × 2 × 2 × 2 × 2 / 3 × 3 × 3 × 3 × 3 × 3= 64/729

When base raised to the negative power value; x-1 = 1/x.
Examples: 5-6= 1/56, 4-3= 1/43, 7-5= 1/75

## 2. What is the value of (12)-4

#### Explanation:

(12)-4
Here 12 is the base and -4 is exponent.
For (12)-4mean 1/ (12)4
(12)-4=1/ (12)4
=1/12 × 1/12 × 1/12 × 1/12= 1 × 1 × 1 × 1 / 12 × 12 × 12 × 12 = 1/ 20736

## Cube Example

A cube shaped container is one third full with kerosene oil. Each side of the container is 9 inches long. What is the volume of kerosene in cubic inches?

#### Explanation :

Volume of a cube= (side)3
Volume of a cube container = 93= 9×9×9= 729.
Volume of a cube container is 729 cubic inches.
The volume of kerosene in cubic inches
= 729 /3= 243 cubic inches

## Rule for multiplication when the base is the same

When we multiply the numbers having the same base, the base of the product remains the same and exponents are added to get the index of the product.

For example: 84× 85= 84+5=89

#### Note:

•  If a is any rational number and m and n are any positive integers then am× an = am + n

## Simple Example

#### Explanation:

75×74
= (7 × 7 × 7 × 7 × 7)×( 7× 7 × 7 × 7)
= (7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7)
= the number 7, (5 + 4) times=79
So that, 75×74=79

(7/5)3×(7/5)8
= (7/5)3+8
= (7/5)11

(y)4×(y)6
= (y)4+6
= (y)10

## Project Completion Example

A group of employers were given a project has three parts. There are 32 ways to complete the 2 documents in Part A. There are 32 ways complete the 2 documents in Part B. How many ways are there to complete the project all 4 documents?

There are 81 ways to complete the project.

#### Explanation :

There are 32 ways to complete the 2 documents in Part A. There are 32 ways complete the 2 documents in Part B. The number of ways to complete the project all 4 documents is the product of 32and 32
The product of 32and 32
= 332
= 32+2
= 34
= 81

## Rule for division when the base is the same

If a is any rational number other than 0 and m and n are any positive integers such that m>n, then am÷ an = am-n

#### Note:

•  When we divide the numbers having the same base, the base remains the same and exponents are subtracted.
•  When the index of the dividend is greater than that of divisor, subtract the exponent of the divisor from that of the dividend.

= 89-4
= 85
84÷ 89
= 1/89-4
= 1/85

## 1. 68÷64?

#### Explanation:

68÷64
= (6×6×6×6×6×6×6×6) ÷ ( 6×6×6×6)
= ( 6×6×6×6)
= 1296

## 1. (2/3)8÷ (2/3)3?

#### Explanation:

= (2/3)8-3
=(2/3)5
= (2/3)× (2/3)× (2/3)× (2/3)× (2/3)
= 2×2×2×2×2 / 3×3×3×3×3
= 32/243

(a) 4×(a) 6
=1/(a) 6-4
=1/(a) 2

## Survey Based Example

There is a survey taken in society about working from home jobs for women and man. It is found that 76 women were doing jobs from home but only 72 men were doing jobs from home. How many times more women were doing jobs from home?

#### Explanation :

To finding more women were doing jobs from home, we have to divide these two expressions shown for women and men. That is 76 and 72
76/72
= 76-2
= 74
=7×7×7×7
= 2401

## Rule for index of a number in index form

If x is any rational number and m and n are positive integers then
(xm )n= (x)m × n

If we say, (23 )5
(23 )5
Multiply the index.
(23 )5= (2)3 × 5= 215

(a4 )5
(a4 )5
= a4 × 5
= a20

## 1. Find the value [(-3/4)2]2

#### Explanation:

[(-3/4)2]2
We will multiply the index numbers.
= (-3/4)2×2
= (-3/4)4
= -3× -3× -3× -3/ 4× 4 × 4× 4
= 81/256

## Rule for the index of a product

If x and y are any rational number and m is any positive integers then
(x× y) m = xm × ym .

If we say, (4×3)7.
(4×3)7= (4)7× (3)7

## Simple Example

(a×b)5
(a×b)5 = (a)5× (b)5

## 1. Find the value (5×2)3

#### Explanation:

(5×2)3
= (5)3× (2)3
= (5×5×5) ×(2×2×2)
= 1000

## 1. Find the value ( 1/3× 2/5 )2

#### Explanation:

( 1/3× 2/5 ) 2
= (1/3)2 × (2/5 )2
= 1×1×2×2/ 3×3×5×5
= 4/225

## Rule for index of a number in the numerator-denominator form

If x / y is any rational number and m is any positive integers then
(x/ y) m = xm/ ym

Let us say,
(3/ 5) 2 = 32/ 52 and (x/y) 2 = x2/ y2

## 1. Find the value (5/2)4

#### Explanation:

(5/2)4
= (5)4/ (2)4
= (5×5×5×5) / (2×2×2×2)
= 625/16

## 1. Find the value (-6/5)2

#### Explanation:

(-6/5)2
= (-6)2/ (5)2
= (-6×-6) / (5×5)
= 36/25

## 1. Find the value (5×2)3 ÷ (5×2)1

#### Explanation:

(5 × 2)3 ÷ (5 × 2)1
We will first subtract the index here.
= (5×2) 3-1
= (5×2)2
= (5)2× (2)2
= 5 × 5 × 2 × 2
= 100

#### Explanation:

We will first add the index here.
= (51 )2+3
= (51 )5
= 51×5
=5 5
= 3125

## 3. Evaluate the expression (a × b) 3 ÷ (a × b) 5 for a=2 and b = 3

#### Explanation:

(a × b) 3 ÷ (a× b) 5
= 1/ (a × b) 5-3
= 1/ (a × b) 2
= 1/ (2 × 3) 2 put the value for a=2 and b = 3
= 1/ (2)2× (3)2
1/ 2 × 2 × 3 × 3
=1/36

## 4.Evaluate the expression (a) 0 ÷ (b) 0 for a=15 and b = 17

#### Explanation:

(a) 0 ÷ (b) 0
= (15) 0 ÷ (17) 0 put the value for a=15 and b = 17
= 1÷ 1
=1

## Fractional index

The square of number is shown by writing the number with exponent 2.
Index ½ represents the square root of the number.
These are known as fractional indices.

Let us say square of 4, we will write as 42.
Now, if we want to write square root of 4, we will write as 41/2.
As well as the cube root of 8 is written as 81/3

#### Note:

• The nth root of a number is shown by the index 1/n.

How to read the numbers as fractional indices?

##### See the following examples:

91/2 means the square root of 9.
10001/3 means the cube root of 1000.
6251/4 means the fourth root of 625.
10001/10 means the square root of 1000.

## 1. What is the fourth root of 256?

#### Explanation:

4 × 4 × 4 × 4 = 256.
Hence, 44 = 256.
Conversely, 4 is the fourth root of 256.

## 1. Find sqaure root of 256.

#### Explanation:

16 × 16 = 256.
Hence, 162 = 256.
Conversely, 16 is the square root of 256.

## Medium Example:

Find the values of the following index numbers.

•  1251/3= Cube root of 125 = 5
•   1441/2= Square root of 144 = 12
•   2161/3= Cube root of 216 = 6
•   1691/2= Square root of 169 = 13

## Multiplication fractional indices

As you know If a is any rational number and m and n are any positive integers then
am× an = am+n
The same rule is applied for fractional indices when we multiply the numbers as index form with the same base.

If we say, 71/2× 71/3
Add the index numbers. 71/2× 71/3 = 71/2+ 1/3
Simplify added index form. 71/2+ 1/3 = 73+2/6
73+2/6= 75/6 means the 5th power of the 6th root of 7.

91/3× 91/3
= 91/3+ 1/3
= 91+1/3
= 92/3

141/4× 141/5
= 141/4+ 1/5
= 145+4/20
= 149/20

x1/3× x1/5
= x1/3+ 1/5
= x5+3/15
= x8/15

## Division fractional indices

If a is any rational number other than 0 and m and n are any positive integers such that m>n, then am÷ an = am-n

The same rule is applied for fractional indices when we multiply the numbers as index form with the same base.
If we say,
71/3÷ 71/3

Subtract the index numbers. 71/3÷ 71/3= 71/3 - 1/3
Simplify index form. = 71/3 - 1/3 = 70
Write the value. 70 = 1

## 1. 131/4÷ 131/5

#### Explanation:

131/4÷ 131/5
= 131/4- 1/5
= 135-4/20
= 131/20
131/20 means the 20th root of 13.

m1/3÷ m1/5
= m1/3 - 1/5
= m5-3/15
= m2/15

81/4÷ 81/6
= 81/4 - 1/6
= 8 3-2/12
= 8 1/12

## Rule for index of a number for fractional indices

If x is any rational number and m and n are positive integers then
(xm )n= xm × n

If we say, (21/3 )1/5
Multiply the index numbers. (21/3 )1/5 = 21/3 ×1/5
Simplify index form. 21/3 ×1/5 = 21/15
Write the value. 21/15

## 1. Simplify [(25)1/2]1/2

#### Explanation:

(25) 1/2 ×1/2

Multiply the index numbers.
= 251/4

## 2. Simplify [ (8)1/2]1/9

#### Explanation:

8 1/2 ×1/9
Multiply the index numbers.
= 81/18

## 3. Simplify [(y)1/2]1/3

#### Explanation:

y 1/2 × 1/3
Multiply the index numbers.
= y1/6

## Rule for the index of a product for fractional indices

If x and y are any rational number and m is any positive integers then
(x× y) m = xm × ym .

We can say,
(5×3)1/7= (5) 1/7× (3) 1/7
Or (a×b) 1/5= (a) 1/5× (b) 1/5

## Simple Example

Simplify (5 × 9) 1/3
(5×9) 1/3
= (5) 1/3× (9) 1/3

## 1. Find the value (125 × 8) 1/3

#### Explanation:

(125×8) 1/3
= (125) 1/3× (8) 1/3
= 5 × 2
= 10

## 2. Find the value (100 × 64)1/2

#### Explanation:

(100 × 64) 1/2
= (100) 1/2× (64) 1/2
= 10 × 8
= 80

## 1. Find the value (625 × 256) 1/2÷ (625 × 256) 1/4

#### Explanation:

(625 × 256) 1/2 ÷ (625 × 256) 1/4
= (625 × 256) 1/2 - 1/4
= (625 × 256) 1/4
= (625) 1/4 × (256) 1/4
= 5 × 4
= 20

## 2. Find the value [(81)1 ]1/2× [(81)1 ]1/4

#### Explanation:

[(81)1 ]1/2 × [(81)1 ]1/4
= [(81)1 ]1/2 + 1/4
= (81 )3/4
Means the 3nd power of the 4th root of 81.
= 27

## Evaluate the expression(a× b) 1/3 ÷ (a× b) 1/5 for a=2 and b = 3

#### Explanation:

(a× b) 1/3 ÷ (a× b) 1/ 5 put the value for a=2 and b = 3
= (2× 3) 1/3 ÷ (2× 3) 1/5
= (2× 3) 1/3- 1/5
= (2× 3) 5-3/15
= (2× 3) 2/15
= 22/15× 3 2/15

### Points to Remember

For any given number make pairs...

