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# Workbench, help decide what to buy

### Help Support The Rocketry Forum:

#### rfjustin

##### Well-Known Member
I relocated from IL to WI, and I'm finally at the point where I can concentrate on some "nice-have" house projects. Can lighting just got completed (29 cans worth with LED trim rings), and now I want to look into getting a new workbench.

I had a plywood 84" by 24" bench at my old house with some pegboard on the wall right behind it. I am looking to get something similar. I'm not trying to duplicate exactly what I had before, but I am looking for functional. Pic of old work area below.

I'm leaning towards a Deluxe Workstation 72" wide and 30" deep from Uline: https://www.uline.com/BL_1498/Deluxe-Workstations These are nice, but spendy once you account for all the custom bells and whistles. Before I go and drop 1200 bucks at Uline, do you have other workbench recommendations?

#### Eric

##### Well-Known Member
I would build it from scratch, over a u-line bench. Thats a lot of money. I bought one nice sheet of thick plywood. Home depot cut it length wise at 27". A couple 4x4's and 2x4's, box of screws and bolts.

With power strips and peg board I think I was under $200. Even put a nice urethane finish on it. I could justify that expense for a nice workbench. My wife would have killed me if I spent$500.

#### jqavins

##### Joseph Avins
TRF Supporter
I'll second DIY. If you're not sure what to build, I'm sure you can find more plans than you can shake a Mean Machine at to get those creative juices flowing. I'd suggest something with a shelf down below to keep the stuff you store there a little off the floor.

#### Cl(VII)

##### Chris Bender, Lab Rat
TRF Sponsor
I have quite a few of the seville classics Ultra-HD benches (https://www.sevilleclassics.com/ultrahd/workbenches-workcenters.html), and cabinets (you can see my setup here: https://www.rocketryforum.com/threads/my-rocket-palace-is-complete.133087/). All are very good quality except the tall cabinet...it doesn't hold square well with stuff on top of it.

My main workbench is described in that thread too, but you may not have room for that kind. If you do, I can't recommend it enough, having access from all sides is very helpful.

#### dr wogz

##### Fly caster
Another for "build it yourself"

I designed & built my own. it came out to be 60 x 36" x 42" high. it has two shelves, and pegboard sides (on the legs, as 'shear webs'). I have an old door on top of the bench, to further extend the width, and i have a piece of sheet-rock / wallboard / gyp rock on top of that, as an added 'sacrificial surface' (in which you can stick pins; great for holding wing spar & ribs & such in place.. my other hobby, R/C airplanes!)

#### PatD

##### Well-Known Member
TRF Supporter
DIY. I built two for my garage. 8 ft long by 24 wide. Doubled 2x6 legs, inset 2x6 top perimeter, 3/4 ply top with 3/4 melamine laminated particle board glued and screwed from the bottom to the ply for the working surface. 2x4 lower perimeter with 3/4 chip board for a shelf. Very flat and stable surface. As Ivan says "Strong like bull!" And a lot less expensive.

#### KILTED COWBOY

##### Well-Known Member
DYI and spend the 1K savings on rockets.

#### neil_w

##### Working on 20K
TRF Supporter
This thread came back to life at a good time for me; my Dad (85) just moved into a smaller house and needs a small, light-duty workbench. He was looking at the one at Harbor Freight: https://www.harborfreight.com/48-in-workbench-with-light-60723.html, which is ridiculously cheap and honestly it looks pretty serviceable. I wish he would have more pegboard space, though; it bugs me that there are several of inches of dead space all around the border of the pegboard.

The Seville Classics (https://www.amazon.com/Seville-Classics-UHD20247B-Workbench-Stainless/dp/B005NAVFEW/) looks to be nicer quality all the way around, solid wood top, and a proper peg board that goes out to the edge, but I'm not sure he'll be interested in spending more than double for it.

#### John Brohm

##### Well-Known Member
I used a set of Ikea base cabinets, and then built the bench top.

John

#### neil_w

##### Working on 20K
TRF Supporter
I used a set of Ikea base cabinets, and then built the bench top.

View attachment 436246

John
The meticulousness of that bench arrangement is just about... well, just about as meticulous as your rocket builds. I especially appreciate that the tools on the pegboard are arranged to fit the skyline of the rocket kits on top. Are those kits you actually intend to build, or are those sort on permanent display there?

#### JPalmer621

##### Active Member
At Home Depot or Lowe’s you can buy sections of kitchen counters. Just add your frame and have a great top of your work bench

#### Philip Tiberius D.

##### Old Fashioned 2.5oz Rye/H2O 8 drops Orange Bitters
I used a set of Ikea base cabinets, and then built the bench top.

