text:
# The Challenge

Write a complete program that writes twice as many bytes to standard output as the length of the program.

# Rules

• The program must write ASCII characters to the standard output.

• The contents of the output doesn't matter.

• The output, measured in bytes, must be exactly twice the length of the program, also measured in bytes, unless you fulfill the bonus.

• Any trailing newline is included in the output's byte count.

# Bonus

Your program can optionally take a number, n, as input. If so, the output must be exactly n * program length bytes. You can assume that n will always be a positive integer. If no input is provided, n must default to 2.

If you do this, you can subtract 25 bytes from your score.

Shortest program wins.

# Restrictions

• No standard loopholes.

• The program must be at least 1 byte long.

• No adding unnecessary whitespace to the source code to change its length. Similarly, comments don't count.

• Unless you fulfill the bonus, the program must accept no input. If you do fulfill the bonus, the integer must be the only input.

Lowest score (program length in bytes - bonus) wins.

The shortest answer for each language wins for that language.

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

# Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

# Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes

var QUESTION_ID=59436,OVERRIDE_USER=41505;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\-?\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="
"+e.body.replace(OVERRIDE_REG,"")+"
")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\-?\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}
 <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b">

Leaderboard

	Author	Language	Size

Winners by Language

Language	User	Score

{{PLACE}}	{{NAME}}	{{LANGUAGE}}	{{SIZE}}	Link

{{LANGUAGE}}	{{NAME}}	{{SIZE}}	Link

• For the bonus, does the output have to be exactly n * program length bytes, or is that a minimum? – xnor Oct 2 '15 at 23:19
• It has to be exact – Daniel M. Oct 2 '15 at 23:20
• Looks like the code snippet has to be modified to handle negative scores. – El'endia Starman Oct 2 '15 at 23:49
• A bonus of -25 is basically mandatory for some languages, since it lets them achieve a negative score. In the future, I'd suggest using a percent bonus, or just making the bonus the question if you really want answers to go for it. Or, just don't have a bonus. – xnor Oct 2 '15 at 23:58
• For "no input is provided", do we assume the empty string is passed in? I can't see how one would deal with the user never typing in an input and the program just waiting. – xnor Oct 2 '15 at 23:59

## Microscript, 1

0

Prints the digit 0, followed by a newline.

# q (bonus), 27 bytes

1#[;"x"]27*2^first"J"$.z.x;  # C#, 104 bytes (79 point with bonus) class a{static void Main(string[] p){System.Console.Write(new System.String('1',104*int.Parse(p[0])));}}  you can run the program by using an argument for example test.exe 2 prints 208 character '1' • I think there's one rule missing: If no input is provided, n must default to 2 – Thomas Weller Oct 7 '15 at 20:30 # Befunge-98, 26-25=1 &:#v_v>1.1-v *d2<2<^_@#:<*  Run it in this interpreter. Apparently, it can't take input. # Befunge-93, 30 - 25 = 5 27-25=2 &:#v_v>1.1-v *93<2<^_@#:<*  There is a trailing space on the first line, and this is done because it's shorter to make 27 than it is to make 26 with Befunge-93. This outputs 27*n 1s in a row. ## Perl, 10 9 bytes Uses @primo's suggestion of $=.

print$=x9  $= is a shortcut for $FORMAT_LINES_PER_PAGE, which defaults to 60. Example: $ cat doubler.pl
print$=x9$ perl doubler.pl
606060606060606060

• print$=x9 for one byte. – primo Oct 3 '15 at 8:35 • I think print$~x3 also works for the same score. On my system, it outputs STDOUTSTDOUTSTDOUT. – PhiNotPi Oct 3 '15 at 19:55

# Stuck, -13 Bytes

i_0>;2?12*N*

This prints a bunch of newlines (N). Empty input is considered to be 0 (or anything less than 0). So, if 4 was given, it will print 48 newlines.

6Rj

will output

123456

Will be giving the bonus a shot.

# C, 25

main(){printf("%*p",50);}

This makes use of UB, but it should work. At least it works with gcc 5.2.0

# MSM, 12 bytes

'...;.;.;...

Outputs ........................

## R, 3 bytes

10;

will print

[1] 10

# QBasic, 2 bytes

?1

Nonnegative numbers in QBasic are output with both a leading and a trailing space. The PRINT command (for which ? is a shortcut) outputs a newline by default. Thus, I count 4 bytes of output: space 1 space newline.

## Julia, 9

warn(⊆)

prints

"WARNING: issubset\n"

• Might want to format your header differently so your score doesn't look like -8 instead of 9 – Patrick Roberts Oct 5 '15 at 4:05

# Insomnia, 1

7

Output:

00

8, A, B are 3 other programs that satisfy the requirement. Their output contains NUL characters, though.

# MUMPS, 4 bytes

w ?8

Well, I'm not sure if this really counts. What this program does is advance the output cursor 8 characters to the right. On every terminal I've used, this is indistinguishable from outputting 8 spaces, but is it really the same thing? I dunno.

# C, 82 bytes (with bonus)

main(int a,char**b){b&&b[1]&&(a=atoi(b[1])-1);a&&main(a-1,0),printf("%.80f",.0);}

Usage:

$wc main.c 1 2 82$ ./a.out | wc
0       1     164
$./a.out 4 | wc 0 1 138$ ./a.out 133475 | wc
0       1 10944950

• I think you can shave some off by modifying b as you go: b&&*++b&&(a=atoi(*b)-1) – Toby Speight Oct 6 '15 at 10:31
• Yes comrade! It does work. – wefwefa3 Oct 6 '15 at 17:43

# DOS, 7 bytes

date /t

outputs di 06-10-2015 (and a newline) on my system, but I'll admit it's locale dependent. So my second best is:

echo %PATH:~0,33%

which outputs C:\WINDOWS\system32;C:\WINDOWS;C: (and a newline).

• With my system settings it prints 2015.10.07\n – SuperJedi224 Oct 7 '15 at 10:12
• Got 6.2 bytes using "type %0". – Star OS Dec 16 '15 at 12:03
• @StarOS That gives me The system cannot find the specified file (roughly translated) – Berend Dec 17 '15 at 7:45
• Berend, it only works in a Batch file. +1 beacuse all of the other things don't use a shell, it uses a file – Star OS Dec 17 '15 at 11:25
• @Berend Note, you are using Windows Batch, not DOS Batch. The string manipulation function is added in Windows, not DOS. – stevefestl Sep 22 '17 at 23:52

printf %24s seq ${1-2}  Score is 24-25 A non-bonus version for +5: seq 5  which produces 1 nl 2 nl 3 nl 4 nl 5 nl. # Bash + GNU coreutils, 26 24 - 25 = 1 -1 byte The x file: yes|head -c$[24*${1:-2}]  (no trailing newline) Running: $ bash x 1 | wc -c
24
$bash x | wc -c 48$ bash x 3 | wc -c
72

Old version with 26 bytes:

The x file:

yes|head -c$((26*${1:-2}))

(no trailing newline)

• @Unihedron I don't see how this violates that rule. – Doorknob Oct 4 '15 at 14:30

# POSIX bc, 4 bytes

10^7

This creates the output 10000000.

# Javascript, 3 bytes!

1e5

Returns 100000. Y'all JS golfers are overthinking it!

# CJam, 1 byte

A

Prints 10. Very straightforward.

# Pyth, 4 bytes

*8"1

will print out

11111111

• Why did someone EDIT my answer??? – Erik the Outgolfer Oct 7 '15 at 15:25
• I like how you rolled back their changes and then added them back in. – Luminous Oct 9 '15 at 12:53
• @Luminous I did it because I want editors to learn not to edit my answers. They can suggest in comments instead! – Erik the Outgolfer Oct 10 '15 at 12:41
• Well the thing in this case was that editor didn't just edit your answer. They changed it to be completely different. They should've made it another answer. Other wise you shouldn't feel like your answer is being violated if someone edits it. We can edit other people's answers for plenty of reasons. – Luminous Oct 11 '15 at 14:43
• @Luminous Yeah, and I added them back in because I got that as a suggestion. But if they do not have rep to comment then they should add just a small not on the bottom of the answer but not EDIT it. – Erik the Outgolfer Oct 12 '15 at 11:33

# [this is a sticky note] This answer has multiple versions.

Due to the fact that I honestly am not quite sure what "input" means here. The first is the one I would use to score myself. Go down to the second to find an explanation. For scoring 1 point, do I tie with the Matmematica one? Or for scoring -1 (-5?) points, do I win? Hmm.

Anywho, this was quite a fun challenge. Maybe I can make my answer a bit better but anyways.

# Python 2.7, 26 - 25 = 1

If input means input from stdin:

print("|"*26*input())[:-1]

# Python 2.7, 24 - 25 = -1

If input means a variable:

n=2;print("|"*24*n)[:-1]

Wuut?

n=2                   # Set a variable n to 2
;                     # Separate statements (like a line break)
print                 # Print..
(                     # This is in 2.7, not 3, where print is a
# statement, not function
"|"                 # Any character works here
*24                 # Multiply said character by 24
*n                  # Multiply resulting string by n
)
[:-1]                 # Remove one character due to the line break
# automatically created by print.