### Quick Tips

•  When we multiply/divide the Indices having same base, the base remains as it is.
•  To write the numbers as a power of 10, the first number must between 1 and 10 and the other is power of 10.

### Think:

•  Which one is greater 73 or 37?
•  What could be the value of (1)1000
•  If 83÷ 32 is given. Can you add the exponents? If No! ‘Why’?

## Worksheets

#### Become a member today!

Register (it’s Free)


text:
Question #48ecb

Feb 18, 2016

${C}_{M} = \left({n}_{\text{total"))/(V_("total}}\right) = \frac{39.1 \times {10}^{- 4} m o l}{37.0 \times {10}^{- 3} L} = 0.106 M$

Explanation:

To find the concentration of chloride $C {l}^{-}$ in the mixture we will first find the total number of mole of $C {l}^{-}$ (${n}_{\text{total}}$) then divide it by the total volume (${V}_{\text{total}}$) of the mixture.

${C}_{M} = \left({n}_{\text{total"))/(V_("total}}\right)$

To find the ${n}_{\text{total}}$ we will calculate the number of mole coming from both salts:

${n}_{L i C l} = {C}_{M} \times V = 0.392 \frac{m o l}{\cancel{L}} \times 2.50 \times {10}^{- 3} \cancel{L} = 9.80 \times {10}^{- 4} m o l$

${n}_{K C l} = {C}_{M} \times V = 84.8 \times {10}^{- 3} \frac{m o l}{\cancel{L}} \times 34.5 \times {10}^{- 3} \cancel{L} = 29.3 \times {10}^{- 4} m o l$

${n}_{\text{total}} = {n}_{L i C l} + {n}_{K C l} = 9.80 \times {10}^{- 4} m o l + 29.3 \times {10}^{- 4} m o l = 39.1 \times {10}^{- 4} m o l$

The total volume is: ${V}_{\text{total}} = 34.5 \times {10}^{- 3} L + 2.50 \times {10}^{- 3} L = 37.0 \times {10}^{- 3} L$

The concentration can the be found by:

${C}_{M} = \left({n}_{\text{total"))/(V_("total}}\right) = \frac{39.1 \times {10}^{- 4} m o l}{37.0 \times {10}^{- 3} L} = 0.106 M$

Therefore, the correct choice is (c) 106 mM.


text:
Home > MC2 > Chapter 7 > Lesson 7.2.2 > Problem7-79

7-79.

Recall that acute angles are angles that measure less than 90°, obtuse angles measure more than 90°, and right angles measure exactly 90°.

Complementary angles are two angles that add up to 90°.
For example, a 30° angle and a 60° angle are complementary because 30°+ 60° = 90°.

Supplementary angles are angles that add up to 180°.
For example, 110° and 70° are supplementary angles because 110° + 70° = 180°.


text:
# Laythe

Laythe is the first of the five natural satellites of Jool. Although mostly covered in seas, Laythe has numerous rocky, sandy islands. Laythe is tidally locked to Jool. Synchronous orbits around Laythe are not possible, as they would lie outside of its SOI, at a radius of 5,186,399 meters. A circular archipelago that is prominent in maps suggests that Laythe suffered a massive impact in its distant past.

Achieving a stable orbit around Laythe from sea level requires a delta-V of ≈2800 m/s.

Laythe has no proper real-world analogue. In the real-world Jupiter system, the moons Ganymede, Europa and Io form a group with an orbital resonance just like KSPs Tylo, Vall and Laythe. But while Tylo and Vall have a certain resemblance to their real-life counterparts, Laythe is very unlike the volcanic, atmosphere-less, dry Io. Though Laythe has liquid water on its surface, the temperature there is below 0 degrees Celsius. This means that Laythe's oceans must contain another compound, most likely ionic, such as salt.

According to developer NovaSilisko, Laythe is planned to have volcanic activity and high radiation levels in later versions, making it a much more hostile moon than now (0.20.2[outdated])[1].

## Atmosphere

Laythe's atmosphere begins at 55,262 meters and is slightly less dense than Kerbin's. It is the only moon in the game which has an atmosphere.

Like all other atmospheres in the game, Laythe's atmosphere fades exponentially with increasing altitude (with a scale height of 4000 m). In general, the atmospheric pressure on Laythe at an altitude expressed in meters is:[2]

${\displaystyle p_{l}=0.8\ e^{-altitude/4000}}$

With proper aerobraking near Laythe and Jool, reaching Laythe may require the least delta-V of all of Jool's moons. When approaching Jool on a Hohmann transfer orbit from Kerbin, aerobraking with a Jool periapsis of 119.5 km will put your apoapsis roughly near Laythe's orbit.

Landing on higher points of Laythe's surface has to be done with a help of retro burns as parachutes alone are not enough to sufficiently lower velocity for landing.

Jet engines work in Laythe's atmosphere as it contains oxygen. However, their air intake efficiency is lower.

## Reference Frames

Time warp Minimum Altitude
Any
5× 50 000 m (above the atmosphere)
10× 50 000 m (above the atmosphere)
50× 60 000 m
100× 120 000 m
1 000× 240 000 m
10 000× 480 000 m
100 000× 600 000 m

## Changes

0.17
• Initial Release


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# Realizing reproducible Machine Learning - with Tensorflow

Sunday, December 22, 2019, 12:00 AM

This is Part 2 - Realizing reproducible Machine Learning - with Tensorflow of technical blog series titled Reproducibility in Machine Learning. Part 1 & Part 3 can be found here & here respectively.

As discussed in Part 1, writing reproducible machine learning is not easy with challenges arising from every direction e.g. hardware, software, algorithms, process & practice, data. In this post, we will focus on what is needed to ensure ML code is reproducible.

## First things first

There are a few very simple things that’s needed to be done to before thinking big and focussing on writing reproducible ML code. These are:

• Code is version controlled

Same input (data), same process (code) resulting into same output is essence of reproducibility. But code keeps evolving, since ML is so iterative. Hence, its important to version control code (including configuration). This allows obtaining same i.e. exact code (commit/version) from source repository.

• Reproducible runtime – pinned libraries

So we version controlled code but what about environment/runtime? Sometimes, non-determinism is introduced not direct by user code but also dependencies. We talked about this at length in software section of Challenges in realizing reproducible ML in Part 1. Taking the example of Pyproj - a geospatial transform library, that I once used to compute geo-location based on some parameters. We changed nothing but just version of pyproj from V1.9.6 to V2.4.0 and suddenly all our calculation were giving different results. The difference was so much that location calculation for San Diego Convention Centre were coming out to be somewhere in Miramar off golf course (see figure 1) issue link. Now imagine ordering pizza delivery on the back of my computation snippet backed with unpinned version of pyproj?

Figure 1: Example of why pinned libraries are important

Challenges like these occur quite often that we would like. That’s why its important to pin/fix dependent runtime either by pinning version of libraries or using versioned containers (like docker).

• Smart randomness

As discussed in Part 1, randomization plays a key role in most ML algorithms. Unseeded randomness is the simplest way to make code non-reproducible. It is also one of the easiest to manage amongst all things gotchas of non-reproducible ML. All we need to do is seed all the randomness and manage the seed via configuration (as code or external).

• Rounding precision, under-flows & overflows

Floating point arithmetic is ubiquitous in ML. The complexity and intensity of floating point operations (FLOPS) are increasing everyday with current needs easily meeting Giga-Flops order of computations. To achieve the efficiency in terms of speed despite complexity, mixed precision floating point operations have also been proposed. As discussed Part 1, accelerated hardware such as (GPGPU), tensor processing unit (TPU) etc. due to their architecture and asynchronous computing do not guarantee reproducibility. In addition, when dealing with floating points, the issues related to overflow and underflow are expected. This just adds to the complexity.

• Dependent library’s behavior aware

As discussed in software section of Challenges in realizing reproducible ML in Part 1, some routines of ML libraries do not guarantee reproducibility. For instance, NVIDIA’s cuda based deep learning library cudnn. Similarly, with Tensorflow, using some methods may result into non-deterministic behaviour. One such example is backward pass of broadcasting on GPUref. Awareness about behaviours of libraries being used and approaches to overcome the non-determinism should be explored.

## Writing reproducible ML

To demonstrate reproducible ML, I will be using Oxford Pet dataset that has labels for pet images. I will be doing a semantic segmentation of pets images and will be using pixel level labelling. Each pixel of the pet image in Oxford Pet dataset is labelled as 1) Foreground (Pet area), 2) Background (not pet area) and 3) Unknown (edges). These labels by definition are mutually exclusive - i.e. a pixel can be only be one of the above 3.

Figure 2: Oxford pet dataset

I will be using convolution neural network (ConvNet) for semantic segmentation. The network architecture is based on U-net. This is similar to standard semantic segmentation example by tensorflow.

Fihure 3: U-net architecture

The reproducible version of semantic segmentation is available in github repository. This example demonstrate reproducible ML and also performing end to end ML with provenance across process and data.

Figure 4: Reproducible ML sample - semantic segmentation of oxford pet

In this post, however, I will be discussing only reproducible ML aspect of it and will be referencing snippets of this example.

### ML workflow

In reality a machine learning workflow is very complex and look somewhat similar to figure 5. In this post, however, we will only discuss data and model training part of it. The remaining workflow i.e. the end to end workflow will be discussed in next post.

Figure 5: Machine learning workflow

### Data

The source dataset is Oxford Pet dataset which contains multitude labels e.g. class outcome, pixelwise label, bounding boxes etc. First step is to process this data to generate the trainable dataset. In the example code, this is done by download_petset.py script.

python download_petset.py  --output /wks/petset

Result sample is show in figure 6.

Figure 6: Pets data partitioned by Pets ID

Post data partition, the entire dataset is divided into 4 set: a) training, b) validation, c) calibration and d) test We would want this set partitioning strategy to be reproducible. By doing this, we ensure that if we have to blow away training dataset, or if accidental data loss occur then the exact dataset can be created.

In this sample, this is achieved by generating hash of petid and partitioning the hash into 10 folds (script below) to obtain partition index of pet id.

partition_idx = int(hashlib.md5((os.path.basename(petset_id)).encode()).hexdigest(), 16) % 10

With partition_idx 0-6 assigned for training, 7 for validation, 8 for calibration, and 9 for test, every generation will result in pets going into their respective partition.

In addition to set partitioning, any random data augmentation performed is seeded with seed controlled as configuration as code. See tf.image.random_flip_left_right used in this tensorflow data pipeline method.

Script for model dataset preparation is located in dataset_gen.py and used as following

python dataset_gen.py --input /wks/petset --output /wks/model_dataset

with results shown as below:

Figure 7: Pets data partitioned into training, validation, calibration and test set

### Modelling semantic segmentation

The model for pet segmentation is based on U-net with backbone of either MobileNet-v2 or VGG-19 (defaults to VGG-19). As per this models network architecture, 5 activation layers of pre-trained backbone network are chosen. These layers are:

• MobileNet
'block_1_expand_relu'
'block_3_expand_relu'
'block_6_expand_relu'
'block_13_expand_relu'
'block_16_project'

• VGG
'block1_pool'
'block2_pool'
'block3_pool'
'block4_pool'
'block5_pool'

Each of these layers are then concatenated with corresponding upsampling layer comprising of Conv2DTranspose layer forming whats known as skip connection. See model code for more info.