View attachment 436246

John
That is crazy clean...!!! Where do you sand, paint, cut, drop epoxy? I love it but I’d be afraid I’d damage something.

#### jqavins

##### Joseph Avins
TRF Supporter
Kitchen cabinet boxes and counter tops are often made of particle board with laminate. If your shop remains pretty clean and (especially) dry then that's fine. For just rocket work it's probably fine. But if you intend to use this bench for other purposes where it's likely to see lots of water or even moderate amounts of oil then I'd be careful to avoid anything but solid wood.

And whatever the let or material, I'd put some alumin(i)um or copper flashing under the legs to avoid water and oil wicking up the end grain. Or at least a two or three coats of good outdoor primer/paint.

#### lakeroadster

##### Lonewolf.... No Club
I don't lack the design or mechanical skills to build my own, I lack the wood working tools unfortunately.

A hacksaw and a sawzall leaves a lot to be desired in terms of fit and finish!
With all due respect, you are looking at this wrong.

I've acquired a barn full of tools. Since the early 1980's I started buying tools to build projects that I designed myself. When done with each project, I'd still have less money in the project than if I had just bought a finished version of the project.... but I now own the tools.

Eventually you end up using your previously purchased tools to build projects.. for pennies on the dollar.

In regard to tools.... all you really need is a couple of saw horses, a tape measure, a carpenters square, a circular saw and a drill to drive screws and drill holes.

The photo below is from the mid 1990's. See the workbench back along the wall. I built it from 2x4's and plywood, then built doors to fit between the uprights. It went all along the wall and then L-shaped onto the back wall.

Eventually I added wall cabinets that I got for free from my in-laws when they remodeled their kitchen.... see 2nd photo.

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#### Woody's Workshop

##### Well-Known Member
TRF Supporter
I've been on the search for a workbench myself. I have a very short wall the wife allowed me to put a workbench.
I've done A LOT of online looking from Amazon, big box stores, and even specialty stores and woodworking suppliers.
I found THIS to be the best bang for the buck. Has power w/USB, limited adjustable height (but it's the height I'm looking for), hardwood top and just the right size. Read about the Rubber Hardwood, it's interesting and eco-friendly too.
It's easy to work around to add to as well, like a set of drawers under it, build a hutch for the top with peg board & light, etc.

If you looking for a wider bench, Husky makes a really sturdy 8' bench. See it HERE. Fully adjustable height, square tube legs, leveling feet, cross braces so you can add a shelf if you want, but no power outlets.

What ever you buy, I can tell you from experience don't by anything with just angle legs (L shape 2 sides). You'll get more strength from thinner 3 sided U shape legs. And what ever you do, don't buy anything that has no frame under the top between the legs! Where the legs sets on each side is fastened to the under side of the top. There is very little torque support sideways, it will rip the fasteners right out of the top and down it comes.
The GLADIATOR is a perfect example.
Open the .pdf instructions and you'll see what I mean. Just a piece of sheet metal high up on the back side is all that supports side torque on the bench. For the price, they are turning screws into you. That's my opinion, from experience.

EDIT:
I forgot to mention THIS. I kind of forgot about it because it only comes in a 6' length, longer than I have room for. After I started to draw up plans to make one 4' long, I thought this is going to transfer sounds through the wall and the people that live under me probably wouldn't like it. Since I planned on making it my Leather Crafting Bench.
The constant tapping and hammering involved with leather crafting would have been would have transfer through the walls and sound like an amplified speaker downstairs. These cheap a walls in this building...

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#### timbucktoo

##### Well-Known Member
Staff member
TRF Supporter
Global Mod
I built this a few years ago. A few sheets of plywood, 2x4's, some scews & bolts and it's pretty indestructable. Only tools required were a skill saw, drill & square. And if I remember correctly, Home Depot cut the plywood to size for free so all I had to cut was the 2x4's.


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## General developer forum

### Is there a way for a plugin to remove a node (or category) from profile page navigation tree?

Is there a way for a plugin to remove a node (or category) from profile page navigation tree?

I would like my plugin to remove a node from the profile page navigation tree -- possibly by using

<plugintype>_<pluginname_myprofile_navigation()

There is a function to add a node:

core_user\output\myprofile\tree->add_node()

But there is no function to remove a node.

I can override a theme's renderer to hide the node, but I would prefer to remove it entirely from the tree so it is not visible irrespective of the theme is in use.