# Python 2.7, 20 - 25 = -5.

If input means a variable, and you don't count n=2; as part of the program:

Simply change 24 to 20:

print("|"*20*n)[:-1]

• It's standard for input to mean STDIN or a function argument. – lirtosiast Oct 27 '15 at 14:46

## Burlesque - 7 Bytes

blsq ) '*14.*Q
**************

It just prints 14 asteriks, while the length of the program is 7 bytes. As for the bonus:

ri12.*'*j.*Q

is 12 bytes long and prints 12*n (number provided on STDIN) asteriks.

• Does your bonus version have a default value? – Titus Jan 27 '17 at 6:32
• No. ri just converts a string to integer. – mroman Jan 29 '17 at 15:50

## Minkolang 0.9, 11-25 = -14 bytes

This language was created after this challenge, but not for it.

nd1+?2["d].

Try it here.

### Explanation

This takes an integer from input, pushing a 2 on the stack if it's -1 (i.e., the input is empty). Then I use the clever quine trick: the " pushes the whole program (except the ") onto the stack. To make up for the ", I duplicate the top of stack with d. This is printed out n or 2 times and then the program stops. In the case where there is no input, there is indeed a -1 on the stack. However, Minkolang outputs nothing for negative numbers, so it does not add anything.

## Perl 6, 3 points

say 'a'x 21 # 11

print 'a'x 28*(@*ARGS[0]//2)
# 28 - 25 = 3

# Sass, 30 bytes

codegolf is absolutely awesome

if you try to compile this with sass the result is the following error

Invalid CSS after "...olutely awesome": expected "{", was ""
/* 60 bytes */

• I get invalid top-level expression on line 1 at column 1, but I also don't think this is a language as it's not turing-complete and it only compiles to CSS. – cat Dec 16 '15 at 13:04
• @cat I'm pretty sure I could implement /// in Sass using just string manipulation and a while loop, and /// is turing complete. – Sparr Dec 17 '15 at 10:05
• @Sparr fair enough – cat Dec 17 '15 at 11:57

# PlatyPar, 1 byte

#

# starts a number, but since no number is found after it, it is substituted with 59. Here's a "real" answer:

77^

Prints the result of 7^7, or 823543, which is of length 6.

# Z80 machine code, 16 bytes

3E 1F 0E 02 1E 2A F5 CD 05 00 F1 3D F2 02 01 C7

This was made on an Osborne Executive running CP/M version 3.0 using SID. Here is a disassembly (with comments following # characters for readability)

MVI A,1F     # Set register A equal to 0x1F (iterations-1)
MVI C,2      # Write is BDOS call 2 (CP/M's system calls)
MVI E,2a     # I'm writing '*' as output. Its ASCII value is 0x2a
PUSH PSW     # Push operates on register pairs. This pair contains A.
CALL 5       # Do the BDOS call (which destroys some registers)
POP PSW      # But we can get A back from the stack.
DCR A        # Decrement the loop counter
JP 102       # If last result is non-negative, go to address 0x102
RST 0        # Otherwise, exit

All programs are loaded at a fixed address of 0x100, so the jump to 0x102 is well-defined. After running this, CP/M wants to print a newline character. If we count that as output from the program, change the second byte from 1F to 1E.

# Fuzzy Octo Guacamole, 8 bytes

42*![o;]

Also could use:

8  ![o;]

Or

81*![o;]

But that is cooler.

Prints 8 "8"s and 8 newlines, including a trailing one.

### Explanation:

4: Push 4 to the stack.

2: Push 2 to the stack.

*: Pop and multiply the top 2 items on the stack and push the result (8)

!: Set the loop counter to the top. Is now 8.

[: Start a loop that lasts 'loopcounter' (8) iterations.

o: Peek at the top of the stack and push it to the temp variable.

;: Print the temp var.

]: End the loop.

# Pylongolf, 2 bytes

.;

. - Reset the stack
; - Debugally print both the stack and the variables.

The interpreter I use prints debugally by converting an 2 arrays into a string which has that string begin with [ and end with ] which prints:

[][]


text:
# Counting power-free values inside in bounded regions in $\mathbb{R}^3$

I apologize if this question is simple and the answer is obvious, but I was not able to find a reference that could be used immediately.

Let $k_1, k_2, k_3$ be three integers which are each at least $2$. Let $K \subset [-1,1]^3$ be a compact set with positive volume, and let $K(B) = \{\mathbf{x} \in \mathbb{R}^3 : \mathbf{x} = B \mathbf{y} \text{ for some } \mathbf{y} \in K\}$. Put $N(K(B)) = \#\{\mathbf{x} \in \mathbb{Z}^3 \cap K(B)\}$. Davenport showed that if $K$ is not too skew, then $N(K(B)) \sim V(K) B^3$, where $V(K)$ denotes the volume of $K$. Now put $N_{k_1, k_2, k_3}(K(B))$ to count the triples $(x_1, x_2, x_3) \in \mathbb{Z}^3$ in $K(B)$ for which $x_i$ is $k_i$-free, meaning if $p | x_i,$ then $p^{k_i} \nmid x_i$. Does it follow that

$$\displaystyle N_{k_1, k_2, k_3} (K(B)) \sim \frac{1}{\zeta(k_1) \zeta(k_2) \zeta(k_3)} V(K) B^3?$$

Here $\zeta(s)$ denotes the Riemann zeta function. The answer to this question is clear if $K = [-1,1]^3$.

More generally, I would like to know when the $k_i$-free conditions are not independent. For instance, a question I am interested in is to count those triples $(x_1, x_2, x_3)$ inside $K(B)$ for which there does not exist an integer $a$ greater than one such that $a^{k_i} | x_i, i = 1, 2,3$. Again, if $K = [-1,1]^3$, then it is easy to see that there are asymptotically

$$\displaystyle \frac{1}{\zeta(k_1 + k_2 + k_3)} V(K) B^3$$

such triples. I suspect the same is true for all `nice' sets $K$.

Any answer or reference is much appreciated.

format: plain


text:
I wanted to check my ancient down votes to see if things have improved:

I tried to make a query on data.se to find my downvotes:

select Name as Type, PostId as [Post Link]
join VoteTypes on VoteTypeId = VoteTypes.Id
where UserId = ##UserId##

but it only shows favorites. I guess it makes sense that we can't view other people's votes, but since we can make data.se accounts and it already looks up our userid based on email hash, can it not also check that we are only looking at our own votes?

Alternatively, is there some other way to see on which questions I still have down votes?

It's hard to check if someone has fixed their answer / question if I don't know what their answer / question is.

You can figure that out by going to https://math.stackexchange.com/reputation and looking for "(-1)".

More details here: How do I audit my reputation?

• If someone has been active for many days and wants to find an old downvote this can be a real hassle.
– Asaf Karagila Mod
Oct 24, 2011 at 22:29
• @AsafKaragila: Jack wants a list, and this gives it. As to your concern, a little bit of scripting would help cut down the hassle, once you have the complete list, I suppose. Oct 24, 2011 at 22:39
• In fact you can find out the classification of the reputation change. Downvoting is 3 (it is in that same page you link)
– Asaf Karagila Mod
Oct 24, 2011 at 22:41
• @AsafKaragila: Right, searching for (-1) rather than 3 is probably faster though... Oct 24, 2011 at 22:45
• True, but vote type 3 is more useful for an SQL query :-)
– Asaf Karagila Mod
Oct 24, 2011 at 22:46
• I don't think data.se will give you any info about individual downvotes, as that is sensitive information. So, unfortunately, the reputation link might be the only way. Oct 24, 2011 at 22:48
• That sure as hell sounds reasonable. Shame :-)
– Asaf Karagila Mod
Oct 24, 2011 at 22:50
• Thanks! I added a simple javascript/bookmarklet filter below. Oct 25, 2011 at 1:09

Using Aryabhata's hint, I put a bookmarklet at http://www.ms.uky.edu/~jack/mse.html -- to use it drag the "My Downvotes" link to the toolbar right below the address bar on your browser. It pulls out the downvotes you cast on answers (as opposed to the upvotes and accepts you did, and the downvotes done to you) and links to the answers. It doesn't give names, but it is very fast and simple.

It does not display downvotes on deleted answers, or on any questions. As far as I can tell, those downvotes are permanent and/or lost.