The training script train.py can be used as following:

python train.py --input /wks/model_dataset --output /wks/model --checkpoint_path /wks/model/ckpts --tensorboard_path /wks/model

#### 1. Seeding randomness

All methods exploiting randomness is used with appropriate seed.

Global seed is set for any hidden methods that may be using randomness by calling in set_seed(seed) which sets seed for used libraries:

def set_seeds(seed=SEED):
os.environ['PYTHONHASHSEED'] = str(seed)
random.seed(seed)
tf.random.set_seed(seed)
np.random.seed(seed)

#### 2. Handing library behaviours

##### 2.1 CuDNN

CuDNN do not guarantee reproducibility in some of its routine. Environment variable TF_DETERMINISTIC_OPS & TF_CUDNN_DETERMINISTIC can be used to control this behavior as per this snippet (figure 8) from cudnn release page.

Figure 8: NVIDIA release page snippet reproducibility

As discussed in Part 1, using high parallelism with compute intensive workflow may not be reproducible. Configurations for interref and intraref operation parallelism should be set if 100% parallelism is desired.

In this example, I have chosen 1 to avoid any non-determinism arising from inter operation parallelism. Warning setting this will considerably slow down training.

tf.config.threading.set_inter_op_parallelism_threads(1)

#### 2.3 Tensorflow determinism

Following are some of the application of Tensorflow that are not reproducible:

• Mention that GPU reductions are nondeterministic in docslink
• Problems Getting TensorFlow to behave Deterministicallylink

Duncan Riach, along with several other contributors have created tensorflow_determinism package that can be used to overcome non-reproducibility related challenges from tensorflow. It should be used in addition to above measures we have discussed so far.

If we combine all the approaches discussed above (aside from using seeded randomness), they can be wrapped into a light weight method like one below:

def set_global_determinism(seed=SEED, fast_n_close=False):
"""
Enable 100% reproducibility on operations related to tensor and randomness.
Parameters:
seed (int): seed value for global randomness
fast_n_close (bool): whether to achieve efficient at the cost of determinism/reproducibility
"""
set_seeds(seed=seed)
if fast_n_close:
return

logging.warning("*******************************************************************************")
logging.warning("*** set_global_determinism is called,setting full determinism, will be slow ***")
logging.warning("*******************************************************************************")

os.environ['TF_DETERMINISTIC_OPS'] = '1'
os.environ['TF_CUDNN_DETERMINISTIC'] = '1'
from tfdeterminism import patch
patch()

which can then be used on top of ML algorithm/process code to generate 100% reproducible code.

### Result

What happens if we dont write reproducible ML? What kind of difference we are really talking about? Last two column of figure 9 shows results obtained by model trained on exact same dataset, exactly same code with EXACTLY one exception. The dropout layer used in the network were unseeded. Every other measures discussed above were taken into account.

Figure 9: Effect of just forgetting to set one seed amidst many

Looking at the result of first pet which is very simple case, we can see subtle difference in outcome of these two models. Second pet case is slightly complicated due to shadow and we can see obvious differences in the outcome. But what about the third case, this is very hard case for pre-trained frozen backbone model we are using but we can see major differences in result from the two models.

If we were to use all the measures discussed above then 100% reproducible ML can be obatined. This is shown in the following 2 logs obtained by running the following:

python train.py \
--input /wks/model_dataset \
--hyperparam_fn_path best_hyper_parameters.json \
--output logs \
--checkpoint_path "logs/ckpts" \
--tensorboard_path logs

wherein content of best_hyper_parameters.json are:

{
"batch_size":60,
"epochs":12,
"iterations":100,
"learning_rate":0.0018464290366223407,
"model_arch":"MobileNetV2",
"steps_per_epoch":84
}

Run attempt 1:

WARNING:root:******* set_global_determinism is called, setting seeds and determinism *******
TensorFlow version 2.0.0 has been patched using tfdeterminism version 0.3.0
Input: tf-data, Model: MobileNetV2, Batch Size: 60, Epochs: 12, Learning Rate: 0.0018464290366223407, Steps Per Epoch: 84
Train for 84 steps, validate for 14 steps
Epoch 1/12
2019-11-07 12:48:27.576286: I tensorflow/core/profiler/lib/profiler_session.cc:184] Profiler session started.
84/84 [==============================] - 505s 6s/step - loss: 0.8187 - iou_score: 0.4867 - f1_score: 0.6092 - binary_accuracy: 0.8749 - val_loss: 1.0283 - val_iou_score: 0.4910 - val_f1_score: 0.6297 - val_binary_accuracy: 0.8393
Epoch 2/12
84/84 [==============================] - 533s 6s/step - loss: 0.6116 - iou_score: 0.6118 - f1_score: 0.7309 - binary_accuracy: 0.9150 - val_loss: 0.6965 - val_iou_score: 0.5817 - val_f1_score: 0.7079 - val_binary_accuracy: 0.8940
Epoch 3/12
84/84 [==============================] - 527s 6s/step - loss: 0.5829 - iou_score: 0.6301 - f1_score: 0.7466 - binary_accuracy: 0.9197 - val_loss: 0.6354 - val_iou_score: 0.6107 - val_f1_score: 0.7312 - val_binary_accuracy: 0.9038
Epoch 4/12
84/84 [==============================] - 503s 6s/step - loss: 0.5733 - iou_score: 0.6376 - f1_score: 0.7528 - binary_accuracy: 0.9213 - val_loss: 0.6192 - val_iou_score: 0.6227 - val_f1_score: 0.7411 - val_binary_accuracy: 0.9066
Epoch 5/12
84/84 [==============================] - 484s 6s/step - loss: 0.5566 - iou_score: 0.6461 - f1_score: 0.7599 - binary_accuracy: 0.9241 - val_loss: 0.5827 - val_iou_score: 0.6381 - val_f1_score: 0.7534 - val_binary_accuracy: 0.9156
Epoch 6/12
84/84 [==============================] - 509s 6s/step - loss: 0.5524 - iou_score: 0.6497 - f1_score: 0.7629 - binary_accuracy: 0.9247 - val_loss: 0.5732 - val_iou_score: 0.6477 - val_f1_score: 0.7605 - val_binary_accuracy: 0.9191
Epoch 7/12
84/84 [==============================] - 526s 6s/step - loss: 0.5439 - iou_score: 0.6544 - f1_score: 0.7669 - binary_accuracy: 0.9262 - val_loss: 0.5774 - val_iou_score: 0.6456 - val_f1_score: 0.7590 - val_binary_accuracy: 0.9170
Epoch 8/12
84/84 [==============================] - 523s 6s/step - loss: 0.5339 - iou_score: 0.6597 - f1_score: 0.7710 - binary_accuracy: 0.9279 - val_loss: 0.5533 - val_iou_score: 0.6554 - val_f1_score: 0.7672 - val_binary_accuracy: 0.9216
Epoch 9/12
84/84 [==============================] - 518s 6s/step - loss: 0.5287 - iou_score: 0.6620 - f1_score: 0.7730 - binary_accuracy: 0.9288 - val_loss: 0.5919 - val_iou_score: 0.6444 - val_f1_score: 0.7584 - val_binary_accuracy: 0.9148
Epoch 10/12
84/84 [==============================] - 506s 6s/step - loss: 0.5259 - iou_score: 0.6649 - f1_score: 0.7753 - binary_accuracy: 0.9292 - val_loss: 0.5532 - val_iou_score: 0.6554 - val_f1_score: 0.7674 - val_binary_accuracy: 0.9218
Epoch 11/12
84/84 [==============================] - 521s 6s/step - loss: 0.5146 - iou_score: 0.6695 - f1_score: 0.7789 - binary_accuracy: 0.9313 - val_loss: 0.5586 - val_iou_score: 0.6581 - val_f1_score: 0.7689 - val_binary_accuracy: 0.9221
Epoch 12/12
84/84 [==============================] - 507s 6s/step - loss: 0.5114 - iou_score: 0.6730 - f1_score: 0.7818 - binary_accuracy: 0.9317 - val_loss: 0.5732 - val_iou_score: 0.6501 - val_f1_score: 0.7626 - val_binary_accuracy: 0.9179

Run attempt 2:

WARNING:root:******* set_global_determinism is called, setting seeds and determinism *******
TensorFlow version 2.0.0 has been patched using tfdeterminism version 0.3.0
Input: tf-data, Model: MobileNetV2, Batch Size: 60, Epochs: 12, Learning Rate: 0.0018464290366223407, Steps Per Epoch: 84
Train for 84 steps, validate for 14 steps

Epoch 1/12
2019-11-07 10:45:51.549715: I tensorflow/core/profiler/lib/profiler_session.cc:184] Profiler session started.
84/84 [==============================] - 549s 7s/step - loss: 0.8187 - iou_score: 0.4867 - f1_score: 0.6092 - binary_accuracy: 0.8749 - val_loss: 1.0283 - val_iou_score: 0.4910 - val_f1_score: 0.6297 - val_binary_accuracy: 0.8393
Epoch 2/12
84/84 [==============================] - 515s 6s/step - loss: 0.6116 - iou_score: 0.6118 - f1_score: 0.7309 - binary_accuracy: 0.9150 - val_loss: 0.6965 - val_iou_score: 0.5817 - val_f1_score: 0.7079 - val_binary_accuracy: 0.8940
Epoch 3/12
84/84 [==============================] - 492s 6s/step - loss: 0.5829 - iou_score: 0.6301 - f1_score: 0.7466 - binary_accuracy: 0.9197 - val_loss: 0.6354 - val_iou_score: 0.6107 - val_f1_score: 0.7312 - val_binary_accuracy: 0.9038
Epoch 4/12
84/84 [==============================] - 515s 6s/step - loss: 0.5733 - iou_score: 0.6376 - f1_score: 0.7528 - binary_accuracy: 0.9213 - val_loss: 0.6192 - val_iou_score: 0.6227 - val_f1_score: 0.7411 - val_binary_accuracy: 0.9066
Epoch 5/12
84/84 [==============================] - 534s 6s/step - loss: 0.5566 - iou_score: 0.6461 - f1_score: 0.7599 - binary_accuracy: 0.9241 - val_loss: 0.5827 - val_iou_score: 0.6381 - val_f1_score: 0.7534 - val_binary_accuracy: 0.9156
Epoch 6/12
84/84 [==============================] - 494s 6s/step - loss: 0.5524 - iou_score: 0.6497 - f1_score: 0.7629 - binary_accuracy: 0.9247 - val_loss: 0.5732 - val_iou_score: 0.6477 - val_f1_score: 0.7605 - val_binary_accuracy: 0.9191
Epoch 7/12
84/84 [==============================] - 506s 6s/step - loss: 0.5439 - iou_score: 0.6544 - f1_score: 0.7669 - binary_accuracy: 0.9262 - val_loss: 0.5774 - val_iou_score: 0.6456 - val_f1_score: 0.7590 - val_binary_accuracy: 0.9170
Epoch 8/12
84/84 [==============================] - 514s 6s/step - loss: 0.5339 - iou_score: 0.6597 - f1_score: 0.7710 - binary_accuracy: 0.9279 - val_loss: 0.5533 - val_iou_score: 0.6554 - val_f1_score: 0.7672 - val_binary_accuracy: 0.9216
Epoch 9/12
84/84 [==============================] - 518s 6s/step - loss: 0.5287 - iou_score: 0.6620 - f1_score: 0.7730 - binary_accuracy: 0.9288 - val_loss: 0.5919 - val_iou_score: 0.6444 - val_f1_score: 0.7584 - val_binary_accuracy: 0.9148
Epoch 10/12
84/84 [==============================] - 531s 6s/step - loss: 0.5259 - iou_score: 0.6649 - f1_score: 0.7753 - binary_accuracy: 0.9292 - val_loss: 0.5532 - val_iou_score: 0.6554 - val_f1_score: 0.7674 - val_binary_accuracy: 0.9218
Epoch 11/12
84/84 [==============================] - 495s 6s/step - loss: 0.5146 - iou_score: 0.6695 - f1_score: 0.7789 - binary_accuracy: 0.9313 - val_loss: 0.5586 - val_iou_score: 0.6581 - val_f1_score: 0.7689 - val_binary_accuracy: 0.9221
Epoch 12/12
84/84 [==============================] - 483s 6s/step - loss: 0.5114 - iou_score: 0.6730 - f1_score: 0.7818 - binary_accuracy: 0.9317 - val_loss: 0.5732 - val_iou_score: 0.6501 - val_f1_score: 0.7626 - val_binary_accuracy: 0.9179

