text:
## Tuesday, March 29, 2016

### 3/29 Euler Characteristic

And like that we're into the Spring session of Math Club. As I eluded to last week I have 5 new students including four new girls. So I brainstormed a lot about what I wanted to start with this week to draw everyone in. Basically, my goal was to find a game, that would be fun, participatory, and not require too much in the way of supplies. Along the way I found an interesting snap-cube challenge which I think I'll pick up in a future week now that I've found a teacher I can borrow enough of them from. Long term, I'm trying to decide which manipulatives to invest in; dice, checkers, snap cubes etc. for next year.  The problem is that I need bulk amounts for 18 kids and I want to buy something that offers the most possibilities for reuse. In the end I found a fun paper and pencil exercise: Criss Cross to try out.

But before I could start into the fun stuff we had to do introductions. I had everyone go around say their names, teachers and either why they had joined if they were new or their favorite part of math club if continuing. I also went over my rough charter points again: Charter.  The ones I tend to emphasize the most are respecting the classroom and each other, listening when others are working on the whiteboard and what to do if you get stuck. (Hint: its not about giving up)

The Criss-Cross game comes from the http://minerva.msri.org/files/circleinabox.pdf  (See page 77) and involves investigating the Euler Characteristic.  To summarize the rules, you break everyone up into pairs (or triplets) and have the kids draw the 3 corners of a triangle plus some dots in the center. Then they take turns connecting two dots with a straight line (no intersections are allowed and only 1 segment can be added between 2 vertices) until all the open moves are used up. Whomever can draw a line last wins. I picked two kids to do a sample game on the whiteboard and then drew a chart up with vertices, 1st player wins, and 2nd player wins for the columns.  The kids split up and starting playing games and generating data. This was a lot of fun for the room. There was a constant stream of kids coming up to add some results to the whiteboard. However, as the games progressed I noticed our results were particularly inaccurate.  At first, I asked for more games for a particular size (say 6 dots) to compensate for the mistakes. But the errors were making it hard for anyone to spot the pattern about which player always wins.

So I called for a halt and said "I think I can see some mistakes in our data lets try again to redo our data chart as a group." This time I  picked two kids at a time and had them play on the whiteboard with the whole group counting and keeping track of mistakes and we started with 5 vertices and added one more each time.  The second attempt was the charm,  The data was consistent now (the 1st player wins if there is an odd number of vertices and the 2nd if its even)  So I was able to have them also count faces and edges after defining what they meant to look for more patterns.

Very quickly I had one boy notice that the number of faces rose by 2 each time, and the number of edges by 3. We were running out of time so I talked about V - E + F and we checked if the characteristic was constant for the game. I then had the kids try out a square on their own.  By this point all the parents had arrived so I plan to follow up and work on the basic idea about why this works next week in the beginning.

The proof is  recursive and fairly simple. Each new vertex must end up in a pre-existing triangle and
follows the 2nd case above adding 3 new edges and 2 new faces. If I repeat again I'll also remember how easy it is to make mistakes and do more work as a group.

For the problem of the week: I chose another @five_triangles problem: (This involves tiling again which is a good tie-in with some our previous problems)

## Friday, March 25, 2016

### Looking for a more elegant approach

This is a continuation of my geometry problem solving posts. I spent the last few days thinking about the above problem from @five_triangles. This ends up serving as an example of a brute force approach versus a more elegant one.

My first step was to find that $\angle DBC$ is 45$^\circ$  and $\angle ABD$ is $15^\circ$ through a quick angle chase.  That immediately made me think of  trisecting $\angle DBC$ and  reflecting the triangle $\triangle$ BED across the segment $\overline{BD}$. This can be reflected two more times to produce this:

Note: $\triangle ABH$ is half of the big $\triangle ABC$ and the desired ratio is at least 3:1 since there are 3 reflections of $\triangle EBD$ within $\triangle DBC$ plus some other bits.

From here I added the parallel line GD to EO which makes a trapezoid and some similar triangles. I was hoping that the remaining similar triangles could be easily combined to form a multiple of EBD but while they all had the same altitude this wasn't the case.

I then experimented with the similar triangles starting from $\triangle{BEO}$  That almost worked I could find BGD and BDC in terms of  BEO.  Using the ratio between GD and EO I  could even break the trapezoid apart to find the remaining piece but was left with a  ratio   $\frac{\overline{GD}}{\overline{EO}}$ embedded in my expression.

So I went back to the drawing board and started with the top $\triangle{AGD}$ which is 30-60-90.

1. Let AD =  x Its possible to then find DG,  and ultimately the entire side AC in terms of x.
2. AHC is also 30-60-90 and since you have AC. You can find AH and HC in terms of x.
3. OH is in a 45-45-90 triangle so it has the same length as HC which is known.
4. The smallest triangle at the the top AGI is also 30-60-90 and similar to AEO.  Its one edge AG is already expressed in terms of x as well and we know AO from the previous step (AO = AH - OH)
5. From that you can finally find GE using the similar triangles.
6. Putting all of that together you have altitudes and bases for the two triangles all in terms of AD and it simplifies down to 1:4.

This is mechanical but fairly messy and has a lot of expression with radicals.

#### Trig Approach

After looking at my first two ideas I didn't like the complexity and I felt the ratio could be visualized more cleanly i.e. something should be quartered.  So I actually asked my friend Dan to try the same problem.  Here's his clever idea which utilizes some trig to quickly find the altitudes and cleanly shows where 1/4 comes in.
Call the intersection point in the middle Q.
Draw QC.
BQC is a 45-degree right triangle.
So the height of DBC is QC which is also BQ.
Angle QBE is 15 degrees.
BE is BQ * cos 15.
Altitude from E to BD is BE * sin 15.
So altitude of triangle EBD is altitude of triangle DBC * cos 15 * sin 15 = altitude * ½ sin 30
Ratio of altitudes is 4, so ratio of areas is also 4.

#### Final Improved Version

Playing with Dan's idea a bit more you can remove the need for trig.