• I don't have a toolbar below the address bar? Oct 25, 2011 at 7:51
• @Jyrki: What browser are you using? Oct 25, 2011 at 8:29
• @J.M. Firefox ver 7. This is probably configurable. I vaguely recall getting rid of a bunch of toolbars, because they took up too much space on the small laptop screen. As things are now, there is nothing between the address bar and the viewing area. Oct 25, 2011 at 8:59
• Clicking Jack's script only gives me the usual (?) listing from the reputation page with all my reputation changes sorted by date and (presumably) ID# of the question. Oct 25, 2011 at 9:02
• @Jyrki: Oh yeah, the bookmark toolbar is disabled by default in version 7; you can enable it in View > Toolbars and then checking "Bookmarks Toolbar". Oct 25, 2011 at 9:02
• @J.M. Ok, I could now drag Jack's link to that area. The output of that script didn't change though. Oct 25, 2011 at 9:05
• @Jyrki: I need to click the bookmarklet twice if I'm not already on the reputation summary page. The first time I get to the usual view, the second time Jack's script reads out the answers I downvoted and provides links to them. (I'm using FF, too). Maybe this simple thing does the job for you, too?
– t.b.
Oct 25, 2011 at 13:28
• Thanks @t.b. Didn't Jack initially tell us to click twice? I was confused and assumed that he told us to "doubleclick". Apparently there is a difference :-) Oct 25, 2011 at 13:55
• @Jyrki: Yes he did. If it's any consolation, I first had the same misinterpretation :)
– t.b.
Oct 25, 2011 at 13:57
• Nice! It worked for me. Oct 26, 2011 at 21:19


text:
# Show that when the temperature is such that T Debye temperature, the specific

## Homework Statement

Consider phonons propagating on a one-dimensional chain of N identical atoms of mass M interacting by nearest-neighoour spring constants of magnitude C.

a) Show that the Debye frequency can be written as $\omega$$_{D}$=$\pi$(C/M)$^{1/2}$.
b) Show that when the temperature is such that T<<$\Theta$$_{D}$, where $\Theta$$_{D}$=$\hbar$$\omega$$_{D}$/k$_{B}$ is the Debye temperature, the specific heat can be written as C$_{V}$$\propto$Nk$_{B}$(T/$\Theta_{D}$)

## The Attempt at a Solution

I have done part a), but for part b):

my notes say that for T<<$\Theta$$_{D}$, C$_{V}$$\approx$($\frac{T}{\Theta_{D}}$)$^{3}$

so how can C$_{V}$$\propto$Nk$_{B}$(T/$\Theta_{D}$)

when there is a power of 3? And where do the N and Boltzmann's constant come from?

I have also looked in other places, but nowhere has told me why the specific heat can be written in the form they ask you to show it can be written in.

Thanks if you help.


text:
# Draw an ogive for the following distribution. Determine the median graphically and verify your result by mathematical formula. Height (in cms.) No. of students 145 − 150 2 150 − 155 5 155 - Mathematics and Statistics

Graph

Draw an ogive for the following distribution. Determine the median graphically and verify your result by mathematical formula.

 Height (in cms.) No. of students 145 − 150 2 150 − 155 5 155 − 160 9 160 − 165 15 165 − 170 16 170 − 175 7 175 − 180 5 180 − 185 1

#### Solution

To draw a ogive curve, we construct the less than cumulative frequency table as given below:

 Height (in cms) No. of students (f) Less than cumulative frequency (c.f.) 145 – 150 2 2 150 – 155 5 7 155 – 160 9 16 160 – 165 15 31 165 – 170 16 47 170 – 175 7 54 175 – 180 5 59 180 – 185 1 60 Total 60

The points to be plotted for less than ogive are (150, 2), (155, 7), (160, 16), (165, 31), (170, 47), (175, 54), (180, 59) and (185, 60).

N = 60

∴ "N"/2=60/2 = 30

∴ We take the value 30 on the Y-axis and from this point, we draw a line parallel to X-axis. From the point where this line intersects the less than ogive, we draw a perpendicular on X-axis. Foot of the perpendicular gives the value of median.
∴ Median ≈ 164.67
Now, let us calculate the median from the mathematical formula.
∵ "N"/2 = 30
The median lies in the class interval of 160 – 165.
∴ L = 160, h = 5, f = 15, c.f. = 16

Median = "L"+"h"/"f"("N"/2-"c.f.")

= 160 + 5/15 (30 - 16)

= 160 + 1/3 xx 14

= 160 + 4.67
= 164.67

Concept: Graphical Location of Partition Values
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 1 Partition Values
Miscellaneous Exercise 1 | Q 10 | Page 21


text:
# Get standard error of exponentiated coefficient in cox regression

When I run coxph, R supplies me with the coefficients, exponentiated coefficient, and standard error. The standard error is for the regression coefficient, $\beta$. How can I get the standard error of the exponentiaed regression coefficient, $exp(\beta)$?

R code:

library("survival")
data("lung") #From the survival package
res.cox <- coxph(Surv(time, status) ~ sex, data = lung)
summary(res.cox)

Which gives me (sorry for the horrible formatting):

Call: coxph(formula = Surv(time, status) ~ sex, data = lung)

n= 228, number of events= 165

coef exp(coef) se(coef) z Pr(>|z|)
sex -0.5310 0.5880 0.1672 -3.176 0.00149 **

se(coef) is giving me the standard error of "sex". I would like the standard error of exp("sex").

I can find the variance using the delta method, but can R give me the results without doing the calculations myself?

You can do it manually by calculating $se(sex) \cdot \exp(sex)=0.1672 \cdot 0.5880=.0983136$ since the derivative of $\exp{}$ is $\exp{}$ itself, or like this using svycontrast() in the survey package:

library("survival")
library("survey")
data("lung") #From the survival package
res.cox <- coxph(Surv(time, status) ~ sex, data = lung)
summary(res.cox)
svycontrast(res.cox, quote(exp(sex)))

which yields

         nlcon     SE
contrast 0.588 0.0983


text:
Find Paper, Faster
Example：10.1021/acsami.1c06204 or Chem. Rev., 2007, 107, 2411-2502
Effect of Brownian rotation on the drift velocity of a nanorod
Journal of Aerosol Science  (IF3.433),  Pub Date : 2021-06-03, DOI: 10.1016/j.jaerosci.2021.105821
George W. Mulholland, Charles Hagwood

A model set of equations including Brownian rotation dynamics is proposed to investigate the role of rotation on the drift velocity of nanorods in the direction of an external field (such as electrical). Both inertial and non-inertial Brownian rotation are considered and the more realistic non-inertial shows more correlation between the rotational and translational motion. The average drift velocity is computed from the spatial trajectory for each path and the ensemble average and its confidence interval are computed from the average of 300 trajectories. The key parameter for the trajectories is the maximum rotational stopping distance, θSt. For a value of about 0.6 radians and a length of about 60 nm, the nanorod velocity is midway between the limiting value for the large and small value of θSt. The reduced velocity is found to depend on the length of the nanorod but not the diameter. A method of estimating the 3D observed mobility from the 1D simulations based on the reduced velocity is proposed. The experimental results of Gopalakrishnan et al. (2015) for 200 nm gold nanorods are consistent with an enhancement from rotation coupling for the gold nanorods $\ge$ 200 nm long in the free molecule limit. The effect of an elastic collision boundary condition is also considered.


text:
# Can linear combinations of any Gaussian random variables be independent?

Suppose that $X=[X_1\; X_2]^t$ is Gaussian vector. My question is whether $U=a_1X_1+a_2X_2$ and $V=b_1X_1+b_2X_2$, where $a_1b_2-a_2b_1\ne 0$, can be independent Gaussian random variables, if $a_1a_2b_1b_2 \neq 0$.

• I believe the answer is yes. Somebody posted an example in a question a couple days ago. I believe it was $X,Y,Z$ are independent standard normals, then $Z-Y+X$ and $X+Y$ are independent. You can prove it easily using the bilinearity property of covariance. That's not exactly what you are looking for because it involves a combination of three instead of two, but I imagine an example using two is not hard to come by – Gregory Grant Jun 5 '15 at 20:34

Let $X=[X_1\; X_2]^t$ be a jointly Gaussian vector and $K_X=E\{XX^t\}$ be its covariance matrix. Since any covariance matrix is a positive semi-definite, it can be decomposed as $$K_X=W\Lambda W^t,$$ where $W=[w_1|w_2]$ is a unitary matrix and $\Lambda$ is a diagonal matrix. Then, since $w_1$ and $w_2$ are orthogonal, i.e., $w_1^tw_2=0$, we can prove that $U=w_1^tX$ and $U=w_2^tX$ are uncorrelated because $$E\{UV\}=E\{UV^t\}=E\{w_1^tXX^tw_2\}=w_1^tE\{XX^t\}w_2=w_1^tK_Xw_2=\Lambda.$$
On the other hand, we know that if two jointly Gaussian random variables are uncorrelated they are independent. Therefore, if $X_1$ and $X_2$ are jointly Gaussian, one can choose $[a_1 \; a_2]=w_1^t$ and $[b_1 \; b_2]=w_2^t$. Note that the elements of $w_1$ and $w_2$ can be non-zero in general.
If $X_1$ and $X_2$ are independent normal random variables, then at least after normalizing the random variables to have equal variance (for simplicity) you can choose any two linear combinations $a_1 X_1 + a_2 X_2$ and $b_1 X_1 + b_2 X_2$ such that $a$ and $b$ are orthogonal, i.e. $a_1b_1 + a_2b_2 = 0$, and the resulting linear combinations will be independent. You can check this by computing the covariance between the two linear combinations, noting that $X_1$ and $X_2$ are independent and have equal variance. For example, you can choose $a_1 = 1, a_2 = 1, b_1 = 1, b_2 = -1$. This satisfies your conditions on the coefficients.


text:
# Quotients of Hurwitz group

Since my question at Simplicity of infinite groups was not answered (well, at least, my second question), instead of trying to find the isomorhism type of those groups, I will instead try to find the quotients of $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$. I know of three quotients: PSL(2,41), the Janko group J(1), and the Janko group HJ. What other finite quotients are there?

Edit: I have done some research, and this paper http://www.heldermann-verlag.de/gcc/gcc02/gcc028.pdf seems to suggest that the chevalley group G(2,5) of order 5859000000 is a quotient of G. So I update my question to: What other simple quotients are there?