So we have 100% reproducible ML code now but saying training is snail-ish is an understatement. Training time has increased (CPU based measures) from 28 minutes vs 1 hr 45 minutes as we give away with inter thread parallelism and also asynchronous computation optimization. This is not practical in reality. This is also why reproducibility in ML is more focussed around having a road map to reach the same conclusions- Dodge. This is realized by maintaining a system capable of capturing full provenance over everything involved in ML process including data, code, processes and infrastructure/environment. This will be the focus of part 3 of this blog series.

Next part of technical blog series, Reproducibility in Machine Learning, is End-to-end reproducible Machine Learning pipelines on Kubernetes.


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With COVID-19 after the vaccination is before the vaccination. Now that most people in the developed countries have been vaccinated the question arises of how much boost is in the booster shot. We are here to help you understand the real power (or lack thereof) of the booster, so read on!

Very recently a new study on the matter was published in The New England Journal of Medicine: Protection of BNT162b2 Vaccine Booster against Covid-19 in Israel:

At least 12 days after the booster dose, the rate of confirmed infection was lower in the booster group than in the nonbooster group by a factor of 11.3 […] the rate of severe illness was lower by a factor of 19.5 […].

Wow, 20% risk reduction. That is really impressive, is it not? But 20% of what exactly?

To bring some light into the matter renowned statistician Katharina Schüller did the maths in the article Dritte Covid-19-Impfung – Ein kleiner Boost:

• With two vaccine doses, the risk of infection was around 85 cases per 100,000 person-days and with three doses around 8 cases per 100,000 person-days.
• With two doses, the risk of severe disease was around 6 cases per 100,000 person-days and with three doses around 0.3 cases per 100,000 person-days.

We go one step further here and illustrate the point visually to get an intuitive feel for the situation.

We already talked about the efficacy of the COVID-19 vaccine on this blog and explained the concepts of relative vs. absolute risk reduction (and its dangers): COVID-19 vaccine “95% effective”: It doesn’t mean what you think it means! There we saw that personograph plots (also called Kuiper-Marshall plots) are an excellent way to communicate risks (the personograph package can be installed from CRAN).

Let us start with the risk of infections without and with an additional booster shot, based on 2,500 fully vaccinated people:

library(personograph)
## Loading required package: grImport
## Loading required package: grid
## Loading required package: XML

n <- 2500
inf_wo_boost <- 64 / n
inf_w_boost <- 6 / n
inf_wo_boost / inf_w_boost # rel. risk reduction
## [1] 10.66667

data <- list(first = inf_w_boost, second = inf_wo_boost-inf_w_boost, third = 1-inf_wo_boost)
personograph(data, colors = list(first = "red", second = "orange", third = "lightgrey"),
fig.title = "Infections w/o booster: < 64/2500, with booster: ~ 6/2500",
draw.legend = FALSE, n.icons = n, dimensions = c(25, 100),
plot.width = 0.97)

All persons are fully vaccinated, orange and red persons are the unlucky ones who still get an infection (a so-called breakthrough infection), while the orange cases can be prevented by a booster dose.

What is the absolute risk reduction in percentage points?

100 * ((n - inf_wo_boost*n) / n - (n - inf_w_boost*n) / n) # perc. points risk reduction
## [1] -2.32

The initial 11% relative risk reduction became only about two percentage points absolute risk reduction. Now the same analysis for the even more important severe cases:

ev_wo_boost <- 5 / n
sev_w_boost <- .25 / n
sev_wo_boost / sev_w_boost # rel. risk reduction
## [1] 20

data_2 <- list(first = sev_w_boost, second = sev_wo_boost-sev_w_boost, third = 1-sev_wo_boost)
personograph(data_2, colors = list(first = "red", second = "orange", third = "lightgrey"),
fig.title = "Severe cases w/o booster: < 5/2500, with booster: < 1/2500",
draw.legend = FALSE, n.icons = n, dimensions = c(25, 100),
plot.width = 0.97)

The remaining risk of a severe illness after the booster shot is too small to be plotted. We would have to add at least another 7,500 persons to get one red one!

Again, what is the absolute risk reduction in percentage points?

100 * ((n - sev_wo_boost*n) / n - (n - sev_w_boost*n) / n) # perc. points risk reduction
## [1] -0.19

So, the initial 20% relative risk reduction became a meager 0.2 percentage points absolute risk reduction! This is not because the efficacy of the booster is so bad but because the standard vaccination without the booster is already so powerful that it can only add a comparatively small advantage. The huge relative risk reduction shown in the article above is not wrong per se but sure it is misleading.

This squares well with another result I found together with my colleagues Dr. Tobias Krebs and Professor Michael Möckel (both also Technical University Aschaffenburg), where we compared different vaccine regimes. We concluded that “vaccinations of unvaccinated people is more effective than booster vaccinations for already fully vaccinated people”. You can read the full preprint here: COVID-19 scenarios for comparing the effectiveness of age-specific vaccination regimes, exemplified for the city of Aschaffenburg (Germany).

Please share your thoughts on this in the comments and stay tuned for more interesting insights!

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text:
# On the distribution of ranked heights of excursions of a Brownian bridge

August, 1999
Report Number:
563
Authors:
Jim Pitman and Marc Yor
Citation:
Annals of Probability vol. 29, pages 362-384 (2001)
Abstract:

The distribution of the sequence of ranked maximum and minimum values attained during excursions of a standard Brownian bridge is described. The height of the $j$th highest maximum $M_j$ over a positive excursion of the bridge has the same distribution as $M_1/j$, where the distribution of $M_1$ is given by L\'evy's formula $P( M_1 > x ) = e^{-2x^2}$. The probability density of the height of the $j$th highest maximum of excursions of the reflecting Brownian bridge is given by a modification of the known $\theta$-function series for the density of the maximum absolute value of the bridge. These results are obtained from a more general description of the distribution of ranked values of a homogeneous functional of excursions of the standardized bridge of a self-similar recurrent Markov process.

PDF File:
Postscript File:


text:
distance - Maple Help

geom3d

 distance
 find the distance between two given objects

 Calling Sequence distance(A, B) distance(l1, l2) distance(p1, p2) distance(A, l1) distance(A, p1) distance(l1, p1)

Parameters

 A, B - points l1, l2 - lines p1, p2 - planes

Description

 • The routine computes the distance between two given objects: points, lines, and planes, or any combination of these.
 • The command with(geom3d,distance) allows the use of the abbreviated form of this command.

Examples

 > $\mathrm{with}\left(\mathrm{geom3d}\right):$

Distance from a point to a plane

 > $\mathrm{assume}\left(l\ne 0\right)$
 > $\mathrm{plane}\left(\mathrm{pp},lx+my+nz+p=0,\left[x,y,z\right]\right):$
 > $\mathrm{point}\left(A,\left[\mathrm{x1},\mathrm{y1},\mathrm{z1}\right]\right):$
 > $\mathrm{distance}\left(A,\mathrm{pp}\right)$
 $\frac{\left|{\mathrm{x1}}{}{\mathrm{l~}}{+}{\mathrm{y1}}{}{m}{+}{\mathrm{z1}}{}{n}{+}{p}\right|}{\sqrt{{{\mathrm{l~}}}^{{2}}{+}{{m}}^{{2}}{+}{{n}}^{{2}}}}$ (1)
 > $l≔'l':$

Find the distance from a point to a straight line

 > $\mathrm{point}\left(A,6,6,-1\right),\mathrm{line}\left(l,\left[2+t,1+2t,-3-t\right],t\right):$
 > $\mathrm{distance}\left(A,l\right)$
 $\frac{\sqrt{{14}}{}\sqrt{{6}}}{{2}}$ (2)

Find the two points on the line $x$=$2y$=$3z+6$ at a distance of 7 units from the plane $2x+y-2z$=5.

 > $\mathrm{point}\left(o,t,\frac{1}{2}t,-2+\frac{t}{3}\right):$
 > $\mathrm{plane}\left(p,2x+y-2z=5,\left[x,y,z\right]\right):$
 > $d≔\mathrm{distance}\left(o,p\right):$
 > $\mathrm{ans}≔\left[\mathrm{solve}\left(d=7,t\right)\right]:$
 > $\mathbf{for}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}i\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{to}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{nops}\left(\mathrm{ans}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}\mathrm{ans}≔\mathrm{subsop}\left(i=\mathrm{subs}\left(t=\mathrm{op}\left(i,\mathrm{ans}\right),\mathrm{coordinates}\left(o\right)\right),\mathrm{ans}\right)\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathbf{end}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}:$

coordinates of the two points are

 > $\mathrm{ans}$
 $\left[\left[{12}{,}{6}{,}{2}\right]{,}\left[{-}\frac{{120}}{{11}}{,}{-}\frac{{60}}{{11}}{,}{-}\frac{{62}}{{11}}\right]\right]$ (3)


text:
0
Explain in detail how to generate temperature independent references.
band gap references • 1.9k  views
0
0
• Reference voltages or current that exhibit little dependence on temperature prove essential in many analogue circuits. It is interesting to note that, since most process parameters vary with temperature, if a reference is independent, and then it is usually process independent.
• To generate a quantity that remains constant with temperature, postulate that if two quantities having opposite temperature coefficients (TCs) I added with proper weighting, the result display a zero TC.
• For example, for two voltages V1 and V2 that vary in opposite directions with temperature α1 and α2 such that

• Obtaining a reference voltage VREF = α1V1 + α2V2 , with zero TC.

• Among various device parameters in semiconductor technologies, the characteristics of bipolar transistor have proven the most reproducible and well defined quantities that can provide positive and negative TCs. Even though many parameters of MOS devices have been considered for the task of reference generation, bipolar devices still forms the core of such circuits.