1. First note: $\triangle{EBD}$ and $\triangle{DBC}$ share a common base $\overline{BD}$.
2. Angle chase to find $\angle OBC$ is 45$^\circ$  and $\angle ABO$ is $15^\circ$.
3. Reflect $\overline{BO}$ and you get a right triangle BOC so $\overline{OC}$ is the altitude of $\triangle{DBC}$.
4. Drop another altitude $\overline{EG}$ for $\triangle{EBD}$
5. Now the trick is to find the ratio between the two altitudes. This is accomplished through following a series of isosceles triangles.
6. First extend $\overline{EO}$ to F. [I experimented with this auxiliary line because the 2 overlapping 30-60-90 triangles were so useful in my first try. I thought I would directly use its side ratios but as soon as it was there all of the following observations were visible one after another.] $\angle{EFC}$ is 150 so $\angle{FOC}$ is 15 and $\triangle{OFC}$ is isosceles.  Therefore $\overline{FC}$ = $\overline{OF}$.
7. Next angle chase again $\angle{DOF}$ is also 75 so $\triangle{ODF}$ is isosceles and
$\overline{DF}$ = $\overline{OF}$.
8. Then look at $\triangle{AOF}$ which is also isoscleses. So $\overline{AO}=\overline{OF}$
9. $\triangle{AEO}$ is a 30-60-90 so $\overline{EO}$ is 1/2 $\overline{AO}$
10. Its easy to verify then that $\triangle{OEG}$ and $\triangle{ODC}$ are similar 15-75-90 triangles.
11. Since  $\triangle{OEG}$  has a hypotenuse (EO) 1/4 of the hypotenuse of  $\triangle{ODC}$  (DC). The altitudes must be in the same ratio, and also the areas.

## Thursday, March 24, 2016

### 3/22 Winter Game Day

I had to hold back a laugh this week. While waiting for everyone to arrive in the cafeteria, one of the kids asked if we were going to have pie again. "If you join Math Club next year during the spring, I'll be serving pie for 2017 Pi Day." I replied. Never underestimate the power of  baked goods.

More seriously this was our last session for the Winter. We switch over without a break next week to the new term. I'm very pleased that my reaching out to the teachers and every parent I knew seems to have paid off. The number of girls will rise to 1/3 of the club. I will continue to do some recruiting from now on before the enrollment periods.

We started the session with a quick survey of what activities the kids liked during this term. Many of the weeks project were mentioned but by far the winner was: Fold And Cut. From there we reviewed the problem of the week:

The question is what is the area of the shaded region?  My volunteers all found variants on the same idea. If you draw a parallel line through the center dot. Then you can find the area of the two sub parallelograms since its twice the area of the embedded triangles.

The games choices for today were a bit limited by supplies. Right now most of my house is packed up due to a house renovation so I chose dots again, a logic puzzle from http://www.puzzlersparadise.com/article1013.html, the game of 24 cards  and to try out checkers stacks. See: https://mathenchant.wordpress.com/2015/08/12/the-life-of-games/

We didn't really have enough checkers for everyone to try at once so for this session I just had the kids try out the basic rules and think about strategies. My plan is to return on a further day and start talking about the optimization strategy and if that goes well to branch into surreal numbers properly.

## Sunday, March 20, 2016

### 3/15 Pi Day 2016 (more or less)

One of these years Pi Day will actually occur on the day Math Club meets. Until that happens it also serves as a demonstration of  approximation for the kids. Luckily, this year one of my parents volunteered to bring in the pie so the only extra item  I took with me this week was a pie server.  We started with pie in the cafeteria and while the parent handed out the apple pie I frantically tried to get my laptop and video projector working. As it turned out a recent update had broken my wifi so I made an emergency dash up to the fourth floor. Fortunately one of the fourth grade teachers generously let us use her projector.

In the end, I was able to show the following Buffon's Needle video from numberphile:

I personally still find it fairly amazing how pi can show up in a monte carlo simulation just involving sticks. The real advantage of a video here is doing this in club would take half of our time and not be nearly as likely to produce as good of an estimate. Hopefully it was as interesting for the fourth and fifth graders. I ended up cutting it a bit short at the calculus section and just talked about the principles involved in general terms.

We then walked down to our regular space and continued with the problem of the day discussion.  See:  Temple Riddle The most important part of this discussion was bringing out the strategies the kids used. So I emphasized every time kids used various charts and notation to dissect the cases.  I'd definitely like to do some more casework type logic puzzles later this year.

Next I  recapped our circumference discussion from two weeks ago (see here) and asked what other basic formula involving pi could they think of. After one false start, a student volunteered the  area of a circle = pi * r^2.  Like the last session, I broke the kids up into groups and asked them to brainstorm for a few minutes for reasons why this was true. Of all the kid's ideas, my favorite one was to take all the rings from center out to the circumference and add them together. The boy with the idea didn't know how to carry through with that and I ended up telling him that it would definitely work but we'd also need calculus to show it. On reflection while an integral would work, its also not necessary.   You can imagine unrolling the circumferences of the circle from the center outwards. The one by the center would 0 in length and the one at the edge 2 * r * pi.  Effectively if you wave your hands a bit about the rate change being regular you'd find yourself with a triangle with a height of 2 * pi * r and a base of r. So I'll come back to the idea this week (I really want to give credit for some good thinking too).

What I actually used next instead is a visual proof I also really like where you cut the circle into pi slices and approximate a rectangle.

For the back half of the session I ended up handing out a pi worksheet from Math Counts . (Which I unfortunately can't link to). Finally for the problem of the week I used a recent problem from @five_triangles.

The problem is to find the missing triangle's area. This is a great one for my kids since it doesn't need anything more than some logic and knowledge of the triangle and parallelogram area formulas.

## Friday, March 11, 2016

We finally reached the last Math Olympiad for the year. Earlier in the week I had received a mail from MOEMS giving some corrections for the third problem. So I tried to explain the ambiguity to the kids at the start of the test and totally flubbed it. I managed to completely mis-describe the problem basically the opposite of what the clarification email indicated and it took a followup question from one of the kids for me to realize what I had done.  I recovered and fixed my mistake but it was a bit mortifying. Moral of the story: not enough prep this week. And the secondary lesson, sometimes things go south and you have to recover and keep going.

Overall this was the trickiest of the five tests. I haven't checked the results yet but I expect them to be a bit lower. I'm also delayed because a followup email from MOEMS indicated they are doing an appeal process about the problem question and to suspend entering scores until they decide what to do.

For the quicker students I brought another medium kenken which I actually had printed a few weeks earlier and not used yet.  This was accidental but actually is a fairly useful practice. Keeping a few fun extra activities in your back pocket (or math bag in my case) gives you a little insurance if an activity goes faster than expected.

I really liked our whiteboard session afterwards when we discussed the answers. The kids surprised me with 5 different strategies for the first problem, a new record.  I'm also having pretty good luck getting everyone to listen through 20 minutes of the other kids showing how to work the problems. The one thing now is I sometimes see a kid patiently waiting their turn to show their ideas but not really listening to whether someone before them does the same thing. I usually try to link these explanations together afterward. For example, "That's great ___ we've now seen 2 different people using the distributive law to get problem b"  I'm not sure whether there is anything to do about this phenomena.