-
These are the only finite quotients of order up to $5 \times 10^8$ (I am trying to extend that, but it is getting harder.) I know also that $q=41$ gives the only finite quotient of form ${\rm PSL}(2,q)$. –  Derek Holt Sep 9 at 11:20
Ah, that would explain why those were the only ones I found. All I did was go through my (short) list of interesting elements that I could not relate the order to some earlier element. For each one I tried to set the order of it. If you want, I'll give you my presentation for the second Janko group. I was quite surprised that HJ was a quotient, since the relations given on the other sites have the order of [a,b] equal to 12. –  Thomas Sep 9 at 11:31
Sorry, I meant those are the only known finite simple quotients. The kernel of the homomorphism onto ${\rm PSL}(2,41)$ has an infinite free abelian quotient of rank 42, so there are arbitrarily large finite quotients. –  Derek Holt Sep 9 at 13:41
Wow, rank 42?! So, what other non-simple quotients are there? –  Thomas Sep 9 at 13:58
All direct products of distinct simple quotients of a group are themselves quotients of that group. –  Derek Holt Sep 10 at 15:17

Just to summarize, here is all I know about the group $G = \langle x,y \mid x^2=y^3=(xy)^7=[x,y]^{10}=1>$ and its quotients.

It has (essentially unique) homomorphisms onto the simple groups ${\rm PSL}(2,41)$, $J_1$ and ${\rm HJ} = J_2$. There are no other finite simple quotients of order up to $2 \times 10^9$ and no others of type ${\rm PSL}(2,q)$. All direct products of these three groups, like $J_1 \times J_2$ or ${\rm PSL}(2,41) \times J_1 \times J_2$ are also quotients.

Computationally, we can also examine the kernels of the homomorphisms onto the three simple groups. (In principle we could study the kernels onto the direct products, but that would be much harder, because the images are too large.)

The abelianization of the kernel $K$ of the map onto ${\rm PSL}(2,41)$ is free abelian of rank $42$ (this proves that $G$ is infinite) and the associated rational module for the action of ${\rm PSL}(2,41)$ is irreducible. However, on reduction mod 2, this module has submodules of dimensions $1$ and $21$. We can also study the $p$-quotients of $K$ for primes $p$, and it appears to have class $2$ $p$-quotients of order $p^{495}$ for all $p$, so it looks as though there are many very large $p$-quotients!

The abelianization of the kernel of the map onto $J_1$ is elementary abelian of order $11^{14}$ and the action of $J_1$ on this quotient is irreducible. There is also a class $2$ quotient of order $11^{28}$ and a class $3$ quotient of order $11^{42}$, which suggests a pattern.

I have not managed to compute the abelianization of the kernel of $G$ onto $J_2$ yet, but it certainly involves a large elementary abelian 2-group. Update: it is elementary abelian of order $2^{41}$.

These computations work by using the Reidemeister-Schreier algorithm to compute a presentation of a subgroup of finite index in a given finitely presented group. This presentation is initially on the Schreier generators and, for a 2-generator group, the number of these is roughly equal to the index of the subgroup. The presentation can be simplified by eliminating redundant generators, but this tends to make the group relations longer, and so the larger the index of the subgroup, the more complex is the computed presentation. This limits our ability to perform computations with the subgroup, like computing its abelianization. (There are alternative approaches to computing subgroup presentations, but in my experience they all ultimately have similar limitations.)

Added later: It turns out that the group $G$ also has quotients isomorphic to (at least) one of the simple groups $G_2(p)$, $G_2(p^2)$, $G_2(p^4)$ for almost all primes $p$. These are finite quotients of the images of 7-dimensional representations over number fields of degree 4 over ${\mathbb Q}$, constructed by Plesken and Souvignier. See the discussion in Other Quotient of Hurwitz group

-
I didn't know that a semidirect product of non-abelian simple groups was isomorphic to the direct product before. It explains a lot. I conjecture that those are the only simple groups. Now all we have to do is find the kernel of the homomorphism for each product (lets start with J1) and we're good. Is there a way to find the presentation of the direct product of two quotients of a group with trivial intersection? That would speed things up a lot. –  Thomas Sep 11 at 13:23
It's true that a semidirect product of nonabelian simple groups is isomorphic to the direct product (that's by the Schreier Conjecture, which was proved true by the classification of finite simple groups), but I don't see how that is relevant, and we do not need that for the result about quotients that are direct products. I will have to think about the problem of computing presentations of the larger quotients. They would not be very nice presentations, so there may not be much point in doing that. –  Derek Holt Sep 11 at 15:54
I was just confused about why the quotients were direct products instead of semidirect products, but that explains it. I guess we don't need the presentations, but it would have been nicer, and it might have made it easier to prove that there wasn't any more simple quotients. –  Thomas Sep 12 at 5:02
It might not be so hard to prove that there aren't any more simple groups. You have already proved that there are no more PSL(2,q) quotients, so that is a start. We could try to prove it for more general classes of simple groups. Also, it is surprising that the action of J1 on 11^14 is irreducible, because J1 has a representation in terms of 7x7 matrices over the field of order 11. –  Thomas Sep 12 at 5:19
Is there a way for me to investigate this, or has it gotten so complicated that we can only do it with a computer program? –  Thomas Sep 12 at 8:25


text:
# Proof of subspace, finding a basis in polynomials

Let W = {(f(x)∈ P2[R] : f '(x) + xf(0) = 0}

i) Prove that W is a subspace of P2[R].

ii) Find a basis for W.

Here's what I have so far:

i) I have to verify that the subspace (1) contains the zero space, (2) is closed under addition, and (3) is closed under scalar multiplication.

i)
(1) Contains the zero space
f'(x) + xf(0) = 0
f'(0) + (0)f(0) = 0
0 = 0,
so 0 ∈ W

I'm stuck on numbers 2 and 3, because I'm not sure how to represent f '(x) + xf(0) = 0 with examples that I can just plug in and show that it is closed under addition and scalar multiplication.

I generally do these problems with vectors or the like - for example, if the question was

Let Let W = {(x, y, z)∈ R^3 : x + y - 2z = 0}

and I wanted to show if it was a subspace, I would just show from some vectors v1 = (a1, a2, a3) and v2 = (b1, b2, b3) that

(a1 + b1) + (a2 + b2) - 2(a3 + b3) = (a1 + a2 - 2a3) + (b1 + b2 - 2b3), thus is closed under addition

and

c(a1 + a2 - 2a3) = ca1 + ca2 - 3ca3, thus is closed under scalar multiplication.

and I would find the basis by just finding the kernel of x + y - 2z = 0

I'm not sure how to deal with the problem when it is converted into polynomials. Thanks.

-
You don't need to check that the point $0$ is in this set. You need to check the function $(x\mapsto 0)$ is in this set (that's the zero vector in this vector space). – Bryan Jul 21 '14 at 3:51
Keep in mind that the 'basis vectors' of a polynomial ring are the monomials $1,x,x^2,\ldots$ themselves. So you could just as well write a polynomial $1+3x-x^2$ as the vector $(1,3,-2)$. – Semiclassical Jul 21 '14 at 3:52
Wait, because verified that the POINT 0 is in the set instead of the vector 0 is in the set, did I do the first step incorrectly? – Ted Jul 21 '14 at 4:12
Yeah, actually your proof of (1) needs revising. You don't replace $x$ by 0. You replace $f$ by the function $0$, and the desired equality follows for every $x$. – Silvio Levy Jul 21 '14 at 5:47

Proving (2) goes like this: Suppose that $f$ and $g$ both belong to $W$. Ask yourself why $f+g$ also belongs to $W$. It's easy if you write out the equation that $f$ satisfies, namely $f'(x)+x f(0)=0$, then the equation that $g$ satisfies (same equation, with $f$ replaced by $g$) and then add the two equations and apply the distributive property to get stuff times $(f+g)$.

You will need to use the fact that $(f+g)(x)=f(x)+g(x)$, which is the definition of the sum of two functions, and also the fact that $f'+g'=(f+g)'$, which is an elementary property of the derivative.

Proving (3) is even easier.

-
What I did, which I'm not sure is correct, is h, g ∈ W h'(x) + x h(0) = 0 g'(x) + x g(0) = 0 h'(x) + g'(x) + xh(0) + xg(0) = (h'(x) + xh(0)) + (g'(x) + xg(0)), c(g'(x) + xg(0)) = cg'(x) + xg(0) – Ted Jul 21 '14 at 4:09
That's basically right except that you should justify the steps ($h'(x) + x h(0) = 0$ is an assumption, etc.) and you also need to clinch the argument: you've written down $$h'(x) + g'(x) + xh(0) + xg(0) = (h'(x) + xh(0)) + (g'(x) + xg(0)),$$ but the important thing is that this shows that $(h+g)'(x)+x(h+g)(0)=0$, and therefore $(h+g)$ satisfies condition (2). – Silvio Levy Jul 21 '14 at 4:58
I think I got it now for the proof part. Thank you! – Ted Jul 21 '14 at 5:31

Let $\phi: P_2 \to P_2$, be given by $\phi(p)(t) = p'(t)+t p(0)$. It is easy to verify that $\phi$ is linear.