#### 1. Negative TC voltage:

• The base-emitter voltage of bipolar transistor exhibits a negative TC. The expression for TC is as follows-
• For a bipolar device we can write,

• The saturation current, Is α μ KT ni^2 where μ denotes mobility of minority carriers and ni is the intrinsic minority carrier concentration of silicon. The temperature dependence of these quantities is represented as,

• Where, b is proportionality factor. Rearranging equation (1), we can write,

• To find temperature coefficient, partially differentiate VBE with respect to T. Assume that Ic is constant with respect to T.

• From equation (3) we have,

• From equation (5) and (6) we can write,

• This is the expression for TC of VBE voltage.

#### 2. Positive TC voltage:

• It is observed that if two bipolar transistors operate at unequal current densities, then the difference between their base-emitter voltages is directly proportional to the absolute temperature. Consider an example, as shown in figure below-

• Here, two identical transistors (with reverse saturation currents, IS1 = IS2) are biased at collector currents of nI0 and I0 and their base currents are negligible then,

• Therefore, temperature coefficient of ΔVBE will be,

• Therefore, temperature coefficient of ΔVBE is positive.

0


text:
Expii

# Length — Definition & Measurement - Expii

The standard SI unit of measurement for length is the meter.


text:
Hot network questions incorrectly use MathJax [duplicate]

Obviously the question on Webmasters.SE is using a \$ sign in the title to talk about money, but MathJax uses it to render in math mode.


text:
# summation of Newton symbol product [duplicate]

Anyone has idea how to simplify this summation: $$\sum\limits_{k=0}^{N}\binom{M-1+k}{k}\binom{M-1+N-k}{N-k}$$

## marked as duplicate by Dietrich Burde, user186170, user91500, Aaron Maroja, PragabhavaOct 13 '16 at 15:55

Note that for $|x|<1$, and for any positive integer $m$, $$\frac{1}{(1-x)^{m}}=\sum_{i=0}^{\infty}\binom{-m}{i}(-x)^i =\sum_{k=0}^{\infty}\binom{m-1+k}{k}x^k.$$ Hence, by the Cauchy product of series, $$\sum\limits_{k=0}^{N}\binom{M-1+k}{k}\binom{M-1+N-k}{N-k}$$ is the coefficient of $x^N$ of the product $$\frac{1}{(1-x)^{M}}\cdot \frac{1}{(1-x)^{M}}=\frac{1}{(1-x)^{2M}}.$$ By the above formula, such coefficient is equal to $$\binom{2M-1+N}{N}.$$


text:
# Properties

 Label 135.2.e.b Level $135$ Weight $2$ Character orbit 135.e Analytic conductor $1.078$ Analytic rank $0$ Dimension $6$ CM no Inner twists $2$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$135 = 3^{3} \cdot 5$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 135.e (of order $$3$$, degree $$2$$, not minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$1.07798042729$$ Analytic rank: $$0$$ Dimension: $$6$$ Relative dimension: $$3$$ over $$\Q(\zeta_{3})$$ Coefficient field: 6.0.954288.1 Defining polynomial: $$x^{6} - x^{5} - 2 x^{4} + 3 x^{3} - 6 x^{2} - 9 x + 27$$ Coefficient ring: $$\Z[a_1, \ldots, a_{4}]$$ Coefficient ring index: $$3$$ Twist minimal: no (minimal twist has level 45) Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{5}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( -\beta_{4} + \beta_{5} ) q^{2} + ( -2 - \beta_{1} - \beta_{2} + 2 \beta_{3} ) q^{4} + ( 1 - \beta_{3} ) q^{5} + ( -2 \beta_{3} - \beta_{4} + \beta_{5} ) q^{7} + ( -1 - \beta_{2} + \beta_{4} ) q^{8} +O(q^{10})$$ $$q + ( -\beta_{4} + \beta_{5} ) q^{2} + ( -2 - \beta_{1} - \beta_{2} + 2 \beta_{3} ) q^{4} + ( 1 - \beta_{3} ) q^{5} + ( -2 \beta_{3} - \beta_{4} + \beta_{5} ) q^{7} + ( -1 - \beta_{2} + \beta_{4} ) q^{8} -\beta_{4} q^{10} + ( \beta_{1} - \beta_{3} ) q^{11} + ( -1 + \beta_{1} + \beta_{2} + \beta_{3} ) q^{13} + ( -4 - \beta_{1} - \beta_{2} + 4 \beta_{3} - 2 \beta_{5} ) q^{14} + ( -\beta_{3} + 2 \beta_{4} - 2 \beta_{5} ) q^{16} + ( 1 + \beta_{2} ) q^{17} + ( 1 - \beta_{2} ) q^{19} + ( -\beta_{1} + 2 \beta_{3} ) q^{20} + ( 1 + \beta_{1} + \beta_{2} - \beta_{3} - 2 \beta_{5} ) q^{22} + ( 2 + 2 \beta_{1} + 2 \beta_{2} - 2 \beta_{3} + \beta_{5} ) q^{23} -\beta_{3} q^{25} + ( 1 + \beta_{2} ) q^{26} + ( 3 + \beta_{2} + 3 \beta_{4} ) q^{28} + ( -2 \beta_{1} - \beta_{3} + 2 \beta_{4} - 2 \beta_{5} ) q^{29} + ( -1 + \beta_{1} + \beta_{2} + \beta_{3} + 4 \beta_{5} ) q^{31} + ( 6 - 6 \beta_{3} + \beta_{5} ) q^{32} + ( -\beta_{1} + \beta_{3} - 2 \beta_{4} + 2 \beta_{5} ) q^{34} + ( -2 - \beta_{4} ) q^{35} + ( 3 + \beta_{2} + 2 \beta_{4} ) q^{37} + ( \beta_{1} - \beta_{3} ) q^{38} + ( -1 - \beta_{1} - \beta_{2} + \beta_{3} + \beta_{5} ) q^{40} + ( -4 - \beta_{1} - \beta_{2} + 4 \beta_{3} + 2 \beta_{5} ) q^{41} + ( \beta_{1} - 3 \beta_{3} + 2 \beta_{4} - 2 \beta_{5} ) q^{43} + ( 7 + \beta_{2} - 2 \beta_{4} ) q^{44} + ( -2 + \beta_{2} - 4 \beta_{4} ) q^{46} + ( \beta_{1} + 5 \beta_{3} + 3 \beta_{4} - 3 \beta_{5} ) q^{47} + ( -1 - \beta_{1} - \beta_{2} + \beta_{3} - 4 \beta_{5} ) q^{49} -\beta_{5} q^{50} + ( \beta_{1} + 3 \beta_{3} - 2 \beta_{4} + 2 \beta_{5} ) q^{52} -2 \beta_{4} q^{53} + ( -1 - \beta_{2} ) q^{55} -3 \beta_{3} q^{56} + ( 6 - 6 \beta_{3} + \beta_{5} ) q^{58} + 2 \beta_{5} q^{59} + ( \beta_{1} + 2 \beta_{4} - 2 \beta_{5} ) q^{61} + ( -15 - 3 \beta_{2} ) q^{62} + ( -6 - \beta_{2} - 2 \beta_{4} ) q^{64} + ( \beta_{1} + \beta_{3} ) q^{65} + ( -5 - 3 \beta_{1} - 3 \beta_{2} + 5 \beta_{3} - \beta_{5} ) q^{67} + ( -7 - \beta_{1} - \beta_{2} + 7 \beta_{3} + 2 \beta_{5} ) q^{68} + ( -\beta_{1} + 4 \beta_{3} + 2 \beta_{4} - 2 \beta_{5} ) q^{70} + ( 3 - 3 \beta_{2} + 2 \beta_{4} ) q^{71} + ( -4 - 4 \beta_{4} ) q^{73} + ( \beta_{1} - 7 \beta_{3} - 4 \beta_{4} + 4 \beta_{5} ) q^{74} + ( 3 - \beta_{1} - \beta_{2} - 3 \beta_{3} - 2 \beta_{5} ) q^{76} + ( -1 - \beta_{1} - \beta_{2} + \beta_{3} - 2 \beta_{5} ) q^{77} + ( -2 \beta_{3} - 4 \beta_{4} + 4 \beta_{5} ) q^{79} + ( -1 + 2 \beta_{4} ) q^{80} + ( -9 - 3 \beta_{2} + 5 \beta_{4} ) q^{82} + ( -6 \beta_{3} - 3 \beta_{4} + 3 \beta_{5} ) q^{83} + ( 1 + \beta_{1} + \beta_{2} - \beta_{3} ) q^{85} + ( 9 + 3 \beta_{1} + 3 \beta_{2} - 9 \beta_{3} - 4 \beta_{5} ) q^{86} + ( -\beta_{1} + 7 \beta_{3} - 4 \beta_{4} + 4 \beta_{5} ) q^{88} + 3 q^{89} + ( 3 - \beta_{2} ) q^{91} + ( -\beta_{1} + 13 \beta_{3} - \beta_{4} + \beta_{5} ) q^{92} + ( 13 + 4 \beta_{1} + 4 \beta_{2} - 13 \beta_{3} + 4 \beta_{5} ) q^{94} + ( 1 - \beta_{1} - \beta_{2} - \beta_{3} ) q^{95} + ( -4 \beta_{1} + 8 \beta_{3} + 2 \beta_{4} - 2 \beta_{5} ) q^{97} + ( 15 + 3 \beta_{2} + 2 \beta_{4} ) q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$6q + q^{2} - 5q^{4} + 3q^{5} - 5q^{7} - 6q^{8} + O(q^{10})$$ $$6q + q^{2} - 5q^{4} + 3q^{5} - 5q^{7} - 6q^{8} + 2q^{10} - 2q^{11} - 4q^{13} - 9q^{14} - 5q^{16} + 4q^{17} + 8q^{19} + 5q^{20} + 4q^{22} + 3q^{23} - 3q^{25} + 4q^{26} + 10q^{28} - 7q^{29} - 8q^{31} + 17q^{32} + 4q^{34} - 10q^{35} + 12q^{37} - 2q^{38} - 3q^{40} - 13q^{41} - 10q^{43} + 44q^{44} - 6q^{46} + 13q^{47} + 2q^{49} + q^{50} + 12q^{52} + 4q^{53} - 4q^{55} - 9q^{56} + 17q^{58} - 2q^{59} - q^{61} - 84q^{62} - 30q^{64} + 4q^{65} - 11q^{67} - 22q^{68} + 9q^{70} + 20q^{71} - 16q^{73} - 16q^{74} + 12q^{76} - 2q^{79} - 10q^{80} - 58q^{82} - 15q^{83} + 2q^{85} + 28q^{86} + 24q^{88} + 18q^{89} + 20q^{91} + 39q^{92} + 31q^{94} + 4q^{95} + 18q^{97} + 80q^{98} + O(q^{100})$$

Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{6} - x^{5} - 2 x^{4} + 3 x^{3} - 6 x^{2} - 9 x + 27$$:

 $$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$($$$$-\nu^{5} + 4 \nu^{4} - \nu^{3} + 9 \nu^{2} - 21 \nu - 9$$$$)/27$$ $$\beta_{2}$$ $$=$$ $$($$$$-\nu^{5} + 4 \nu^{4} - \nu^{3} - 18 \nu^{2} + 33 \nu - 9$$$$)/27$$ $$\beta_{3}$$ $$=$$ $$($$$$-2 \nu^{5} - \nu^{4} - 2 \nu^{3} + 12 \nu + 36$$$$)/27$$ $$\beta_{4}$$ $$=$$ $$($$$$-2 \nu^{5} - \nu^{4} + 7 \nu^{3} + 9 \nu^{2} + 12 \nu + 9$$$$)/27$$ $$\beta_{5}$$ $$=$$ $$($$$$4 \nu^{5} + 2 \nu^{4} - 5 \nu^{3} + 18 \nu^{2} + 3 \nu - 72$$$$)/27$$
 $$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$($$$$\beta_{5} + \beta_{4} + \beta_{3} + \beta_{2} - \beta_{1} + 1$$$$)/3$$ $$\nu^{2}$$ $$=$$ $$($$$$2 \beta_{5} + 2 \beta_{4} + 2 \beta_{3} - \beta_{2} + \beta_{1} + 2$$$$)/3$$ $$\nu^{3}$$ $$=$$ $$($$$$-2 \beta_{5} + 7 \beta_{4} - 11 \beta_{3} + \beta_{2} - \beta_{1} + 7$$$$)/3$$ $$\nu^{4}$$ $$=$$ $$($$$$2 \beta_{5} + 2 \beta_{4} - 7 \beta_{3} + 8 \beta_{2} + 10 \beta_{1} + 20$$$$)/3$$ $$\nu^{5}$$ $$=$$ $$($$$$7 \beta_{5} - 2 \beta_{4} - 20 \beta_{3} + \beta_{2} - 10 \beta_{1} + 43$$$$)/3$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/135\mathbb{Z}\right)^\times$$.

 $$n$$ $$56$$ $$82$$ $$\chi(n)$$ $$-1 + \beta_{3}$$ $$1$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
46.1
 1.71903 + 0.211943i −1.62241 + 0.606458i 0.403374 − 1.68443i 1.71903 − 0.211943i −1.62241 − 0.606458i 0.403374 + 1.68443i
−1.04307 + 1.80664i 0 −1.17597 2.03684i 0.500000 + 0.866025i 0 −2.04307 + 3.53869i 0.734191 0 −2.08613
46.2 0.285997 0.495361i 0 0.836412 + 1.44871i 0.500000 + 0.866025i 0 −0.714003 + 1.23669i 2.10083 0 0.571993
46.3 1.25707 2.17731i 0 −2.16044 3.74200i 0.500000 + 0.866025i 0 0.257068 0.445256i −5.83502 0 2.51414
91.1 −1.04307 1.80664i 0 −1.17597 + 2.03684i 0.500000 0.866025i 0 −2.04307 3.53869i 0.734191 0 −2.08613
91.2 0.285997 + 0.495361i 0 0.836412 1.44871i 0.500000 0.866025i 0 −0.714003 1.23669i 2.10083 0 0.571993
91.3 1.25707 + 2.17731i 0 −2.16044 + 3.74200i 0.500000 0.866025i 0 0.257068 + 0.445256i −5.83502 0 2.51414
 $$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 91.3 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
9.c even 3 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 135.2.e.b 6
3.b odd 2 1 45.2.e.b 6
4.b odd 2 1 2160.2.q.k 6
5.b even 2 1 675.2.e.b 6
5.c odd 4 2 675.2.k.b 12
9.c even 3 1 inner 135.2.e.b 6
9.c even 3 1 405.2.a.i 3
9.d odd 6 1 45.2.e.b 6
9.d odd 6 1 405.2.a.j 3
12.b even 2 1 720.2.q.i 6
15.d odd 2 1 225.2.e.b 6
15.e even 4 2 225.2.k.b 12
36.f odd 6 1 2160.2.q.k 6
36.f odd 6 1 6480.2.a.bs 3
36.h even 6 1 720.2.q.i 6
36.h even 6 1 6480.2.a.bv 3
45.h odd 6 1 225.2.e.b 6
45.h odd 6 1 2025.2.a.n 3
45.j even 6 1 675.2.e.b 6
45.j even 6 1 2025.2.a.o 3
45.k odd 12 2 675.2.k.b 12
45.k odd 12 2 2025.2.b.m 6
45.l even 12 2 225.2.k.b 12
45.l even 12 2 2025.2.b.l 6

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
45.2.e.b 6 3.b odd 2 1
45.2.e.b 6 9.d odd 6 1
135.2.e.b 6 1.a even 1 1 trivial
135.2.e.b 6 9.c even 3 1 inner
225.2.e.b 6 15.d odd 2 1
225.2.e.b 6 45.h odd 6 1
225.2.k.b 12 15.e even 4 2
225.2.k.b 12 45.l even 12 2
405.2.a.i 3 9.c even 3 1
405.2.a.j 3 9.d odd 6 1
675.2.e.b 6 5.b even 2 1
675.2.e.b 6 45.j even 6 1
675.2.k.b 12 5.c odd 4 2
675.2.k.b 12 45.k odd 12 2
720.2.q.i 6 12.b even 2 1
720.2.q.i 6 36.h even 6 1
2025.2.a.n 3 45.h odd 6 1
2025.2.a.o 3 45.j even 6 1
2025.2.b.l 6 45.l even 12 2
2025.2.b.m 6 45.k odd 12 2
2160.2.q.k 6 4.b odd 2 1
2160.2.q.k 6 36.f odd 6 1
6480.2.a.bs 3 36.f odd 6 1
6480.2.a.bv 3 36.h even 6 1

## Hecke kernels

This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{6} - T_{2}^{5} + 6 T_{2}^{4} - T_{2}^{3} + 28 T_{2}^{2} - 15 T_{2} + 9$$ acting on $$S_{2}^{\mathrm{new}}(135, [\chi])$$.

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$9 - 15 T + 28 T^{2} - T^{3} + 6 T^{4} - T^{5} + T^{6}$$
$3$ $$T^{6}$$
$5$ $$( 1 - T + T^{2} )^{3}$$
$7$ $$9 - 9 T + 24 T^{2} + 21 T^{3} + 22 T^{4} + 5 T^{5} + T^{6}$$
$11$ $$144 + 96 T + 88 T^{2} + 8 T^{3} + 12 T^{4} + 2 T^{5} + T^{6}$$
$13$ $$16 + 16 T + 32 T^{2} - 8 T^{3} + 20 T^{4} + 4 T^{5} + T^{6}$$
$17$ $$( 12 - 8 T - 2 T^{2} + T^{3} )^{2}$$
$19$ $$( 4 - 4 T - 4 T^{2} + T^{3} )^{2}$$
$23$ $$13689 - 3861 T + 1440 T^{2} - 135 T^{3} + 42 T^{4} - 3 T^{5} + T^{6}$$
$29$ $$2601 + 1479 T + 1198 T^{2} - 101 T^{3} + 78 T^{4} + 7 T^{5} + T^{6}$$
$31$ $$219024 + 28080 T + 7344 T^{2} + 456 T^{3} + 124 T^{4} + 8 T^{5} + T^{6}$$
$37$ $$( 4 - 12 T - 6 T^{2} + T^{3} )^{2}$$
$41$ $$9 + 57 T + 322 T^{2} + 241 T^{3} + 150 T^{4} + 13 T^{5} + T^{6}$$
$43$ $$16 + 16 T + 56 T^{2} - 32 T^{3} + 104 T^{4} + 10 T^{5} + T^{6}$$
$47$ $$136161 - 4059 T + 4918 T^{2} - 595 T^{3} + 180 T^{4} - 13 T^{5} + T^{6}$$
$53$ $$( 24 - 20 T - 2 T^{2} + T^{3} )^{2}$$
$59$ $$576 + 480 T + 448 T^{2} + 8 T^{3} + 24 T^{4} + 2 T^{5} + T^{6}$$
$61$ $$5041 - 2627 T + 1298 T^{2} - 179 T^{3} + 38 T^{4} + T^{5} + T^{6}$$
$67$ $$257049 + 19773 T + 7098 T^{2} + 585 T^{3} + 160 T^{4} + 11 T^{5} + T^{6}$$
$71$ $$( 708 - 92 T - 10 T^{2} + T^{3} )^{2}$$
$73$ $$( -128 - 64 T + 8 T^{2} + T^{3} )^{2}$$
$79$ $$576 - 2016 T + 7008 T^{2} - 216 T^{3} + 88 T^{4} + 2 T^{5} + T^{6}$$
$83$ $$6561 - 2187 T + 1944 T^{2} + 567 T^{3} + 198 T^{4} + 15 T^{5} + T^{6}$$
$89$ $$( -3 + T )^{6}$$
$97$ $$1700416 - 46944 T + 24768 T^{2} - 1960 T^{3} + 360 T^{4} - 18 T^{5} + T^{6}$$


text:
# Thread: Finding the value of X for two Supplementary Angles

1. ## Finding the value of X for two Supplementary Angles

suppose angles A and B are supplementary angles. Also suppose the measure of angle A is (2x -30)*degrees* and the measure of angle B is (2/3X + 15)*degrees*.

ok what I did was take angle A + B = 180, and solve for the X value. I had to combine the two X values into one ... then isolate X on one side of the equal sign.

my answer was X= 435/8 which in decimal form is 54.375.

2. You can quickly check your answer, by calculating A and B from X, and checking that A + B = 180.

I think you must have made a small mistake since using your value for X, A + B = 130.

3. Originally Posted by sharkfood4
suppose angles A and B are supplementary angles. Also suppose the measure of angle A is (2x -30)*degrees* and the measure of angle B is (2/3X + 15)*degrees*.

ok what I did was take angle A + B = 180, and solve for the X value. I had to combine the two X values into one ... then isolate X on one side of the equal sign.

my answer was X= 435/8 which in decimal form is 54.375.

$A+B=180^o$
$\displaystyle\ 2x-30^o+\frac{2x}{3}+15^o=180^o\Rightarrow\frac{6x}{3 }+\frac{2x}{3}-30^o+15^o=180^o$
$\Rightarrow\displaystyle\frac{8x}{3}-15^o=180^o$
$\displaystyle\Rightarrow\frac{8x}{3}=180^o+15^o$
but somehow you got $180^o-35^o$


text:
Symbol
Problem
$crc^{-\left(sc}$ Solve the following problems. $1$ If one ream of $oon$ paper costs $\left(3x-4\right)$ pesos, how many reams can you $bayf_{or\left(6x}4-17{x_{+24x}}^{32}-34x$ $+24\right)$ $2$ The diagonal of a rectangle is 8 m longer than its shorter side. If the area of the rectangle is 60 square meters, find its dimensions. $3.7$ The dimensions of a rectangular metal are $3cm$ $50$ 5cm and $8cm$ $l1$ the first two dimensions are increased by the same number of centimeters, while the third dimension remains the same, the new volume is 34cm' more than the original volume. What is the new dimension on the enlarged rectangular metal box $34c$ $4.0$ One dimension of a cube is increased by $1$ inch to form a rectangular block. Suppose that the volume of the new block is $150$ cubic inches, find the length of an edge of the original cube.


text:
# Not so standard binomial

Algebra Level 4

$\frac { 7 }{ 2\times 3 } \left( \frac { 1 }{ 3 } \right) +\frac { 9 }{ 3\times 4 } { \left( \frac { 1 }{ 3 } \right) }^{ 2 }+\frac { 11 }{ 4\times 5 } { \left( \frac { 1 }{ 3 } \right) }^{ 3 }+\ldots$

The sum of the expression above up to 10 terms equal to?