For the problem of the week I chose a riddle from the recent TED talks: Temple Riddle

After transcribing this it turned out to be quite wordy so we'll see how it goes. Also next Tuesday is the day after Pi Day. So I'm gearing up to celebrate. For sure we'll go over the area formula and how to derive it. I think we'll repeat the amazing race themed activity from last year.  If I borrow a projector again, there's a really fun numberphile video about Buffon's needle.

By coincidence I found this awesome area related problem this last week: http://seekecho.blogspot.fr/2016/02/searching-for-rings.html

Oh and there will definitely be pie!

## Wednesday, March 2, 2016

### 3/1 Fold and Cut

Even after two years, I'm still experimenting and searching for new material. Case in point, yesterday was notable for two changes.  I had decided after last week: pi day prep  that I didn't want to immediately attack the area of a circle formula and instead needed something really different. Then over the week I came across some rave reviews of fold and cut activities from @mrhonner. I also remembered looking at a numberphile video on the subject a while back and thinking it was fairly compelling. So I decided that I would go that direction.   This involved a few logistical risks.

1. I don't have enough scissors/hole punches for all the kids. Therefore, I asked ahead for everyone to bring any they had. My parents are great but anytime I do something like this I have a semi-irrational fear that everyone will forget and I will be up the creek.
2. I also had another parent lend me their video projector. So for the first time I could try out using a video in the middle of the session.  However, I didn't have enough time to test the equipment or confirm that the school's WiFi filtering would allow access to youtube.  So I gambled that I had enough to keep the kids busy while I worked through the kinks and a fallback plan if it failed to work.
3. Anything with scissors and paper means extra cleanup

So we started up with Gummi bears in the cafeteria to celebrate reaching our completed problem of the week goal. And then while discussing the the kids' solutions to the tessellation problem from last week I was lucky and the video hookup worked almost immediately.

This was a good whiteboard discussion.  Everyone who demonstrated came up with the idea of finding a tiling pattern that could create the image. We had two competing answers: A hexagon with one black square vs. a hexagon with 2 black squares for the base pattern.  I asked the room for various ideas on which was correct but there was no consensus. I decided to point out/ask that there were 6 black triangles per hexagon and each one was shared between 3 hexagons. Rather than telling them the solution I then asked the kids to test each pattern at home. If the test tiles could be combined to create the floor pattern then they had the correct solution.

Moving on we turned off the lights and watched the following numberphile video on the fold and cut  theorem.

After an annoying and slightly off color 5 second ad, as expected the video went really well. I'm hoping I can borrow the projector again for Pi Day and show another one that would pair really well with the day's activities.

We then broke out the scissors and punches for some experimentation of our own. I preprinted the following templates pdf file  from Joel Hamkins.  These also were a big hit. We worked on the 8 sheets for about 40 minutes. Most kids completed at least the first 5 pages. The hardest issue here was I had to assist in punching holes for some of the thicker patterns. (Who knew math club was also a test of physical grip strength)

This was actually quite messy and you can probably tell from the photos above. So as each parent arrived for pickup I asked to make sure that there child had cleaned up the holes. Despite my trepidation after a few minutes at the end the room was back to normal.

## Welcome

Welcome to everyone who's coming here via the Hoagie's "March Mathness" blog hop. I imagine you have some interest in mathematics especially around your children's individual progress. You're looking for interesting resources to enrich at home. Perhaps you're worried about the school curriculum and are considering after schooling etc. In many ways this was the position I found  myself in a few years ago. I'm deeply interested in math and wanted to pass that on. So as my own children began entering school I started discovering what resources existed in my backyard.

And in many ways what I found was that this is a golden age for mathematics education and enrichment.  One of the earliest discoveries I made was a local Math Circle offering enrichment on the weekends, Then there was my introduction to Numberphile videos.  These were great hits with my sons. More formally, there's the innovative curriculum being developed by the Art of Problem Solving. If you look around you can find amazing stories about what kids are experimenting with and discovering.  What summarizes the moment really well for me is a recent article in the Atlantic: http://www.theatlantic.com/magazine/archive/2016/03/the-math-revolution/426855/

This is where  I was two years ago when I made the transition from a family oriented to group oriented focus.  What I realized then was that I could leverage my own passion and affect more kids. It was definitely a bit scary to step out of my comfort zone in front of 15 unknown fifth graders: first impressions. However, I believe it was and still is ultimately very rewarding. As I've said elsewhere, its tremendously gratifying seeing kids get excited about a concept or making a breakthrough or even just engaging enthusiastically with an activity. There's power in creating a peer group of kids who all are interested in the same subject.  For me this goes full circle back to my own fourth and fifth grade experiences where a key teacher made me realize that I too loved to do Math. I'm hoping at least a few readers take a moment and ask themselves the question "Can I do this too?" You don't have to be perfect or an expert on everything. Just like Math itself, running a club is quite creative and can go many directions. It also doesn't have to start very large. But a small effort can have a huge impact on your community.

## Getting Started

I inherited my group: background which in many ways made things easier. If you're starting from scratch there are actually a few great places with some checklists.

1. One of our local high schools has this manual: http://www.wastudentmath.org/content/clubs/plan/StarterPackComprehensive.pdf
2. Math Circle in a box (one of the first documents I read):  http://www.mathcircles.org/GettingStartedForNewOrganizers_WhatIsAMathCircle_CircleInABox
3. Natural math has a really cool quiz to help clarify what your goals are: http://naturalmath.com/math-circles-1001-leaders-course/
In my mind, these are the actually the easy parts. Things become harder when you actually start meeting and going out to lead a group and try planning out activities. That's the area I mostly focus on in this blog.

Here's a few pointers into some of my past writing:
1. Resource Page: This is where I collect other sites I've found useful: Resources. At the bottom of the page I've embedded links to my activity maps for the year (which are *mostly* up to date)
2. My mega post on  what I learned over the first year: http://mymathclub.blogspot.com/2015/06/the-year-in-review.html. I'm a big fan of improvement through reflection.
3. Looking back through the blog most of the posts with a date in the title reflect a log of what I did at that session and how it went. I try to focus on both mechanics. I.e. this took 20 minutes,  evaluation of how well the activities went and thoughts on what I could personally improve on.

## Some goodies for everyone

Finally here are three of my favorite discoveries from this year so far.

1. The no-rectangles problem: here  and here
2. Match stick (tooth stick in our case) puzzles: here
3. "This is not  a Maths book"   here and here

I am joining conversation on math enrichment and math education with Hoagies Gifted Education Page. Check out Hoagies Gifted website and like Hoagies Gifted on Facebook!


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###### Theorem (Law of Quadratic Reciprocity).