Then $W = \ker \phi$. Since the null space of a linear operator is a subspace, it follows that $W$ is a subspace.

If we use the basis $\{t \mapsto 1, t \mapsto t, t \mapsto t^2 \}$ for $P_2$, we see that $\phi$ has the matrix representation $\Phi (p_0,p_1,p_2)^T = (p_1, 2p_2+p_0,0)^T$. That is, $\Phi = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 0 & 0\end{bmatrix}$. It is straightforward to see that $\dim {\cal R} \Phi = 2$, hence $\dim \ker \Phi = 1$. It is also straightforward to see that $\Phi(2, 0, -1)^T = 0$, hence $\ker \Phi = \operatorname{sp} \{ (2, 0, -1)^T \}$.

Now use the basis above to find a basis for $W$.

-
I'm not sure how to arrived here: If we use the basis $\{t \mapsto 1, t \mapsto t, t \mapsto t^2 \}$ for $P_2$, we see that $\phi$ has the matrix representation $\Phi (p_0,p_1,p_2)^T = (p_1, 2p_2+p_0,0)^T$. – Ted Jul 21 '14 at 4:14
Suppose $p \in P_2$, then $p(t) = p_0+p_1 t + p_2 t^2$, and so $p'(t)+tp(0) = p_1+(p_0+2 p_2)t$. The representation of $p$ in the basis is $(p_0,p_1,p_2)^T$ and the representation of $t \mapsto p'(t)+tp(0)$ is $(p_1, p_0+2 p_2, 0)^T$. – copper.hat Jul 21 '14 at 5:40


text:
# A Complete Classification of the Isolated Singularities for Nonlinear Elliptic Equations with Inverse Square Potentials - Memoirs of the American Mathematical Society (Paperback)

(author)
£67.50
Paperback 85 Pages / Published: 30/01/2014
• We can order this

Usually despatched within 2 weeks

In this paper, the author considers semilinear elliptic equations of the form $-\Delta u- \frac{\lambda}{|x|^2}u +b(x)\,h(u)=0$ in $\Omega\setminus\{0\}$, where $\lambda$ is a parameter with $-\infty0$. The author completely classifies the behaviour near zero of all positive solutions of equation (0.1) when $h$ is regularly varying at $\infty$ with index $q$ greater than $1$ (that is, $\lim_{t\to \infty} h(\xi t)/h(t)=\xi^q$ for every $\xi>0$). In particular, the author's results apply to equation (0.1) with $h(t)=t^q (\log t)^{\alpha_1}$ as $t\to \infty$ and $b(x)=|x|^\theta (-\log |x|)^{\alpha_2}$ as $|x|\to 0$, where $\alpha_1$ and $\alpha_2$ are any real numbers.

Publisher: American Mathematical Society
ISBN: 9780821890226
Number of pages: 85
Dimensions: 254 x 178 mm

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text:
When To Use Pointers?

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dertharino    121

I've just moved from C#/Java to C++ and, as you might expect, a big problem I'm facing is understanding when to use pointers (I know that Java and C# use pointers under the hood but I never needed to understand them to program :P). I understand conceptually what pointers are but I can't seem to think of when to actually use them over just plain objects, except for perhaps arrays.

What makes it worse is that while most people on the internet say to avoid pointers when possible, whenever I read source codes for game engines and for games developed in C++, I see pointers everywhere.

For example, an excerpt from Game Programming Patterns:

class GameObject
{
public:
int velocity;
int x, y;

GameObject(InputComponent* input,
PhysicsComponent* physics,
GraphicsComponent* graphics)
: input_(input),
physics_(physics),
graphics_(graphics)
{}

void update(World& world, Graphics& graphics)
{
input_->update(*this);
physics_->update(*this, world);
graphics_->update(*this, graphics);
}

private:
InputComponent* input_;
PhysicsComponent* physics_;
GraphicsComponent* graphics_;
};

Why shouldn't InputComponent, PhysicsComponent, and GraphicsComponent just be objects?

I guess all I'm asking for is help understanding why and when to use pointers.

Thanks (sorry for all the text).

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kop0113    2453
Posted (edited)

If you passed in Objects rather than pointers to objects, they would be copies. So if you modify them or get data from them, you would be modifying the copies. The original ones (perhaps a core part of you engine) will remain unchanged. This is obviously an issue if you added (i.e) a new Physics object to the *copy* of the PhysicsComponent, the original PhysicsComponent will remain unchanged causing issues later when you query the original expecting the additional Physics object to be registered.

In C#, I see a lot of inexperienced developers make the same mistake for "boxed" value types like structs. They modify a Vector3 for example, pass that Vector3 into some other class but then modify the original Vector3 again, expecting it to change the Vector3 in the other class. In C# *everything* acts a pointer apart from structs/boxed types.

// C# using a library where Vector3 is a struct, value type
Vector3 v = new Vector3();
player.position = v;
v.x = 99;
// player.position.x is not 99

// C++ (avoiding pointers)
Vector3 v;
player->position = v;
v.x = 99;
// player->position.x is not 99

// C# using a library where Vector3 is a class
Vector3 v = new Vector3();
player.position = v;
v.x = 99;
// player.position.x is 99

// C++ (using pointers)
Vector3 *v = new Vector3();
player->position = v;
v->x = 99;
// player->position->x is 99

Perhaps refer to the following complexity in C# to better understand how C++ handles this situation: https://docs.microsoft.com/en-us/dotnet/csharp/programming-guide/types/boxing-and-unboxing

When people say do not use pointers in C++... that simply isn't feasible. Perhaps they mean they use smart pointers with references as parameters instead? Either way, unless they only stick to OpenGL's GLuint type system, I cannot see them avoiding pointers when dealing with 3rd party C libraries such as SDL. Would be cool to be proved otherwise. Can anyone point me towards a largish project where this is the case?

Edited by kop0113

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jpetrie    13159

You use pointers when you want referential semantics. That is, you want to be able to refer to some object without actually needing a complete copy.

I can't speak 100% to the rationale for why pointers are used in that code example you provided, as I didn't write it. I can theorize, however. One reason to use pointers here would be if the components are owned by something else, and the game object just needs to be able to refer to the components it owns.

For example, a typical way to implement a game object system using an "entity and component" approach like this is to have all the actual instances of the PositionComponent objects stored in, say, a PhysicsSystem. This system is responsible for the actual lifetime (creating and destroying) the position components, as well as updating them all in one big cache-friendly batch during the game update.

Individual game objects get pointers to the position component that represents their position, so they can read from it when needed. Pointers enable this design; if the game object just has its own copy of a position component, that copy is distinct from any copies in the position system.

You can implement a game object that stores its position component (et cetera) directly if you want. You don't have to use pointers there. But in the particular example you provided, the pointers are being used to refer to an object that is managed elsewhere, which is in general a common use case for referential semantics.

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Posted (edited)

simply put, a pointer should be used as a memory saving practice. sometimes there are objects that a very large and use a lot of memory. for example in the code above, PhysicComponent could be referring to an entire library of 1000s of lines of code. and a single instance of the Physic Component could be several MBytes. then consider that when you pass the object into another function, that function creates a COPY of that object. therefore with each call to that instance, there could be multiple copies of it at runtime. In come pointers. with a pointer we pass only the address of the Obj to the function instead of its value. now there is only one copy of that instance needed for our code to work at runtime

consider the following:

using namespace std;
#include<hugeLibrary.h>
int main()
{
Library test[100];//an array of objs
for(int x = 0;x<100;x++){

Lbrary lib = hugeLibrary();//2Mb of data
test[x] = lib;
}
//test now is 200Mb
startEngine(test);
doSomething(test);
closeEngine(test);//each call results in a Copy of test being created

}

or-

int main(){
Library* test[100];//an array of pointers
for(int x = 0;x<100;x++){

Lbrary lib* = &(hugeLibrary());//32bits of data
test[x] = lib;
}
//test now is 3kb
startEngine(test);
doSomething(test);
closeEngine(test);//each call results in a Copy of test being created

}

PS. this is an Extremely simplified answer. technically each lib would be also be destroyed after each function lost scope.

Edited by dcoolllx

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dertharino    121

Ohh, that makes a lot more sense, thanks! On a side note, I was wondering on whether it's better to use smart pointers vs raw pointers when developing games. I know lots of people recommend using smart pointers to avoid the hassle of raw pointers, but when it comes to developing games, would having the additional control over when to free the pointers be worth the trouble?

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Lactose    11471
13 minutes ago, dertharino said:

but when it comes to developing games, would having the additional control over when to free the pointers be worth the trouble?

No.

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jpetrie    13159
27 minutes ago, dertharino said:

I know lots of people recommend using smart pointers to avoid the hassle of raw pointers, but when it comes to developing games, would having the additional control over when to free the pointers be worth the trouble?

It's important for you to understand how regular, "non-smart" pointers work, so I do encourage you to learn. But once you learn how, it should illustrate just how much more useful smart pointers are when it comes to the actual management of object lifetimes. There are very few cases where "having additional control over when to free the pointers" is worth the trouble, and none of them are applicable to a beginner.

That said, smart pointers are not a magical bullet and I disagree with the common dogma that says a regular pointer should never appear in one's code. Generally I use smart pointers such as std::unique_ptr (I rarely have a concrete need for std::shared_ptr) where I need to use a pointer to manage lifetime. I use a bare pointer where I need a nullable reference and lifetime management isn't involved.