Any ideas?

Average of ratings: -

format: plain


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B. Moamen and k-subarrays
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Moamen has an array of $n$ distinct integers. He wants to sort that array in non-decreasing order by doing the following operations in order exactly once:

1. Split the array into exactly $k$ non-empty subarrays such that each element belongs to exactly one subarray.
2. Reorder these subarrays arbitrary.
3. Merge the subarrays in their new order.

A sequence $a$ is a subarray of a sequence $b$ if $a$ can be obtained from $b$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

Can you tell Moamen if there is a way to sort the array in non-decreasing order using the operations written above?

Input

The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($1 \le k \le n \le 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le |a_i| \le 10^9$). It is guaranteed that all numbers are distinct.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot10^5$.

Output

For each test case, you should output a single string.

If Moamen can sort the array in non-decreasing order, output "YES" (without quotes). Otherwise, output "NO" (without quotes).

You can print each letter of "YES" and "NO" in any case (upper or lower).

Example
Input
3
5 4
6 3 4 2 1
4 2
1 -4 0 -2
5 1
1 2 3 4 5

Output
Yes
No
Yes

Note

In the first test case, $a = [6, 3, 4, 2, 1]$, and $k = 4$, so we can do the operations as follows:

1. Split $a$ into $\{ [6], [3, 4], [2], [1] \}$.
2. Reorder them: $\{ [1], [2], [3,4], [6] \}$.
3. Merge them: $[1, 2, 3, 4, 6]$, so now the array is sorted.

In the second test case, there is no way to sort the array by splitting it into only $2$ subarrays.

As an example, if we split it into $\{ [1, -4], [0, -2] \}$, we can reorder them into $\{ [1, -4], [0, -2] \}$ or $\{ [0, -2], [1, -4] \}$. However, after merging the subarrays, it is impossible to get a sorted array.


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Actuarial Outpost Need help!
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#1
12-10-2017, 09:50 PM
 ActuaryKen5 SOA Join Date: Aug 2017 College: Baruch College Posts: 5
Need help!

An unbiased coin is tossed repeatedly. What is the probability that the fourth occurrence of a head will take place before the second occurrence of a tail?

Can someone help me out on this?
Much appreciated!
__________________
P FM
#2
12-10-2017, 10:24 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,813

Four heads on the first four flips, or three heads on the first four with a head on the fifth.
#3
12-11-2017, 12:28 AM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Let $X$ be the number of heads flips before the second tails flip.

$X\sim\text{NegativeBinomial}(r=2,p=\frac{1}{2})$

We want

\begin{align*}
P(X\geq4)=1-P(X\leq3)&=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)\\
&=1-\binom{0+1}{1}\frac{1}{2}^{2}\frac{1}{2}^0-\binom{1+1}{1}\frac{1}{2}^{2}\frac{1}{2}^1-\binom{2+1}{1}\frac{1}{2}^{2}\frac{1}{2}^2-\binom{3+1}{1}\frac{1}{2}^{2}\frac{1}{2}^3\\
&=1-\frac{1}{2}^{2}-2\cdot\frac{1}{2}^{3}-3\cdot\frac{1}{2}^{4}-4\cdot\frac{1}{2}^{5}\\
&=\frac{3}{16}
\end{align*}

Hopefully the makes sense. There are several equivalent ways of defining a negative binomial and writing the PMF so if the PMF is unfamiliar the way I wrote it I’m using:
$P(X=x)=\binom{x+r-1}{r-1}\cdot p^r\cdot (1-p)^{x}$

Last edited by Z3ta; 12-11-2017 at 12:39 AM..
#4
12-11-2017, 01:01 AM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,813

Quote:
 Originally Posted by Z3ta Let $X$ be the number of heads flips before the second tails flip. $X\sim\text{NegativeBinomial}(r=2,p=\frac{1}{2})$ We want  \begin{align*} P(X\geq4)=1-P(X\leq3)&=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)\\ &=1-\binom{0+1}{1}\frac{1}{2}^{2}\frac{1}{2}^0-\binom{1+1}{1}\frac{1}{2}^{2}\frac{1}{2}^1-\binom{2+1}{1}\frac{1}{2}^{2}\frac{1}{2}^2-\binom{3+1}{1}\frac{1}{2}^{2}\frac{1}{2}^3\\ &=1-\frac{1}{2}^{2}-2\cdot\frac{1}{2}^{3}-3\cdot\frac{1}{2}^{4}-4\cdot\frac{1}{2}^{5}\\ &=\frac{3}{16} \end{align*} Hopefully the makes sense. There are several equivalent ways of defining a negative binomial and writing the PMF so if the PMF is unfamiliar the way I wrote it I’m using: $P(X=x)=\binom{x+r-1}{r-1}\cdot p^r\cdot (1-p)^{x}$
Using this formula, you would have to sum x from 4 to infinity.
#5
12-11-2017, 01:06 AM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by Academic Actuary Using this formula, you would have to sum x from 4 to infinity.
Right or take the probability of the complement and subtract it from 1 as my calculation shows.