Let $p$ and $q$ be two distinct odd primes. Then:

 $\left(\frac{q}{p}\right)\left(\frac{p}{q}\right)=(-1)^{(p-1)(q-1)/4}$

where $\left(\frac{\cdot}{\cdot}\right)$ is the Jacobi (http://planetmath.org/JacobiSymbol) symbol (or Legendre symbol).

The following is an equivalent formulation of the Law of Quadratic Reciprocity:

###### Theorem (Quadratic Reciprocity (second form)).

Let $p,q$ be distinct odd primes. Then:

1. 1.

$\displaystyle\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)$ if one of $p$ or $q$ is congruent to $1$ modulo $4$;

2. 2.

$\displaystyle\left(\frac{p}{q}\right)=-\left(\frac{q}{p}\right)$ if both $p$ and $q$ are congruent to $3$ modulo $4$.

Title quadratic reciprocity rule QuadraticReciprocityRule 2013-03-22 11:42:27 2013-03-22 11:42:27 alozano (2414) alozano (2414) 33 alozano (2414) Theorem msc 11A15 quadratic reciprocity EulersCriterion CubicReciprocityLaw QuadraticReciprocityForPolynomials LegendreSymbol

format: plain


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# wigner_3j.m

Wigner 3j-symbols.

## Syntax

    w=wigner_3j(j1,m1,j2,m2,j3,m3)

## Description

If physically inadmissible indices are supplied, a zero is returned. Order of elements:

                         /j1 j2 j3\
\m1 m2 m3/

## Arguments

     j1-j3   - integers arranged in the order shown above

m1-m3   - integers arranged in the order shown above

## Outputs

    w       - the resulting 3j-symbol

## Notes

The function calls clebsch_gordan.m, which uses an arbitrary precision library. The output is numerically accurate to about j=10000.


text:
18 of 18 added 9 characters in body

The following is exceedingly speculative and some of the arguments are anthropomorphic, so read at your own tolerance level. It relates essentially to interpreting the physical world in terms of information theory and possibly quantum measurement theory, instead of directly from quantum mechanics.

If we consider space or spacetime as a statistical construction from finite information acquired over time, and there exists a lower discrete limit to time such as Planck time, there must in fact be such a speed limit (perhaps c or some multiple of c) which arises naturally, since the observer cannot perceive objects traveling faster than the finite rate at which he/she can calculate the metric relationships between spacetime points. Traveling faster than this limit would be like trying to have your cake and eat it too...you wouldn't be able to observe a faster than light object because you would not have the time to perceive the space backdrop from information received. Now there might be very interesting loopholes to this idea which could allow FTL in certain circumstances, particularly if space can be created at a rate faster than the speed of light as perhaps occurred in the early universe. One could also argue, FTL is possible just not directly observable in this scenario. If c is the actual speed limit, one experimental effect one might expect is mixing of x,y, and z coordinates at speeds close to c, so that there should also be a y/z contraction as well as a Lorentz x contraction.

Perhaps more interesting than setting a simple speed limit however is that the types of such statistically determined background spaces which could be realistically measured and determined by an observer might have deeper connections with gravity at large open scales and O(N) groups at small closed scales. The Euclidean space we generally observe at intermediate scales between these two extremes has very simple and unique symmetric properties (rotation, translation, inversion invariance) which one might expect to naturally emerge from any statistical construction of all possible spaces much as Feynman many paths merge toward the least action principle. At very large scales, however there are most definitely dimensional (and likely topological) constraints for the perception of such a statistical space that would require asymmetries (and thus curvatures) to be introduced. We can for example approximate an observer who can collect only finite amounts of information about his space over time as a random walker who can observe one space point on a space lattice per unit time. It is a well know fact that on a infinite lattice higher than two dimensions, the observer would only return/observe any one point or transition a finite number of times despite an infinite time for observation, and would thus be unable to statistically determine the metric of such a space! For generally finite (and thus closed and small) spaces, this is not a problem however, and it is perhaps the reason why we get interesting gauge groups like SU(3) etc at at small scales whereas we perceive simple and limited 2D projection of a 3D Euclidean space at larger scales, and require curvatures/gravity at the largest scales.

It is perhaps also a telling anthropomorphism that we perceive open 2D spaces in two different ways, as a 2D screen like projection in front of us, or as a linear horizon-like projection times a radial distance upon the surface of a gravitating body. The later is much less direct and the linearity or lack thereof appears limited and controlled by gravity reaching an absolute on the surface of a black hole where the observer and his reality is completely flattened. If there was a continuous/smooth connection between the two, this could form a new duality.


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# Java console version Conway's Game of Life

I did a super simple version of Conway's Game of Life and would like a review.

public class Conway {
String [][] grid;
int [][] neighbors;
int [][] directions = {{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};
Conway (int gridSize) {
this.neighbors = new int [gridSize][gridSize];
this.grid = new String[gridSize][gridSize];
for (int row = 0; row < this.grid.length; row++) {
for (int cell = 0; cell < this.grid[row].length; cell++) {
this.grid[row][cell] = (Math.random() > .9) ? "x|" : " |";
System.out.print(this.grid[row][cell]);
}
System.out.println();
}
this.getNeihbors();
}
void getNeihbors() {
int i = 0;
for (int row = 0; row < this.grid.length; row++ ) {
for (int col = 0; col < this.grid[row].length; col++) {
for (int drow = 0; drow < this.directions.length; drow++) {
int x = (this.directions[drow][this.directions[drow].length-this.directions[drow].length]);
int y = (this.directions[drow][this.directions[drow].length - 1]);
x = row + x;
y = col + y;
if (x < 0) {
x += this.grid.length;
} else if (x + 1 > this.grid.length) {
x -= x;
}
if (y < 0) {
y += this.grid[row].length;
} else if (y + 1 > this.grid[row].length) {
y -= y;
}
if (this.grid[x][y] == "x|") i++;
}
this.neighbors[row][col] = i;
i=0;
}
}
this.showGrid();
}
void showGrid() {
for (int row = 0; row < this.grid.length; row++) {
for (int cell = 0; cell < this.grid[row].length; cell++) {
if (this.grid[row][cell] == " |" && this.neighbors[row][cell] == 3) {
this.grid[row][cell] = "x|";
} else if (this.grid[row][cell] == "x|" && this.neighbors[row][cell] > 3) {
this.grid[row][cell] = " |";
} else if (this.grid[row][cell] == "x|" && this.neighbors[row][cell] < 2) {
this.grid[row][cell] = " |";
}
System.out.print(this.grid[row][cell]);
}
System.out.println();
}
this.getNeihbors();
}
public static void main(String[] args) {
Conway c = new Conway(80);
}
}