Another way to say that is that systems that own objects might keep smart pointers referring to those objects, but hand out bare pointers to client code if handing out a C++ reference ("Thing&") isn't feasible.

If this seems like too subtle a distinction, then I encourage you to simply use smart pointers until you are more comfortable with the ins and outs of lifetime management in code architecture, and then revisit the topic with your improved understanding and experience at a later time.

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Ryan_001    3476
1 minute ago, jpetrie said:

That said, smart pointers are not a magical bullet and I disagree with the common dogma that says a regular pointer should never appear in one's code. Generally I use smart pointers such as std::unique_ptr (I rarely have a concrete need for std::shared_ptr) where I need to use a pointer to manage lifetime. I use a bare pointer where I need a nullable reference and lifetime management isn't involved.

I was going to post basically the exact same thing and agree completely.

I also use raw pointers in places where the lifetime management is know and handled by something else.  You'll quite commonly find raw pointers in class internals as private members since you know when they are created, when they are accessed, when they are destroyed, and who owns what.  When in doubt use a smart pointer.  Many others will suggest to always use smart pointers, which isn't a terrible thing, but is more a 'rule of thumb' for beginners than a hard and fast rule.

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dertharino    121
23 minutes ago, jpetrie said:

That said, smart pointers are not a magical bullet and I disagree with the common dogma that says a regular pointer should never appear in one's code. Generally I use smart pointers such as std::unique_ptr (I rarely have a concrete need for std::shared_ptr) where I need to use a pointer to manage lifetime. I use a bare pointer where I need a nullable reference and lifetime management isn't involved.

Another way to say that is that systems that own objects might keep smart pointers referring to those objects, but hand out bare pointers to client code if handing out a C++ reference ("Thing&") isn't feasible.

That makes a lot of sense too. Thanks again!

32 minutes ago, felipefsdev said:

I saw this image days ago, I liked how didactic it is:

Haha this is a really good graphic. Thanks for that!

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Khatharr    8812
Posted (edited)

The default case is to work by value:

void doThing(int num) { num++; }

In this case 'num' is a copy of whatever you pass in when calling the function.

int derp = 1;
doThing(derp);
//derp still equals 1 because it was a copy that got incremented

If you want to be able to modify the original value then you can pass by reference:

void doThingByRef(int& num) { num++; }

Now 'num' is effectively a direct alias of whatever you pass in:

int derp = 1;
doThing(derp);
//derp now equals 2

There's another case to consider... What if the thing you're passing in is large, but you don't want to modify it?

You don't want to do an unnecessary copy, so you may use a reference, but that suggests that you're modifying the item within the function. In this case you could use a const reference:

void doThingByConstRef(const BigThing& thing) { ??? }

This lets you read 'thing' but not modify it (you'll get a compile error).

These are the basics of reference semantics. Pointers are similar, but they have some additional properties:

1) A pointer is "nullable". That is, it can be set to 'nulllptr' in order to indicate that it doesn't actually point to anything.

2) A pointer is "reseatable". That is, it can be modified to point at something else.

A note about pointers and constness:

int* num; //the pointer can be changed and the int that it points to can be changed
const int* num; //the pointer can be changed but the int that it points to can not
int* const num; //the pointer can not be changed but the int that it points to can be (probably use a reference instead in this case)
const int* const num; //neither the pointer or the pointed-to int can be changed (probably just use a const reference instead)

Be careful about returning pointers and references from functions. If the thing that you're referring to is destroyed or moved then you end up with a dangling reference/pointer, and that's no fun at all.

So the basic rule of thumb here is:

Use value semantics by default.

Use references when you want references.

Use pointers when you want references that are reseatable and/or nullable.

Cheers.

Edited by Khatharr

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jamespetts    172

My experience of pointer use is perhaps slightly different to some others on this thread, as I work on a fork of Simutrans, which has a very old codebase (1990s), and which uses pointers in a more traditional way: basic pointers are usually used whenever reference semantics are needed (and references used fairly occasionally). "Quickstone" type smart pointers (a variant on the "tombstone" system, the code for the smart pointers being custom written for this specific game) are used in certain special cases where one part of the code cannot easily know whether a pointer to an object in another part of the code has been deleted. Typically, this is used for in-game objects (such as vehicles) which the player may delete at any time. Null pointers (using a preprocessor definition of NULL as 0x000000, as the codebase was written before nullptr was standard) are used frequently for logic operations (e.g., if a player queries a tile of ground that does not have a way such as a road on it, the function that returns a pointer to a way will return NULL, which then has to be checked in the logic for deciding what information to display before attempting to retrieve any data from that object).

This more traditional use of pointers is, I suspect, a little faster but considerably more error-prone than the more modern approach suggested here of using smart pointers in all cases. I do not necessarily recommend that approach in a new project, but it may be useful to know that this is how things were done in the past to put your understanding of the use of pointers into some perspective, and it may well also be useful if you were ever to work on an older codebase where these older techniques are still used.

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Uniform convergence of $\sqrt{x^2+n^2}^{-1}-n^{-1}$

Assume the following sum: $$\sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{x^2+n^2}}-\frac{1}{n}\right),$$ where $x \in \mathbb{R}$. The problem is to determine whether the sum converges uniformly, i.e. whether there exists $s(x)$ such as $\sum f_n(x) \rightrightarrows s(x)$. My approach is following: I show $f_n \rightarrow 0$ for every $x$, as $$f_n \sim \frac{1}{n^3}, \lim_{n \rightarrow \infty} \frac{f_n}{n^{-3}} = -\frac{x^2}{2}.$$ Then I can show $f_n \rightrightarrows 0$, as $$|f_n(x)| \leq \frac{1}{n}, \forall x,$$ thus the sequence $f_n$ converges uniformly to zero, and here I got stuck. Is my approach correct? Could you help me find out on which interval the sum converges uniformly? Is the convergence absolute/local? What is the prescription of $s(x)$?

• It is probably easier to deal with $F(x)$ written this way: $$F(x)=-x^2\sum_{n\geq 1}\frac{1}{\sqrt{n^2+x^2}\left(n+\sqrt{n^2+x^2}\right)}$$ – Jack D'Aurizio Oct 19 '16 at 18:26
• Then you may notice that the $n$-th term of the series is bounded by $\frac{1}{2n^2}$, and $\left\{\frac{1}{2n^2}\right\}_{n\geq 1}$ is a summable sequence, hence we have uniform convergence over any compact subset of the real line. – Jack D'Aurizio Oct 19 '16 at 18:29
• Took me a while to get it, now I see: $f_n = \frac{n-\sqrt{n^2+x^2}}{n\sqrt{n^2+x^2}} = \frac{n-\sqrt{n^2+x^2}}{n\sqrt{n^2+x^2}} \frac{n+\sqrt{n^2+x^2}}{n+\sqrt{n^2+x^2}}= \frac{-x^2}{\sqrt{n^2+x^2}\left(n+\sqrt{n^2+x^2}\right)}$. It seems to be a good trick, I will look at it. – George Oct 20 '16 at 5:48
• Then $f_n(x) = -x^2 g_n(x)$. By the way, the series $g_n$ is bounded as $|g_n| \leq \frac{1}{2n^3}$, isn't it? – George Oct 20 '16 at 18:56
• $g_n$ is bounded because $\sum_{n\geq 1}\frac{1}{2n^2}$ is convergent. – Jack D'Aurizio Oct 20 '16 at 19:08

For fixed $x$,

$$f_n(x)=\frac{1}{n}((1+(\frac{x}{n})^2)^{-\frac{1}{2}})-\frac{1}{n}$$

$$=\frac{1}{n}(1-\frac{1}{2}\frac{x^2}{n^2}(1+\epsilon(n)))-\frac{1}{n}$$

$$=-\frac{x^2}{2n^3}(1+\epsilon(n))$$

thus the series $\sum f_n(x)$ converges.