×

Problem Loading...

Note Loading...

Set Loading...


text:
# Thread: Taylor series question confusion

1. ## Taylor series question confusion

There is this problem that has been appearing a lot lately but I cannot find any examples that help me solve this.

It states:

"Find (at least) the first 3 nonzero terms in the Taylor (power) series for

$f(x) = \frac{x^3}{3+5x}$ centered at $a = 0.$

(This is the same as the Maclaurin series for the function $f$)"

Now I know how to use the Taylor series and get an expansion, but my method involves continuously taking the derivative of the function which can get pretty messy after even the second derivation. And even at that point, I'm still getting zero for my expansion terms.

What is this question asking exactly?

2. Originally Posted by freyrkessenin
There is this problem that has been appearing a lot lately but I cannot find any examples that help me solve this.

It states:

"Find (at least) the first 3 nonzero terms in the Taylor (power) series for

$f(x) = \frac{x^3}{3+5x}$ centered at $a = 0.$

(This is the same as the Maclaurin series for the function $f$)"

Now I know how to use the Taylor series and get an expansion, but my method involves continuously taking the derivative of the function which can get pretty messy after even the second derivation. And even at that point, I'm still getting zero for my expansion terms.

What is this question asking exactly?
I agree that Taylor series can be fairly nasty at time, but sometimes (as in your case you can make life alot easier on yourself)

here you are given x^3/( 3 + 5x) = x^3 * ( 1/(3 + 5x) )

Here we observe that the only component that truely requires a taylor expansion is (1/(3+ 5x)), whose derivatives are far easier to compute than that of the original function.

Once this is obtained you simply multiple the series through by x^3 and viola you have answered the problem with only a 10th of the pain!!

Hope this points you in the right direction,

Let me know if you require any further assistance,

David

ps - make life even easier on yourself by simplifying the function even further, i.e.
1/(3 + 5x) = 1/( 3 (1 + ((5/3)x)) = (1/3)(1/(1 + (5/3)x)) i.e bring the 1/3 out the front
furthermore let t = 5/3x and thus you you need only solve for 1/(1+t) = a0 + a1*t + ....

To then get in back in terms of x, sub t = 5/3x into the the taylor expansion and your done!

3. ## Hi

I tried performing some hand calculations, and got the following result.
$f(x) = \frac{x^3}{3+5x} = x^3 \cdot \frac{1}{3+5x}$

Now I only performing a maclaurin series on the $\frac{1}{3+5x}$
term.
Let $h(x) \, \mbox{be the maclaurin series of }\frac{1}{3+5x}$

Then $f(x) \approx x^3 \cdot h(x)$

If I dont bother with the restterm, then we get something like (see picture).

And the two functions seem to be pretty close around $x=0$

You can see the maclaurin series in the picture...

4. Originally Posted by Twig
I tried performing some hand calculations, and got the following result.
$f(x) = \frac{x^3}{3+5x} = x^3 \cdot \frac{1}{3+5x}$

Now I only performing a maclaurin series on the $\frac{1}{3+5x}$
term.
Let $h(x) \, \mbox{be the maclaurin series of }\frac{1}{3+5x}$

Then $f(x) \approx x^3 \cdot h(x)$

If I dont bother with the restterm, then we get something like (see picture).

And the two functions seem to be pretty close around $x=0$

You can see the maclaurin series in the picture...
Yes, I prefer Twiggy's method.

$\forall{x}\backepsilon|x|<1~\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

And

\begin{aligned}\forall{x}\backepsilon|5x|<1~\frac{ 1}{1+5x}&=\frac{1}{1-(-5x)}\\
&=\sum_{n=0}^{\infty}\left(-5x\right)^n\\
&=\sum_{n=0}^{\infty}(-1)^n5^nx^n
\end{aligned}

\begin{aligned}\therefore~\forall{x}\backepsilon|5 x|<1~\frac{x^3}{1+5x}&=x^3\cdot\frac{1}{1+5x}\\
&=x^3\cdot\sum_{n=0}^{\infty}(-1)^n5^nx^n\\
&=\sum_{n=0}^{\infty}(-1)^n5^nx^{n+3}
\end{aligned}

So now truncate to the number of terms you need for your series.


text:
How To Find The Angle Measure Of A Minor Arc. Multiply this root by the central angle again to get the arc length. In this case we can say the measure of arc a b ab a b is equal to 4 5 ∘ 45^\circ 4 5 ∘.

Multiply the area by 2 and divide the result by the central angle in radians. 180° = π radians 180 ° = π r a d i a n s. A circumscribed angle is the angle made by two intersecting tangent lines to a.

### The Measure Of The Angle Subtended At The Centre By An Arc Is Taken To Be The Measure Of The Arc.

This is less cluttered, but be sure to add the degree mark or it may get confused with the arc length. An arc is a curve made by two points on the circumference of a circle. Then you add the symbol m m m or the word “measure.”.

### Multiply The Area By 2 And Divide The Result By The Central Angle In Radians.

The measure of an arc corresponds t. Identify the arc length given in the diagram. To convert degrees to radians, we take the degree measure multiplied by pi divided by 180.

### It Is Written As M(Arc Ayq) = 70°.

The measure of an arc = the measure of its central angle. In the above figure, the measure of ∠aoq = 70°. To calculate arc length without radius, you need the central angle and the sector area:

### 180° = Π Radians 180 ° = Π R A D I A N S.

The units will be the square root of the sector area units. Let's convert 90 degrees into radians for example: We could also see the angle from the figure.

### A Circumscribed Angle Is The Angle Made By Two Intersecting Tangent Lines To A.

Multiply this root by the central angle again to get the arc length. The measure of = 360° − 85° = 275°. In this case we can say the measure of arc a b ab a b is equal to 4 5 ∘ 45^\circ 4 5 ∘.


text:
# Help with this double integral

## Main Question or Discussion Point

i would need help with this integral:

$$\int_a^{\infty}\int_a^{\infty}dxdyF(y/x)$$

now i make the change of variable y/x=u x=v then what would be the new integration limits?..thanks.

where a can be 0 or 1

Last edited:

dextercioby


text:
# Unable to open .bam file in C++ using SeqAn due to 'seqan::UnknownExtensionError'

I am trying to open .bam files in C++ to extract reads occurring at specific scaffolds and loci. I essentially want to call "samtools view sample.bam -o sample.sam scaffold:pos-pos" from C++. I have tried system("samtools view sample.bam -o sample.sam scaffold:pos-pos") and can explain the issues if you want to know, but for the sake of conciseness I will leave the details out of this post. So, I am using SeqAn. When I run my script, I get the following error:

terminate called after throwing an instance of 'seqan::UnknownExtensionError'
what():  Unknown file extension of copyPolly_SRR6511930.bam: iostream error

I am running my program on my institution's supercomputer clusters (Texas A&M Terra cluster), to which I do not have sudo access. I have loaded the following modules:

GCC/9.3.0
SeqAn/2.4.0-GCCcore-9.3.0
zlib/1.2.11-GCCcore-9.3.0

Here is my minimal script:

#include <seqan/bam_io.h>
#include <zlib.h>

using namespace seqan;

int main(){
CharString bamFileName = "copyPolly_SRR6511930.bam";
// Open input file.
BamFileIn bamFileIn(toCString(bamFileName));
if (!open(bamFileIn, toCString(bamFileName))){
std::cerr << "ERROR: Could not open " << bamFileName << std::endl;
return 1;
}

try{

BamAlignmentRecord record;
while (!atEnd(bamFileIn)){
if (hasFlagUnmapped(record))
}
}
catch (Exception const & e)
{
std::cout << "ERROR: " << e.what() << std::endl;
return 1;
}
return 0;
}

I have tried the following solutions:

-Adding #define SEQAN_HAS_ZLIB (1) to my script's header

-Adding -D SEQAN_HAS_ZLIB to my compilation such that it reads g++ script.cpp -D SEQAN_HAS_ZLIB

-Adding -lz to my compilation such that it reads g++ script.cpp -lz

I am not dead-set on using SeqAn for this project, so if anyone knows of another C++ library which extracts specific reads from .bam files, please share.

Update: I've never used HTSlib and it seems to have a fairly steep learning curve, so if anyone has an answer for the SeqAn error, that would still be very much appreciated.

Htslib is the official library to access alignment files. It supports more formats. The typical way to read through a SAM/BAM/CRAM:

#include "htslib/sam.h"
int main(int argc, char *argv[])
{
samFile *fp = sam_open(argv[1], "r");
bam1_t *b = bam_init1();
while (sam_read1(fp, h, b) >= 0) {
if (b->core.tid < 0) continue;
int end_pos = bam_endpos(b);
printf("%s\t%d\t%d\n", h->target_name[b->core.tid], b->core.pos, end_pos);
}
bam_destroy1(b);
bam_hdr_destroy(h);
sam_close(fp);
}

PS: the above started as a comment for "relevant but minor or transient information to" the original question. I didn't consider htslib a direct answer to the question, but because this had been changed to an answer, I added some details anyway. By the way, seqan2 has largely been discontinued. Use seqan3.

• Thank you for this! I tested this script and it worked. I will try to edit it to suit my needs. Also, I am using SeqAn2 because SeqAn3 is not available on my institution's supercomputers. If HTSlib doesn't pan out or if no one can figure this error out, I will email them to request that it be installed. Jun 29 '21 at 16:22

The helpdesk for my institution's supercomputers showed me the following flags:

g++ seqan_test.cpp -DSEQAN_HAS_ZLIB -lz -lpthread

This fixed the error. I'm now running into different errors and am suspicious that I am getting them because I am using SeqAn2 rather than SeqAn3. I'm going to try HTSlib.

I'm not quite sure why assigning a string to a variable would lead to a complaint about file extensions. From the SeqAn BAM tutorial, it looks like you need to use the BamFileIn object for reading BAM/SAM files:

#include <seqan/bam_io.h>
using namespace seqan;

int main(){
CharString bamFileName = "copyPolly_SRR6511930.bam";
// Open input file, BamFileIn can read SAM and BAM files.
BamFileIn bamFileIn;
if (!open(bamFileIn, toCString(bamFileName))) {
std::cerr << "ERROR: Could not open " << bamFileName << std::endl;
return 1;
}
return 0;
}

Edit: after seeing the updated code... I've got nothing, sorry. It all seems like reasonable code that shouldn't be generating errors.