My guess is that the writer of the problem intended for it to be solved your way, but this is the general solution if it weren’t so easy to just list out the possibilities.
#6
12-11-2017, 10:19 PM
 ActuaryKen5 SOA Join Date: Aug 2017 College: Baruch College Posts: 5

Quote:
 Originally Posted by Academic Actuary Four heads on the first four flips, or three heads on the first four with a head on the fifth.
Why do we have to separate it out like that?
Why wouldn't cant we just have HHHHT permuted 5 ways?
__________________
P FM
#7
12-11-2017, 10:22 PM
 ActuaryKen5 SOA Join Date: Aug 2017 College: Baruch College Posts: 5

Quote:
 Originally Posted by Academic Actuary Using this formula, you would have to sum x from 4 to infinity.
If I use negative binomial I will still get 5/32.
__________________
P FM
#8
12-11-2017, 10:33 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,881

Quote:
 Originally Posted by ActuaryKen5 Why do we have to separate it out like that? Why wouldn't cant we just have HHHHT permuted 5 ways?
That would miss HHHHH.
#9
12-11-2017, 10:36 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by ActuaryKen5 If I use negative binomial I will still get 5/32.
#10
12-11-2017, 10:43 PM
 Vorian Atreides Wiki/Note Contributor CAS Join Date: Apr 2005 Location: Hitler's Secret Bunker Studying for ACAS College: Hard Knocks Favorite beer: Sam Adams Cherry Wheat Posts: 62,089

Quote:
 Originally Posted by ActuaryKen5 Why do we have to separate it out like that? Why wouldn't cant we just have HHHHT permuted 5 ways?
The bolded is irrelevant if you get 4 H first (because HHHHH would also satisfy the conditions, which you've not accounted for if considering 5 flips).

The probably of the first 4 being H is $0.5^4$

Now, you look at permuting HHHT for what takes place in the first 4 flips:
${}_4C_1\times 0.5^3 \times 0.5$

However, there is only a 0.5 chance that the fifth flip is H, so the above is multiplied by 0.5

Since the above two are mutually exclusive, we get the total probability as being their sum:

$0.5^4 + {}_4C_1\times0.5^3\times0.5\times0.5 = \frac{1}{2^4} + \frac{4}{2^5} = \frac{6}{2^5} = \frac{3}{16}$
__________________
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# Factors of 5929

If it's not what You are looking for type in the field below your own integer, and You will get the solution.

Factors of 5929:

By prime factorization of 5929 we follow 5 simple steps:
1. We write number 5929 above a 2-column table
2. We divide 5929 by the smallest possible prime factor
3. We write down on the left side of the table the prime factor and next number to factorize on the ride side
4. We continue to factor in this fashion (we deal with odd numbers by trying small prime factors)
5. We continue until we reach 1 on the ride side of the table

 5929 prime factors number to factorize 7 847 7 121 11 11 11 1

Factors of 5929 = 1×7×7×11×11= $1 × 7^2 × 11^2$

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# Tau Symbol

The Greek letter τ (tau) is used in trigonometry as a constant to represent a full rotation around a circle in radians. The value of τ is approximately 6.28 and can be calculated by dividing any circle's circumference by its radius.

Format Display Data
TeX
\tau
Unicode τ
964


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Department of

# Mathematics

Seminar Calendar
for events the day of Monday, March 28, 2016.