• You tagged this with comparative-review but posted only one solution, what do you want this to be compared to? – I'll add comments tomorrow Oct 24 '16 at 23:41
• How many different ways are there to spell "neighbors" in this program? – Eric Lippert Oct 25 '16 at 18:32
• Spell check really wasn't the answer I was looking for. – Jim Ferguson Oct 26 '16 at 3:29
• Naming does matter, and in programming naming is hard. It's not "spell check", it's a rethorical way of pointing out that deliberately making code look like it's riddled with typos might not be the best of ideas in terms of maintainability and usability. – Mathieu Guindon Oct 26 '16 at 4:46

### Object fields

    String [][] grid;
int [][] neighbors;

I would make both of these private rather than the default package-protected visibility. Unless of course you are actually planning to access these from another class in the package.

    private String [][] grid;
private int [][] neighborCounts;

I also changed the name of neighbors, as I found the original inaccurate. I would expect a variable called neighbors to hold some kind of reference to the neighbors. Instead, this counts the neighbors, which is necessary for Conway's Game of Life. So let's call it what it is.

I'd also consider changing the grid to be some other kind of object rather than String. But that's something of a judgment call. One advantage of doing that would be that you could get rid of neighbors altogether. Instead, each cell would track its own neighbors.

### Class constants

        int [][] directions = {{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};

You make this a regular object field, but it never changes and is the same for all instances of the class.

        private static final int [][] directions
= {{0,1}, {1,1}, {1,0}, {1,-1}, {0,-1}, {-1,-1}, {-1,0}, {-1,1}};

Now only one copy will be initialized for the application.

### Constructor initialization

                this.neighbors = new int [gridSize][gridSize];
this.grid = new String[gridSize][gridSize];

Personally, I think that eight-column indent is a bit much.

            grid = new String[gridSize][gridSize];
neighborCounts = new int[grid.length][grid[0].length];

Four-column indent is more common in Java.

If you do use an eight-column indent, consider adopting the same rule as the Linux source: only two levels of indent inside a function. That will help keep function contents from shooting off the edge. It also helps keep functions simple, without too much nested logic.

I prefer not to use the this. with object fields. Java doesn't require it, so I only use it when necessary to disambiguate from parameters with the same name.

You want neighborCounts to be the same size as grid. So make sure of that by setting it to the same dimensions dependently rather than in parallel. Parallel logic is fragile.

### Don't run inside the constructor

At the end of the constructor, you

                this.getNeihbors();

First, that's misspelled (missing a 'g'). It should be getNeighbors.

Second, it's not idiomatic. I would expect a method called getNeighbors to return the contents of an object field neighbors which holds the neighbors of something. This fills out the neighbors multidimensional array. I would probably call it countNeighbors.

Third, if you start the simulation from the constructor, then you have no way of constructing an object without starting the simulation.

Consider removing all current calls to getNeighbors and showGrid and instead adding to main something like

        while (true) {
c.countNeighbors();
c.showGrid();
}

That also has the advantage of not filling up the stack with recursive calls.

### Simplifying

                                for (int drow = 0; drow < this.directions.length; drow++) {
int x = (this.directions[drow][this.directions[drow].length-this.directions[drow].length]);
int y = (this.directions[drow][this.directions[drow].length - 1]);
x = row + x;
y = col + y;

This could be as simple as

                for (int[] direction : directions) {
int y = row + direction[0];
int x = col + direction[1];

And

                                        } else if (x + 1 > this.grid.length) {
x -= x;

could be

                    } else if (x >= grid.length) {
x -= grid.length;

Now if you change the rules to allow neighbors that aren't adjacent, this will still work.

And we don't have to do an extra, unnecessary math operation.

### Descriptive names

                int i = 0;

I would find this easier to follow as

        int count = 0;

When I see a variable named i, I expect it to be a loop iteration variable. This isn't. We're counting the neighbors. Why not name it based on that?

• Thank you very much. This is exactly what I was looking for. I did this as an exercise to see how quickly I could write a simple implementation of Conway's game without reference to anyone else's code. I considered using a cell object but settled for the neighbors grid because it was the first thing I thought of. (The exercise was to see how fast I could get it working.) – Jim Ferguson Oct 25 '16 at 13:21

Not sure what kind of feedback you are looking for so I'll break it up into sections:

## Overall computation:

Your overall algorithm is sound. Clearly from your code (example else if (x + 1 > this.grid.length)) you have coded a wrap-around such that effects that go off the top wrap around to the bottom, bottom to top, left to right, and right to left.

## Code style:

I'd recommend against storing the space as "x|" and " |". In other implementations I've seen the alive space stored initially as the integer 1 that gets incremented ever time it survives from one generation to the next, whereas a dead space is stored as the integer 0. The printing of "x|" and " |" is part of the "view" and would do well to be decoupled and pulled out into a separate class.

The getNeighbours function could have the inner most loop

for (int drow = 0; drow < this.directions.length; drow++) {

pulled out into a separate function called something like updateNeighbourCountForOneCell. Having a very focused function that just checks one cell will help with unit testing.

## Clarity of code:

The section of code y -= y; and x -= x;, you are really just setting y and x to 0. Declaring that specifically will be clearer. Similarly, your grid is a square of gridSize by gridSize, yet you check the variable grid for its size instead of directly storing gridSize and using it. Using gridSize makes it clear that your grid is a square.

• Thank you for your help. This code was an exercise in speed. Basically it came straight out of my head and into the compiler. Your suggestions will help me with the next exercise of this type. – Jim Ferguson Oct 25 '16 at 13:49
• Curious why I got downvoted. Any feedback would be helpful so I can improve my future responses. – Anton Codes Oct 25 '16 at 14:51
• I did not downvote you. However, I would disagree with "Similarly, your grid is a square of gridSize by gridSize, yet you check the variable 'grid' for its size instead of directly storing gridSize and using it. Using gridSize makes it clear that your grid is a square." The general recommendation is to use grid.length because then it will stay consistent even in the face of other changes, e.g. switching from a square to a rectangle. Your recommendation hardens the requirement that it is a square. It's usually better to be more flexible rather than less. – mdfst13 Oct 25 '16 at 15:27


text:
## Intermediate Algebra (12th Edition)

$-20p^{6}$
We are given the expression $-4p^{2}(5p^{4})$. We can group like terms and then simplify. $-4\times5\times p^{2} \times p^{4}$ Next, we can use the product rule for exponents to multiply the exponents. $-20\times p^{2+4}=-20p^{6}$


text:
## You Already Know Calculus: Derivatives

Calculus is made to be a whole big hoopla in high school and first year university. It’s supposedly the hardest math class in high school, notoriously complicated and unintuitive.