Now let $a>0$.

for large enough $n$

$\forall x\in [-a,a]$

$|f_n(x)|\leq \frac{a^2}{2n^3}$

thus the series converges uniformly at

$[-a,a]$ and also at $[a,b]$ but not on $\mathbb R.$

• You are tacitly assuming that the radius of convergence of the Taylor series of $\frac{1}{\sqrt{1+x^2}}$ at $x=0$ is $+\infty$, that is false. You cannot prove or disprove uniform convergence through that. – Jack D'Aurizio Oct 19 '16 at 18:17
• i used equivalence – hamam_Abdallah Oct 19 '16 at 18:52
• My objection still holds. You used a Taylor expansion at $0$, that does not converge to the actual value of the function if $x$ is too large in absolute value. Equivalence holds in a neighbourhood of zero, not on the whole real line. – Jack D'Aurizio Oct 19 '16 at 18:56
• no, Taylor expansion when $n\to \infty$ – hamam_Abdallah Oct 19 '16 at 19:05
• Still the same objection. $\frac{1}{\sqrt{x^2+n^2}}$ is not an entire function, regarded as a function of $x$ or $n$. – Jack D'Aurizio Oct 19 '16 at 19:08


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# why use cnn for image classification

Anything you can do with a CNN, you can do with a fully connected architecture just as well. I can't find any example other than the Mnist dataset. Use CNNs For: Image data; Classification prediction problems; Regression prediction problems; More generally, CNNs work well with data that has a spatial relationship. It is implemented as an image classifier which scans an input image with a sliding window. Therefore, the literature about how to select and enhance features in the medical image will not be reviewed. They can be found at the core of everything from Facebook’s photo tagging to self-driving cars. The CNN approach is based on the idea that the model function properly based on a local understanding of the image. Follow edited Oct 16 '18 at 7:59. before the Image Net challenge of 2012, researchers in image processing used to design hand made features for solving problems of image processing in general and image classification in particular. I'm trying to look for the classification of images with labels using RNN with custom data. asked Dec 9 '15 at 6:54. They’re most commonly used to analyze visual imagery and are frequently working behind the scenes in image classification. The CIFAR-10 dataset can be a useful starting point for developing and practicing a methodology for solving image classification problems using convolutional neural networks. The more number of filters we have, the more image features get extracted and the better our network becomes at recognizing patterns in unseen images. before the training process). Why use CNN? By feeding the algorithm by many examples of image and their associated labels, we teach the algorithm to find the patterns of each class. share. August 01, 2017. Figure 4: Typical block diagram of a CNN [4] CNNs are used in variety of areas, including image and pattern recognition, speech recognition, natural language processing, and video analysis. Neural Network Model. Any help regarding the classification of images using RNN would be helpful. I want to use Pre-trained models such as Xception, VGG16, ResNet50, etc for my Deep Learning image recognition project to quick train the model on training set with high accuracy. During the pre deep learning era of artificial intelligence i.e. You most likely won't be able to find any meaningful comparison, since CNNs are able to handle image data that is infeasible using only FC layers. The features have been extracted using a convolutional neural network, which will also be discussed as one of our classifiers. If we can organize training images in sub-directories under a common directory, then this function may allow us to train models with a couple of lines of codes only. A rich yet manageable image classification dataset is Caltech-101. Such as many image enhanced methods to enhance the discriminable features for classification . I am not clear the reason that we normalise the image for CNN by (image - mean_image)? While a fully connected network generates weights from each pixel on the image, a convolutional … It stands apart from traditional machine learning algorithms. Thanks! Bare bones of CNN. Nearly every year since 2012 has given us big breakthroughs in developing deep learning models for the task of image classification. We’re not actually “learning” to detect objects; we’re instead just taking ROIs and classifying them using a CNN trained for image classification. CNNs represent a huge breakthrough in image recognition. Now that we’ve discussed regularization in the context of machine learning, let’s look at some code that actually performs various types of regularization. Getting Started with CNN Image Classification. hide. Image classification using regularization with Python and scikit-learn. Another reason why CNN are hugely popular is because of their architecture — the best thing is there is no need of feature extraction. But all the machine learning algorithms required proper features for doing the classification. Let's load these images off disk using the helpful image_dataset_from_directory utility. However, as CNN is an end to end solution for image classification, it will learn the feature by itself. Feature extraction. Now let’s get started with the task of Image Classification with TensorFlow by … when the model starts using irrelevant features for making predictions. The system learns to do feature extraction and the core concept of CNN is, it uses convolution of image and filters to generate invariant features which are passed on to the next layer. This is an essential phase of a supervised algorithm such as CNN for image classification. The rapid progress of deep learning for image classification . By building a neural network we can discover more hidden patterns than just classification. The difference here is that instead of using image features such as HOG or SURF, features are extracted using a CNN. Since we only have few examples, our number one concern should be overfitting. Image classification! deep-learning conv-neural-network image-processing. Convolutional Neural Network(or CNN). To support their performance analysis, the results from an Image classification task used to differentiate lymphoblastic leukemia cells from non-lymphoblastic ones have been provided. Now, Image Classification can also be done by using less complex models provided by Scikit-Learn, so why TensorFlow. Any help like this repository where CNN is used for classification would be grateful. Using FastAI’s library for multi-class classification. Contribute to mesushan/CNN-for-image-Classification development by creating an account on GitHub. Utilizing only the raw pixel intensities of the input image images, we obtained 54.42% accuracy. It uses fewer parameters compared to a fully connected network by reusing the same parameter numerous times. The main goal of the project is to create a software pipeline to identify vehicles in a video from a front-facing camera on a car. report. Can I apply CNN modelling on other regular dataset for classification or prediction? There are a number of reasons that convolutional neural networks are becoming important. Prerequisite: Image Classifier using CNN. Cite. Share. CNN uses convolutional layers which allows us to use filters compared to DNN. To achieve our goal, we will use one of the famous machine learning algorithms out there which is used for Image Classification i.e. Thanks! The major application of CNN is the object identification in an image but we can use it for natural language processing too. Convolution. By manageable I meant, ... that can help us construct a compact utility function for efficiently training a CNN model for an image classification task. Image classification can be accomplished by any machine learning algorithms( logistic regression, random forest and SVM). The pooling layer reduces the number of parameters so the model learns the important features. The sole fact that since AlexNet won ImageNet competition, every neural network that wins it uses CNN component, should be enough to convince you that CNNs are better for image data. This will take you from a directory of images on disk to a tf.data.Dataset in just a couple lines of code. 1 comment. The dataset already has a well-defined train and test dataset that we will use. Overfitting happens when a model exposed to too few examples learns patterns that do not generalize to new data, i.e. Still learning here. This is done by computing filter’s parameters $\theta$ (weight and bias). An alternative … So basically what is CNN – as we know its a machine learning algorithm for machines to understand the features of the image with foresight and remember the features to guess whether the name of the new image fed to … 8 mins read Introduction. Using CNN for classifying dogs and cats images. The top layer in CNN architectures for image classification is traditionally a softmax linear classifier, which produces outputs with a probabilistic meaning. The convolutional neural network (CNN) is a class of deep learnin g neural networks. Many techniques in image classification can also be used on it. We then applied the k-NN classifier to the Kaggle Dogs vs. Cats dataset to identify whether a given image contained a dog or a cat. CNN itself is a technique of classifying images as a part of deep learning. Reasons that I could think right off my head for using CNN over DNN for image classification is because. Improve this question. References; 1. Image Classification. In practice, a CNN learns the values of these filters on its own during the training process (although we still need to specify parameters such as number of filters, filter size, architecture of the network etc. On my Intel Xeon W 3 Ghz processor, applying object detection to a single image took ~4-9.5 seconds, depending on the input image resolution. These outputs can then be used to compute the cross-entropy loss with respect to the ground truth and backpropagate the gradients through the CNN. 4,584 5 5 gold badges 38 38 silver badges 58 58 bronze badges. With these image classification challenges known, lets review how deep learning was able to make great strides on this task. Image classification is a method to classify the images into their respective category classes using some method like : Training a small network from scratch; Fine tuning the top layers of the model using VGG16; Let’s discuss how to train model from scratch and classify the data containing cars and planes. The object detection results are incredibly slow. This article and notebooks demonstrate the main techniques used in setting up an end-to-end workflow training and deploying a Neural Network in production on Azure. At present there is no image classification algorithms in CNN. For example, the Image Category Classification Using Bag of Features example uses SURF features within a bag of features framework to train a multiclass SVM. PIL.Image.open(str(tulips[1])) Load using keras.preprocessing. Why? Image classification plays an important role in remote sensing images and is used for various applications such as environmental change, agriculture, land use/land planning, urban planning, surveillance, geographic mapping, disaster control, and object detection and also it has become a hot research topic in the remote sensing community [1]. Instead of reviewing the literature on well-performing models on the dataset, we can develop a new model from scratch. It has a lot of what I call PPPPI, Power per Parameter per Input. Ferdi. In this blog post, we reviewed the basics of image classification using the k-NN algorithm. Correct me if I'm wrong. By using TensorFlow we can build a neural network for the task of Image Classification. save. Why do we use it then? The right tool for an image classification job is a convnet, so let's try to train one on our data, as an initial baseline. Is CNN supports only image classification applications? Zhi Lu Zhi Lu. The CNN input is traditionally two-dimensional, a field or matrix, but can also be changed to be one-dimensional, allowing it to develop an internal representation of a one-dimensional sequence. Image classification is one of the use-case which can be solved by CNN. Before we discuss how, we should first understand why. Image classification using SVM . The Use of Convolutional Neural Networks for Image Classification. ( str ( tulips [ 1 ] ) ) why use cnn for image classification using keras.preprocessing it for natural language too! Be reviewed help like this repository where CNN is an end to end solution for image classification can be! Of a supervised algorithm such as HOG or SURF, features are extracted using a CNN, you can with! Rnn with custom data starts using irrelevant features for doing the classification of with! Images with labels using RNN would be helpful for developing and practicing a methodology for solving image.! Examples, our number one concern should be overfitting [ 1 ] ) Load... It is implemented as an image but we can discover more why use cnn for image classification patterns than just classification idea the. Using a CNN but we can use it for natural language processing too to new data, i.e used it. Cnn approach is based on a local understanding of the famous machine learning algorithms ( regression... Badges 38 38 silver badges 58 58 bronze badges to mesushan/CNN-for-image-Classification development by creating an account on.. Learning for image classification using the helpful image_dataset_from_directory utility given us big breakthroughs developing... The cross-entropy loss with respect to the ground truth and backpropagate the gradients through the CNN regression, random and! During the pre deep learning have been extracted using a convolutional neural network, which will be... Right off my head for using CNN for image classification tagging to self-driving cars with respect to the truth. With custom data classification algorithms in CNN architectures for image classification dataset is Caltech-101 take from! To compute the cross-entropy loss with respect to the ground truth and backpropagate the gradients the. As many image enhanced methods to enhance the discriminable features for classification or prediction the literature about how select. Done by using less complex models provided by Scikit-Learn, so why TensorFlow, image problems... To end solution for image classification discuss how, we obtained 54.42 % accuracy many image enhanced methods enhance. The feature by itself is the object identification in an image classifier scans! Present there is no need of feature extraction popular is because methods to enhance the discriminable features for would... Learning for image classification using the helpful image_dataset_from_directory utility progress of deep learnin g neural networks meaning! And enhance features in the medical image will not be reviewed popular is because of image... Will not be reviewed these images off disk using the k-NN algorithm can then be used analyze. Pixel intensities of the input image with a CNN linear classifier, which will also be used it. Top layer in CNN the dataset already has a lot of what i PPPPI. Just classification of deep learning for image classification, it will learn the by! … using CNN for image classification to look for the task of image classification challenges known, lets review deep! And enhance features in the medical image will not be reviewed a convolutional neural networks are becoming important it implemented! Discuss how, we obtained 54.42 % accuracy s photo tagging to self-driving.... Of parameters so the model function properly based on the idea that the model starts using irrelevant for... In image classification is traditionally a softmax linear classifier, which will also done! Modelling on other regular dataset for classification not be reviewed it has a train. As well 'm trying to look for the classification on disk to a fully connected by. Model from scratch which will also be used on it with these classification! Any help regarding the classification of images with labels using RNN with custom data - mean_image ) been extracted a... Popular is because the gradients through the CNN approach is based on a local understanding of the input with! Era of artificial intelligence i.e could think right off my head for using CNN for classifying and. Utilizing only the raw pixel intensities of the famous machine learning algorithms ( regression... To a fully connected architecture just as well ’ s parameters $\theta$ ( weight and bias ) lot! The CIFAR-10 dataset can be accomplished by any machine learning algorithms ( logistic regression, forest... Image but we can use it for natural language processing too 54.42 % accuracy %... Not clear the reason that we normalise the image PPPPI, Power per parameter input. By computing filter ’ s parameters $\theta$ ( weight and bias ) am clear... Raw pixel intensities of the image for CNN by ( image - mean_image ) reviewed basics... To new data, i.e examples, our number one concern should be.... Model learns the important features let 's Load these images off disk using the helpful image_dataset_from_directory utility few! Obtained 54.42 % accuracy frequently working behind the scenes in image classification challenges known, lets review how learning. Raw pixel intensities of the image for CNN by ( image - )! Are becoming important ) is a technique of classifying images as a part deep.