• Sorry about that, I did not include all of my code. I have updated my post with the rest of the code which led to my error. Jun 29 '21 at 16:05


text:
On the Balanced Case of the Brualdi-Shen Conjecture on 4-cycle Decompositions of Eulerian Bipartite Tournaments

Rafael Del Valle Vega

Abstrak

The Brualdi-Shen Conjecture on Eulerian Bipartite Tournaments states that any such graph can be decomposed into oriented 4-cycles. In this article we prove the balanced case of the mentioned conjecture. We show that for any $2n\times 2n$ bipartite graph $G=(U\cup V, E)$ in which each vertex has $n$-neighbors with biadjacency matrix $M$ (or its transpose) there is a proper edge coloring of a column permutation of $M$ denoted $M^{\sigma}$ in which the nonzero entries of each of the $first$ $n$ columns are colored with elements from the set $\{n+1, n+2, \ldots, 2n\}$ and the nonzero entries of each of the $last$ $n$ columns are colored with elements from the set $\{1, 2, \ldots, n\}$ and if the nonzero entry $M^{\sigma}_{r,j}$ is colored with color $i$ then $M^{\sigma}_{r,i}$ must be a zero entry. Such a coloring will induce an oriented 4-cycle decomposition of the bipartite tournament corresponding to $M$. We achieve this by constructing an euler tour on the bipartite tournament which avoids traversing both pair of edges of any two internally disjoint $s$-$t$ 2-paths consecutively, where $s$ and $t$ belong to $V$.

Metrics

• 237 kali dilihat
• 113 kali diunduh

Jurnal

Electronic Journal of Graph Theory and Applications

The Electronic Journal of Graph Theory and Applications (EJGTA) is a refereed journal devoted to ... tampilkan semua


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/">

Complement Rule for Probability

%
Progress
Practice Complement Rule for Probability
Progress
%
Complement Rule for Probability

Have you ever played a basketball game at an amusement park? Well, Jeff is going to do just that.

Jeff stopped going on rides to play a basketball game at the amusement park. In this game, Jeff had ten chances to make a basket. Kyle stopped by to support Jeff in his attempt.

"I have to make four out of ten," Jeff told Kyle.

"Well, I'm pretty good at basketball. I bet I have a 75% chance of getting the four in."

If Jeff has a 75% chance of making the basket, what is the probability that he won't make the basket?

This is an situation with complementary events. You will learn how to answer this question during Concept.

Guidance

What happens when we know the likelihood that something will happen? Well, we can determine or base our actions on that event happening.

If there is a 10% chance of rain, what is the probability that it will be sunny? We can say that there is a 90% chance that it will be sunny.

If someone only knew that there was a 10% chance that it would be rainy, and that if it wasn't rainy the only other option was to be sunny, could they tell the chance of it being sunny? To figure this out, we have to figure out what the chances are of something not happening. This is called a complementary event .

If there is a 10% chance that it will not be sunny, then there is a 90% chance that it will be sunny.

Write the complementary event for the probability shown here.

There is a 50% chance that Mary will be coming over on Saturday.

To write the complementary event, we look at the opposite probability. There is a 50% chance that Mary will be coming over, so there is a 50% chance that she will not be coming over.

There is a 50% chance that Mary will not be coming over.

We can write complementary events as fractions, decimals and percents. Use whatever form is used in the example and have the complementary event match that form.

We can also predict how likely an event is to happen or not to happen based on common sense. Some things we can know for certain and some things are left up to chance.

The sunrise is one of those events. We know that the sun will rise tomorrow. Sometimes we won’t see it due to weather, but it certainly will rise.

You can also catch yourself arguing about this too. How do we know that it will rise? You could find yourself debating this with a friend for a long time. However, we need to use common sense when we are thinking about these things and not just figuring the probability using numbers.

Don’t get too caught up!

Predict whether each event is likely, impossible, unlikely or certain or write a complementary event for each situation.

Example A

The team lost its last four games, it is __________ that they will win tonight.

Solution: Unlikely

Example B

On her fifth birthday, Joanna turned five years old.

Solution: Joanna is five years old now.

Example C

There is a 20% chance that it will snow tonight.

Solution: There is an 80% chance that it will not snow.

Here is the original problem once again.

Jeff stopped going on rides to play a basketball game at the amusement park. In this game, Jeff had ten chances to make a basket. Kyle stopped by to support Jeff in his attempt.

"I have to make four out of ten," Jeff told Kyle.

"Well, I'm pretty good at basketball. I bet I have a 75% chance of getting the four in."

If Jeff has a 75% chance of making the basket, what is the probability that he won't make the basket?

To answer this question, we use 100% as our total and subtract 75% from 100%. This is the part of the whole not represented in the first percent.

$100 - 75 = 25$

Jeff has a 25% chance of not getting the basket.

Guided Practice

Here is one for you to try on your own.

Use likely, not likely or uncertain to describe the following event.

Mary will eat chocolate ice cream this week.

This is an uncertain event. We haven't been given any other information about Mary and her likes and dislikes when it comes to ice cream. Therefore, this is uncertain.

Video Review

Here is a video for review.

Explore More

Directions: Use common sense and make a prediction, use likely, impossible unlikely or certain to describe each statement.

1. Our team has a perfect record. It is _________ that we will win on Saturday.

2. A baby born will either be a boy or a girl.

3. A pig will fly through the sky.

4. A cat will like a dog.

5. There is an 85% chance it will rain. It is ________ that it will rain.

Directions: Find the complement.

6. $A$ and $B$ are complements. $P (B) = 15\%$ . Find $P (A)$ .

7. $C$ and $D$ are complements. $P (C) = 80\%$ . Find $P (D)$ .

8. $G$ and $H$ are complements. $P (H) = 49\%$ . Find $P (G)$ .

9. $T$ and $S$ are complements. $P (T) = 33\%$ . Find $P (S)$ .

10. $L$ and $K$ are complements. $P (K) = 70\%$ . Find $P (L)$ .

11. $A$ and $B$ are complements. $P (B) = 12\%$ . Find $P (A)$ .

12. $N$ and $M$ are complements. $P (N) = 96\%$ . Find $P (M)$ .

13. $Q$ and $Z$ are complements. $P (Q) = 10\%$ . Find $P (Z)$ .

Directions: Write complementary or not complementary .

14. Percentage of votes that 2 candidates get in a 2-candidate election

15. Percentage of votes that 2 candidates get in a 3-candidate election

16. Winning a game or losing a game

Vocabulary Language: English

complement

complement

A mutually exclusive pair of events are complements to each other. For example: If the desired outcome is heads on a flipped coin, the complement is tails.
Complement rule

Complement rule

The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event A, P(A) + P(A') = 1.
enumerate

enumerate

Enumerate means to catalogue or list members independently.
Probability

Probability

Probability is the chance that something will happen. It can be written as a fraction, decimal or percent.
Venn diagrams

Venn diagrams

A diagram of overlapping circles that shows the relationship among members of different sets.


text:
Writing MATLAB with the original font in modernCV [closed]

I'm wondering if I can write MATLAB with the original font in my CV using class moderncv:

closed as off-topic by user156344, user170109, Raaja, TeXnician, Andrew SwannApr 23 at 15:06

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

• "This question does not fall within the scope of TeX, LaTeX or related typesetting systems as defined in the help center." – Community, Andrew Swann
If this question can be reworded to fit the rules in the help center, please edit the question.

• Do you have a license of the commercial typeface required to typeset this logo? – Henri Menke Apr 23 at 10:58
• what you try so far? if you know and have original font, than \textsc{Matlab}\textsuperscript{\circledR} can help you. – Zarko Apr 23 at 10:59
• I'm voting to close this question as off-topic because this is a just-do-it-for-me question. – Raaja Apr 23 at 11:31

Well, if you have a look to the german page of MATLAB about Trademarks you will find:

As you can see it the registered mark set with only uppercase letters. About the font was nothing said, but it is a font without serif!

Because an cv is usualy set in non serif font (class option sans) you can use the following code:

\documentclass[10pt,a4paper,sans]{moderncv} % <=========================

\moderncvtheme{classic}

\usepackage[top=1.1cm, bottom=1.1cm, left=2cm, right=2cm]{geometry}
\usepackage{textcomp} % <===============================================

\name{Joe}{Doe}

\begin{document}

MATLAB\textsuperscript{\textregistered} % <=============================
\end{document}

In this way you respect the layout on the homepage but use your font in the cv. On the homepage of MATLAB the name of the software is printed without registered mark, but uppercase. So it seems okay to use only MATLAB in your cv.

The printed logos with class moderncv are then:

That is near enouph to the original logo so one would recognize the used software as MATLAB. If you realy want to use the completely original logo you need to contact the owner of the logo, and ask for the specifications for the logo:

1. font,
2. fontsize,
3. color,
4. needed space around the logo
5. other specifications they have to allow using there logo to be used.

That could get problematic if you want to add more than one logo to your cv in its original form.

Please note that it is not very good to use a lot of different fonts perhaps in different font sizes in one document ...

Please note that there can be -- depending on the place you live -- a legal problem using logos in the original font: Often logos are set in commercial fonts, you need to own to be allowed to use them in your cv. That can be get very explensive soon ... Or you simply are not allowed to use the original logo ...

At last: to be sure ask a layer what is allowed in the place you live ...

• @GennaroArguzzi well, in your image the font uses serifes, but that is not the way the logo is shown on the official web page of MATLAB. Where do you have this logo/image? Do they have informations about the used color and font? Could it be you show us an outdated one? Are you allowed to use that logo? – Mensch Apr 30 at 21:12
• if I want to use the logo in the normal text, have I to ask the permission to Matlab company? – Gennaro Arguzzi Apr 30 at 21:35


text:
# How large can you draw an island on a map?

A cartographer friend asked me this question: could you classify (shapes of) islands by how much space they occupy on a map (comparatively to how much space is occupied by water) if you draw them as large as possible? He had something in mind like: if

$$\frac{\mathrm{area}(\mathrm{island})}{\mathrm{area}(\mathrm{water} + \mathrm{island})} < 50\%$$

then the island is "thin", otherwise the island is "fat". I thought about it a little bit but could not come up with anything enlightening.

Here is a mathematical translation: let $$P$$ be a bounded domain in the plane (I suppose all you need is $$P$$ to be a measurable set). Define the "fatness" constant $$f(P) = \max_D \frac{\mathrm{area}(P)}{\mathrm{area}(D)}$$ where the max is taken over all Euclidean disks $$D$$ containing $$P$$.

I don't have a very precise question, I just wonder: is there anything interesting to be said about $$f$$? Has it been studied? Does it have notable properties? Is it computable in some cases? etc.

Note that this "fatness constant" could be generalized in several ways. For starters one could take Euclidean rectangles instead of disks. Another example of generalization would be to ask the question on the round sphere rather than in the plane. Here is another idea that comes to my mind: what if the island lives on the sphere but you want to draw it in the plane? Could we ask that same question over all representations of the island that are conformal (or satisfy some other property)?

NB: As has been pointed out, the words "thin" and "fat" I used here are probably a poor choice of terminology, but I did not have a better idea.

• I feel that you have already provided a good answer to your question by your $f(P)$. Could you please indicate why you're unsatisfied by that answer. – André Henriques Nov 22 '14 at 18:21
• Well, okay I have defined $f(P)$. Now I would like to know if I can say something about it. – seub Nov 22 '14 at 20:21
• If the island lies in a hemisphere then the minimal disc is uniquely defined (it is easy to proof). – Anton Petrunin Nov 22 '14 at 20:36
• One possible generalization would be to take the convex hull of your domain, rather than the bounding disk/rectangle. – Jeremy Salwen Nov 23 '14 at 0:30
• A notion of a fat region is used in computational geometry: e.g., On fat partitioning, fat covering etc. – Joseph O'Rourke Nov 23 '14 at 2:54