I blame this on bad education, not just because I’ve observed so much bad math education at the high school level, but because I can’t see any other way anyone could conclude that calculus is difficult. Because I believe that everyone already knows calculus. They just never connect what they already know to the symbols they’re manipulating in math class.

(I’m writing this series of posts — yes, this is only the first of a number of posts! — in a didactic form, but I think they may be of interest to mature mathematicians. While they are (will be?)  informal and unrigorous, they provide intuitive reasons for why everything in basic calculus is true. The results in it are certainly valuable to me: they’re the result of me spending some time trying to answer ‘why’ everything is true at a compelling level. If nothing else, they may be useful in teaching calculus.)

So what is calculus? It’s the mathematical study of rates of change, nothing more and nothing less. We call the rate of change of a function its derivative.

The fact of the matter is that you deal with derivatives all the time. They’re part of the way you think. For example, the speed of your car is the derivative of the distance you’ve traveled and acceleration is the derivative of the speed.

A large part of the calculus curriculum is concerned with the “rules of differentiation” which are just a couple of identities regarding derivatives that are useful in calculating them for a function. With the possible exception of one, you already know all of them, though some may take a tiny bit of prodding to draw out. We shall begin with a few rules that you not only know, but intuitively use on a regular basis.

Sum rule is an identity for when two functions are added together. For example, consider a person walking on a boat: their distance from the shore is the distance of the center of the boat from the shore plus the amount they’ve walked away from the shore relative to the mast..

Well, their speed is the speed of the boat plus their speed relative to the boat! Obvious enough, and that’s sum rule. To write it out, we will use the prime notation, with $f'$ representing the derivative of $f$. Then the sum rule can be stated as $(a+b)' = a' + b'$. Just don’t  let the symbolic version make you forget the real thing!

But suppose they’d been walking in the opposite direction. Then their position would have been the distance from the shore to the boat’s mast minus the distance they’ve walked from the mast towards the shore. And obviously their speed would be the speed of the boat minus their relative speed! That’s difference rule: $(a-b)' = a' - b'$.

As another example, consider the amount of water in a bath tub: It’s the amount that has gone in minus the amount that has gone out. And clearly the derivative of the amount of water in the tub, that is the rate at which the amount of water in the tub is changing, is the rate at which it is entering minus the rate at which it is leaving!

Now lets consider a very different scenario. You’re laying back on your couch and watching a movie. A runner dashes through the screen!

Now things are slightly tricky because movie time can flow differently than normal time (fastforward, rewind, etc). So at any point in real time t, the movie player is at a certain point in movie time, $m(t)$. Normally video time and real time flow at the same rate — one movie second passes per real second — so the derivative of movie time, $m'(t)$, is 1. But when we fastforward, it might be $4$ and when we rewind it might be $-4$.

Now consider our runner friend. Let’s call him Bob. Bob is running at $5 m/s$ in the movie, and normally that is what we observe him doing as well. But when we hit fastforward he  moves (to us) at $20 m/s$, even though in movie time he is only going at $5m/s$. That’s perfectly natural, the speed Bob seems to be going at is the speed he moves at in the movie times the rate movie time is passing relative to real time.

Let’s try and formalize this a little bit, let Bob’s position be describe by a function $p$. $p(m(t))$ is his position at a point in real time. $p'(m(t))$ is Bob’s in-movie speed he at a point in real time. $m'(t)$ is the rate movie time passes with respect to real time. And, as we agreed, his in-reality speed is those two multiplied: $p'(m(t))*m'(t)$.

More generally, the rate at which $f(g(x))$ changes (ie. $f(g(x))'$) is the rate at which $f$ changes at $g(x)$ (ie. $f'(g(x))$) times the rate $g(x)$ changes (ie. $g'(x)$).

This is what we call chain rule: $f(g(x))' = f'(g(x))g'(x)$

(If you don’t immediately get the chain rule, don’t worry! While you already know the underlying idea, it is more complicated to formalize than the previous ones. Go back over the last few paragraphs and pay special attention to what each part of the formula means. )

Now, when you were watching Bob, you were fairly into the movie and so you were thinking about his movement in terms of in-movie meters. But now imagine you are somewhat more detached. He’s moving across the screen at 2 (real) cm/s. Now you go watch the same scene at your friend Sally’s house because her television screen is twice as big (on both axis, not area! So it’s a massive difference!). And — jolly gosh! — Bob now moves 4 (real) cm/s. Imagine that!

This seemingly obvious fact is given the (extremely creative) name ‘constant multiple rule’ and goes $(af(x))' = af'(x)$. If you multiply a function by a constant amount, you multiply its rate by the same amount. If the world becomes twice as big, everything goes twice as fast!

(Together with sum rule, the constant product rule means that differentiation is linear. That is to say $(af(x) + g(x))' = af'(x) + g(x)$.)

Another question: at what rate does something that doesn’t change change? How fast does a person who is standing still move? If you answered, nada, it wasn’t changing, zero, they didn’t move, zero meters per second or any similar answers, you were correct. The derivative of a constant is zero. That is to say, if $k$ is a constant, $k'=0$. This fact is called constant rule, and it is definitely in the running for the most obvious rule of differentiation.

Now, lets consider something else entirely. You may not know this, since the way we teach trigonometry approaches being brain-dead, but cosine and sine are, respectively, the horizontal and vertical components of the point at angle theta on the unit circle. (As far as I’m concerned, this is how we should define cosine and sine. I can’t really go into detail with this, though, since it is beyond the purposes of this post.)

Our units are radians, the distance we’ve swept out on the circle. So if we consider our point on the circle as a function of the angle, we’re moving at unit speed.

So now, imagine yourself running along the perimeter of a circle with a radius of one meter. You’re going at 1 m/s. At every point, you’re moving tangent to the circle, at a speed of one meter a second.

That arrow is the derivative! Imagine moving it so that it’s base was at the center of the circle.

So you see, the derivative of moving in a circle is the position a quarter of a turn in front of you (with appropriate changes in units). If we split up the axes we get:

$\cos'(\theta) = \cos(\theta + \frac{\tau}{4}) = \cos(\theta + \frac{\pi}{2}) = -\sin(\theta)$

$\sin'(\theta) = \sin(\theta + \frac{\tau}{4}) = \sin(\theta + \frac{\pi}{2}) = \cos(\theta)$

If your having trouble following, focus on the underlying idea: If you walk around a circle with a radius of one meter, at a speed of one meter a second, you are walking tangent to the circle (and, get this, your speed is one meter a second!). If you imagine this as an arrow pointing out from the center of the circle, it lands on the circle a quarter turn ahead of you.