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Complex Dynamical Systems: The Mathematics Behind the Mandelbrot and Julia Sets

Edited by: Robert L. Devaney Boston University
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AMS Member Price: $60.00 Click above image for expanded view Complex Dynamical Systems: The Mathematics Behind the Mandelbrot and Julia Sets Edited by: Robert L. Devaney Boston University Available Formats:  Hardcover ISBN: 978-0-8218-0290-8 Product Code: PSAPM/49  List Price:$50.00 MAA Member Price: $45.00 AMS Member Price:$40.00
 Electronic ISBN: 978-0-8218-9264-0 Product Code: PSAPM/49.E
 List Price: $47.00 MAA Member Price:$42.30 AMS Member Price: $37.60 Bundle Print and Electronic Formats and Save! This product is available for purchase as a bundle. Purchasing as a bundle enables you to save on the electronic version.  List Price:$75.00 MAA Member Price: $67.50 AMS Member Price:$60.00
• Book Details

Proceedings of Symposia in Applied Mathematics
Volume: 491994; 209 pp
MSC: Primary 58; 30;

In the last fifteen years, the Mandelbrot set has emerged as one of the most recognizable objects in mathematics. While there is no question of its beauty, relatively few people appreciate the fact that the mathematics behind such images is equally beautiful. This book presents lectures delivered during the AMS Short Course entitled “Complex Dynamical Systems: The Mathematics Behind the Mandelbrot and Julia Sets”, held at the Joint Mathematics Meetings in Cincinnati in January 1994. The lectures cover a wide range of topics, including the classical work of Julia and Fatou on local dynamics of analytic maps as well as recent work on the dynamics of quadratic and cubic polynomials, the geometry of Julia sets, and the structure of various parameter spaces. Among the other topics are recent results on Yoccoz puzzles and tableaux, limiting dynamics near parabolic points, the spider algorithm, extensions of the theory to rational maps, Newton's method, and entire transcendental functions. Much of the book is accessible to anyone with a background in the basics of dynamical systems and complex analysis.

Graduate students and researchers in all areas of pure and applied mathematics.

• Articles
• Robert L. Devaney - The complex dynamics of quadratic polynomials [ MR 1315532 ]
• Bodil Branner - Puzzles and para-puzzles of quadratic and cubic polynomials [ MR 1315533 ]
• Linda Keen - Julia sets of rational maps [ MR 1315534 ]
• Adrien Douady - Does a Julia set depend continuously on the polynomial? [ MR 1315535 ]
• Paul Blanchard - The dynamics of Newton’s method [ MR 1315536 ]
• John H. Hubbard and Dierk Schleicher - The spider algorithm [ MR 1315537 ]
• Robert L. Devaney - Complex dynamics and entire functions [ MR 1315538 ]
• Reviews

• The book is a useful introduction to a survey of this field of active research. It is illustrated with many beautiful pictures of Julia sets, the Mandelbrot set, and other sets related to the theory.

Mathematical Reviews
• This collection of lectures … contains interesting survey articles on the classical work of Julia and Fatou as well as on the more recent work on the synamics of quadratic and cubic polynomials, on Yoccoz puzzles and tableaux, on the spider algorithm and on the dynamics of entire transcendental functions. Much of the book is accessible to anyone with a background on complex analysis. Several impressive color plates are included.

Monatshefte für Mathematik
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Volume: 491994; 209 pp
MSC: Primary 58; 30;

In the last fifteen years, the Mandelbrot set has emerged as one of the most recognizable objects in mathematics. While there is no question of its beauty, relatively few people appreciate the fact that the mathematics behind such images is equally beautiful. This book presents lectures delivered during the AMS Short Course entitled “Complex Dynamical Systems: The Mathematics Behind the Mandelbrot and Julia Sets”, held at the Joint Mathematics Meetings in Cincinnati in January 1994. The lectures cover a wide range of topics, including the classical work of Julia and Fatou on local dynamics of analytic maps as well as recent work on the dynamics of quadratic and cubic polynomials, the geometry of Julia sets, and the structure of various parameter spaces. Among the other topics are recent results on Yoccoz puzzles and tableaux, limiting dynamics near parabolic points, the spider algorithm, extensions of the theory to rational maps, Newton's method, and entire transcendental functions. Much of the book is accessible to anyone with a background in the basics of dynamical systems and complex analysis.

Graduate students and researchers in all areas of pure and applied mathematics.

• Articles
• Robert L. Devaney - The complex dynamics of quadratic polynomials [ MR 1315532 ]
• Bodil Branner - Puzzles and para-puzzles of quadratic and cubic polynomials [ MR 1315533 ]
• Linda Keen - Julia sets of rational maps [ MR 1315534 ]
• Adrien Douady - Does a Julia set depend continuously on the polynomial? [ MR 1315535 ]
• Paul Blanchard - The dynamics of Newton’s method [ MR 1315536 ]
• John H. Hubbard and Dierk Schleicher - The spider algorithm [ MR 1315537 ]
• Robert L. Devaney - Complex dynamics and entire functions [ MR 1315538 ]
• The book is a useful introduction to a survey of this field of active research. It is illustrated with many beautiful pictures of Julia sets, the Mandelbrot set, and other sets related to the theory.

Mathematical Reviews
• This collection of lectures … contains interesting survey articles on the classical work of Julia and Fatou as well as on the more recent work on the synamics of quadratic and cubic polynomials, on Yoccoz puzzles and tableaux, on the spider algorithm and on the dynamics of entire transcendental functions. Much of the book is accessible to anyone with a background on complex analysis. Several impressive color plates are included.

Monatshefte für Mathematik
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# Math & Pressbooks

Hey @oct22-b-cohort As promised, I’ve found some resources for you and your teams around math equations in Pressbooks:

@james.boffenmyer- turns out LaTeX is compatible with Pressbooks! There is an OER all about Using LaTeX in Pressboks that you may find beneficial to skim through.

The Pressbooks User Guide also has a section around adding mathematical notations (with both LaTex and MathJax)

I also wanted to point your attention to an Introduction to (Ethical) Statistics OER: Numbers Don’t Lie (but People do). I encourage you all to take a look at the team’s Project Summary.

As always, if you have any questions or need any assistance in finding something in particular, don’t hesitate to reach out!

-Kaitlin


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, 02.07.2018 17:55 alyssasnyderrr

# He equation of the graphed line in point-slope form is , and its equation in slope-intercept form is

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