And thats it for the post! We just went through about a third of a high school calculus course, and you knew it all!

In the next post, I will introduce an alternative way to look at derivatives, differential forms, in light of which the remaining rules of differentiation will become obvious.

Continued in: You Already Know Calculus: Differential (One) Forms

Some Notes

Please leave a comment below if there was something you couldn’t follow or a question you’d like to ask. I’m far from a perfect author, and my explanations often leave much to be desired.

This first post didn’t have the most interesting content, but things should get better. My current plans for the series is:

1. Derivatives (Sum/Difference rule, chain rule, trigonometric derivatives)
2. Differential Forms (Product Rule, Power Rule, exponentials)
3. Integration (from differential forms perspective)
4. Partial Derivatives, Jacobian, and Determinants
5. n-forms
6. Stokes Theorem

Several of these are mostly written and just need some editing. The first three were originally going to be a single post, but I was persuaded that shorter is better.

I also want to talk about gradients, but I’m going to do that in a different context: the geometry of implicit functions. (I have plans for a CAD program based on this that I’m working on implementing.)

Regarding whether people already know calculus… I find it difficult to imagine it not being the case. People deal with rates all the time.

But I have to admit that I really don’t have much insight into how most people think. Most people aren’t really interested in talking about this kind of stuff with me (even when they want me to help them with their homework, they usually want to finish the problem set, not understand the concept) which leaves me with only my friends to probe, and they aren’t exactly representative of society at large.

For now, I’m going to stick with my claim. I certainly think it is true for the large majority of the population. And if nothing else, it’s an awesome title!

Tags: ,

### 17 Responses to “You Already Know Calculus: Derivatives”

1. Anonymous Says:

Is f'(g(x)) the same as f(g(x))’ ?

• George G. Says:

No. Suppose f(x):=x^2, g(x):=3x

then:

f'(g(x))=2(3x)=6x and f(g(x))’=3*2*(3x)=18x.

• christopherolah Says:

Nope, not quite. f’(g(x)) is the value of (the rate f is changing) at g(x), where as f(g(x))’ is the rate f(g(x)) is changing at x.

• Anonymous Says:

Thanks for the clarification. It was not clear if the ‘ placement before or after the function arguments was just a stylistic choice.

• christopherolah Says:

And thank *you* for pointing out to me that this wasn’t obvious to a reader unfamiliar with the notation. I’ve added the paragraph:

>More generally, the rate at which $f(g(x))$ changes (ie. $f(g(x))'$) is the rate at which $f$ changes at $g(x)$ (ie. $f'(g(x))$) times the rate $g(x)$ changes (ie. $g'(x)$).

Which will hopefully clarify things.

If you could give some more in depth comments on how you found this essay, it would be greatly appreciated.

2. tolomea Says:

+1 for the use of tau

Also you have two obvious errors in the notes, the first is “my explanations often leave ??? to be desired”, somthing has to be left there, generally “much”.

The other is “exponentials_” that “_” should be a “)”

3. George G. Says:

Nice entry! +epsilon for using tau.

I am especially intrigued now that you’ve promised to write about differential forms and Stokes theorem, some of my favorite subjects. (On a somewhat unrelated note, you might find this book useful: http://arxiv.org/abs/math/0306194 it seems to be similar in tone and approach to your post.)

Of course, I thought your was post clear and perfectly intuitive. But the real question is: what would a [poor, miserable] person that doesn’t know calculus make of it?

• George G. Says:

Oh, I forgot to ask: got any nifty, intuitive explanations for
(fg)’=f’g+g’f and (f/g)’=f’g-g’f/g^2? I can think one, but it’s not truly all that clear.

• christopherolah Says:

I have a very nice one for product rule, in the context of differential forms. And quotient rule follows simply from product rule.

You’ll see in the next post :)

• christopherolah Says:

> +epsilon for using tau.

Yay Tau!

>On a somewhat unrelated note, you might find this book useful…

Oooh, just a quick glance and I like this textbook. In the course I took at UofT we used Munkres’ Analysis On Manifolds, which I’ve come to rather dislike. I already knew a substantial part of the content and that book confused me — it took hours to figure out the connection between alternating tensors and differential forms, which were treated as a simple matter of definition. (On the other hand, I liked Munkres’ Topology, though that may say more about the previous textbooks I used…)

> what would a [poor, miserable] person that doesn’t know calculus make of it?

This is indeed the question. Once I have a few more posts up, I’m going to post this here and there. With luck, I’ll be able to attract a few such people.

4. Sacha Chua Says:

Thanks for the teaching tips! I’ll see if I can sneak calculus into the everyday math we’ve been teaching J-, who’s 13. =)

5. Ivan Avery Frey Says:

Is there a syntactical mistake above when you wrote “And, as we agreed, his in-reality speed is those two multiplied: p'(m(t)) * m'(t)”?

Didn’t you mean to say his apparent speed?

• christopherolah Says:

I’m trying to contrast in-movie and in-reality speed. I guess I could also have described it as apparent speed, but then what’s his real speed — the one in the movie?

In any case, I don’t see anything grammatically wrong with the way I handled it. Could you be more explicit with where you see the syntax mistake?

• Ivan Avery Frey Says:

Hi Chris,
I’m sorry if I wasn’t explicit enough. Upon reflection what I found confusing was the use of the terms of in-movie speed and in-reality speed.

In movie-speed obviously refers to the Bob’s speed when the movie is played back at normal speed and in-reality speed refers to Bob’s speed when the movie is played back at four times normal speed.

I would have preferred to use Bob’s apparent speed when the movie is played back at normal speed and Bob’s apparent speed when the movie is played back at four times normal speed. Maybe the terms B1x and B4x could be used.


text:
## BDM mixed methods for a nonlinear elliptic problem.(English)Zbl 0819.65129

The author introduces a mixed formulation for a nonlinear two-dimensional elliptic Dirichlet problem. The formulation is based on a mixed finite element introduced by F. Brezzi, J. Douglas jun. and L. D. Marini [Numer. Math. 47, 217-235 (1985; Zbl 0599.65072)] but includes an auxiliary vector variable which plays the role of Lagrange multiplier. This auxiliary variable is approximated by discontinuous polynomials and so can be eliminated by static condensation with little practical increase in computational effort.
The author proves existence and uniqueness of solutions to the mixed formulation as well as optimal error estimates in $$L^ 2$$, $$L^ \infty$$, and $$H^{-s}$$ of the approximate solutions. Two numerical examples are presented to illustrate the results.

### MSC:

 65N30 Finite element, Rayleigh-Ritz and Galerkin methods for boundary value problems involving PDEs 65N15 Error bounds for boundary value problems involving PDEs 35J65 Nonlinear boundary value problems for linear elliptic equations

Zbl 0599.65072
Full Text:

### References:

 [1] Brezzi, F.; Douglas, J.; Marini, L. D., Two families of mixed finite elements for second order elliptic problems, Numer. Math., 47, 217-235 (1985) · Zbl 0599.65072 [2] Chen, Z., On the existence, uniqueness and convergence of nonlinear mixed finite element methods, Mat. Apl. Comput., 8, 241-258 (1989) · Zbl 0709.65080 [3] Chen, Z., Analysis of mixed methods using conforming and nonconforming finite element methods, RAIRO Model. Math. Anal. Numér., 27, 9-34 (1993) · Zbl 0784.65075 [4] Chen, Z.; Douglas, J., Approximation of coefficients in hybrid and mixed methods for nonlinear parabolic problems, Mat. Apl. Comput., 10, 137-160 (1991) · Zbl 0760.65093 [5] Douglas, J.; Dupont, T.; Serrin, J., Uniqueness and comparison theorems for nonlinear elliptic equations in divergence form, Arch. Rational Mech. Anal., 43, 157-168 (1971) · Zbl 0222.35017 [6] Douglas, J.; Roberts, J. E., Global estimates for mixed methods for second order elliptic equations, Math. Comp., 44, 39-52 (1985) · Zbl 0624.65109 [7] Gilbarg, D.; Trudinger, N., Elliptic Partial Differential Equations of Second Order, 224 (1977), Springer: Springer Berlin, Grundlehren Math. Wiss. · Zbl 0361.35003 [8] Lewy, H.; Stampacchia, G., On existence and smoothness of solutions of some noncoercive variational inequalities, Arch. Rational Mech. Anal., 41, 241-253 (1971) · Zbl 0237.49005 [9] Milner, F., Mixed finite element methods for quasilinear second-order elliptic problems, Math. Comp., 44, 303-320 (1985) · Zbl 0567.65079 [10] Morrey, C. B., Multiple Integrals in the Calculus of Variations, 130 (1966), Springer: Springer New York, Grundlehren Math. Wiss. · Zbl 0142.38701 [11] Raviart, P. A.; Thomas, J. M., A mixed finite element method for 2nd order elliptic problems, (Proc. Conf. on Mathematical Aspects of Finite Element Methods, 606 (1977), Springer: Springer Berlin), 292-315, Lecture Notes in Math. · Zbl 0362.65089
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Just how can I modify.xfdl documents? (Update # 1)

The .XFDL documents expansion recognizes XFDL Formatted Document documents. These come from the XML-based record and also layout format criterion. This layout is specifically like the XML documents layout nonetheless, has a degree of security for usage in safe and secure interactions.

I recognize just how to watch XFDL documents making use of a documents visitor I located here. I can additionally change and also conserve these documents by doing File:Save/Save As. I would certainly such as, nonetheless, to change these documents on the fly. Any kind of pointers? Is this also feasible?

Update # 1: I have actually currently efficiently translated and also unziped a .xfdl right into an XML documents which I can after that modify. Currently, I am seeking a means to re-encode the changed XML documents back right into base64-gzip (making use of Ruby or the command line)

0
2019-05-04 05:02:40
Source Share

You do not need to leave Ruby to do this, can make use of the Base64 component in Ruby to inscribe the record similar to this:

irb(main):005:0> require 'base64'
=> true

irb(main):007:0> Base64.encode64("Hello World")
=> "SGVsbG8gV29ybGQ=\n"

irb(main):008:0> Base64.decode64("SGVsbG8gV29ybGQ=\n")
=> "Hello World"

And you can call gzip/gunzip making use of Kernel #system:

system("gzip foo.something")
system("gunzip foo.something.gz")

0
2019-05-12 02:19:37
Source

If the encoding is base64 after that this is the remedy I've come across on the internet link :

" Decoding XDFL documents conserved with 'encoding= base64'. Documents conserved with :

application/vnd.xfdl;content-encoding="base64-gzip"

are straightforward base64-encoded gzip documents. They can be conveniently recovered to XML by first decoding and afterwards unzipping them. This can be done as adheres to on Ubuntu :

sudo apt-get install uudeview uudeview -i yourform.xfdl gunzip -S "" < UNKNOWN.001 > yourform-unpacked.xfdl

The first command will certainly install uudeview, a plan that can translate base64, to name a few. You can miss this action as soon as it is mounted.

Thinking your kind is conserved as 'yourform.xfdl', the uudeview command will certainly translate the materials as 'UNKNOWN.001', given that the xfdl documents does not have a documents name. The '- i' alternative makes uudeview uninteractive, remove that alternative for even more control.

The last command gunzips the translated documents right into a documents called 'yourform-unpacked. xfdl'. "

An additional feasible remedy - here

Side Note : Block priced estimate < code > does not benefit lengthy strings of code

0
2019-05-07 21:29:09
Source

The only solution I can consider now is - read the guidebook for uudeview.

As high as I would love to aid you, I am not a specialist around, so you'll need to await a person extra knowledgable ahead down below and also aid you.

At the same time I can offer you web links to some records that could aid you :

Sorry if this does not aid you.

0
2019-05-07 19:38:07
Source


text:
## Invitation to Computer Science 8th Edition

The numbers from $1\ to\ n$, where $n$ is $even$, can be grouped into $n/2$ pairs of the form $1 + n = n + 1$ $2 + (n - 1) = n + 1$ $…$ $n /2 + (n/2 + 1) = n + 1$ giving a sum of $(n/2)(n + 1)$. This formula gives the correct sum for all cases shown, whether $n$ is $even$ or $odd$.
The numbers from $1\ to\ n$, where $n$ is $even$, can be grouped into $n/2$ pairs of the form $1 + n = n + 1$ $2 + (n - 1) = n + 1$ $…$ $n /2 + (n/2 + 1) = n + 1$ giving a sum of $(n/2)(n + 1)$. This formula gives the correct sum for all cases shown, whether $n$ is $even$ or $odd$.


text:
# Chapter 11 - Section 11.4 - Conservation of Angular Momentum - Conceptual Example - Page 202: 11.1

Boy: Increases $\omega$ Girl: Can't tell

#### Work Step by Step

Boy: The boy is on the merry-go-round, so angular momentum will be constant. Thus, when he runs to the center and decreases the moment of inertia, the angular velocity must increase for angular momentum to be conserved. Girl: She increases angular momentum by applying a torque when she jumps on. However, she also increases the moment of inertia, so we do not have enough information to tell what will happen.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.