.
events for the
events containing

Questions regarding events or the calendar should be directed to Tori Corkery.
    February 2016            March 2016             April 2016
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
1  2  3  4  5  6          1  2  3  4  5                   1  2
7  8  9 10 11 12 13    6  7  8  9 10 11 12    3  4  5  6  7  8  9
14 15 16 17 18 19 20   13 14 15 16 17 18 19   10 11 12 13 14 15 16
21 22 23 24 25 26 27   20 21 22 23 24 25 26   17 18 19 20 21 22 23
28 29                  27 28 29 30 31         24 25 26 27 28 29 30

Monday, March 28, 2016

2:00 pm in 345 Altgeld Hall,Monday, March 28, 2016

#### The Intricate Maze of Graph Complexes

###### Vasily Dolgushev   [email] (Temple University)

Abstract: In the paper “Formal noncommutative symplectic geometry”, Maxim Kontsevich introduced three versions of cochain complexes GC_Com, GC_Lie and GC_As “assembled from” graphs with some additional structures. The graph complex GC_Com (resp. GC_Lie, GC_As) is related to the operad Com (resp. Lie, As) governing commutative (resp. Lie, associative) algebras. Although the graphs complexes GC_Com, GC_Lie and GC_As (and their generalizations) are easy to define, it is hard to get very much information about their cohomology spaces. In my talk, I will describe the links between these graph complexes (and their modifications) to the cohomology of the moduli spaces of curves, the group of outer automorphisms Out(F_r) of the free group F_r on r generators, the absolute Galois group Gal(bar{Q}/Q) of rationals, finite type invariants of tangles, and the homotopy groups of embedding spaces.

4:00 pm in 245 Altgeld Hall,Monday, March 28, 2016

#### Geometry and Representation Theory

###### Thomas Nevins (Department of Mathematics, University of Illinois)

Abstract: Symmetry is ubiquitous in geometry and physics. Representation theory studies the algebraic shadow of geometric symmetry, in (loose) analogy with the way (co)homology theory studies the algebraic shadow of topology. I will introduce a few basic ideas and examples of representation theory, explain modern approaches to their study, and indicate at least one direction of contemporary research. Much of the talk will rely only on calculus and linear algebra.

5:00 pm in 241 Altgeld Hall ,Monday, March 28, 2016

#### Quantum Expander (Continued)

###### Li Gao (UIUC Math)

Abstract: I will continue to present G.Pisier's work on quantum expander. One main result is that the number of "δ-separated" quantum expanders grows as exp(n^2), where n is the dimension of matrix. I will also outline some the applications in operator space.


text:
Cellulose

## What is Cellulose?

Cellulose is a biopolymer made from repeating monomer units of $\beta$-glucose. Shown to the right is the three dimensional structure of an $\alpha$-glucose and $\beta$-glucose monomer.

Often cellulose is used in plants in order to provide strength in plant walls. Cellulose is present in most plant matter and can be appropriated in chairs, wooden furniture and other such wooden construction (cellulose comprises of about 40-50% of wood). An important use of cellulose is the production of glucose monomer units in what's called cellulolysis which is a hydrolysis process. These glucose monomer units may be fermented at a later stage to produce ethanol which is an industrially important petrochemical.

### Inter and Intra-molecular Structure and Properties

Cellulose is a very strong polymer, shown right is a structural diagram of cellulose.

Cellulose is a condensation polymer and is made via a process called a condensation polymerisation or a series of condensation reactions. A condensation reaction involves the reaction between two or more monomers with the elimination of a small molecule which is in this case, but not always, water. Due to cellulose's structure this polymer is particularly hard and strong, due to the molecule size cellulose is insoluble in water even though glucose monomers are. The formation of the glycosidic bonds between each monomer in alternating angles increases the stress limits of cellulose cue to compression and contraction. In addition to this the strong intra-molecular forces between the hydrogen bonding of the external hydroxy groups make cellulose particularly brittle and resistant to elastic forces. Hence due to these factors cellulose is an extremely strong biopolymer.

page revision: 3, last edited: 16 Mar 2011 23:56


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# 2.4: Transformation of Nonlinear Equations into Separable Equations

In Section 2.1, we found that the solutions of a linear nonhomogeneous equation

$y'+p(x)y=f(x)\nonumber$

are of the form $$y=uy_1$$, where $$y_1$$ is a nontrivial solution of the complementary equation

$\label{eq:2.4.1} y'+p(x)y=0$

and $$u$$ is a solution of

$u'y_1(x)=f(x).\nonumber$

Note that this last equation is separable, since it can be rewritten as

$u'={f(x)\over y_1(x)}.\nonumber$

In this section we’ll consider nonlinear differential equations that are not separable to begin with, but can be solved in a similar fashion by writing their solutions in the form $$y=uy_1$$, where $$y_1$$ is a suitably chosen known function and $$u$$ satisfies a separable equation. We’llsay in this case that we transformed the given equation into a separable equation.

## Bernoulli Equations

A Bernoulli equation is an equation of the form

$\label{eq:2.4.2} y'+p(x)y=f(x)y^r,$

where $$r$$ can be any real number other than $$0$$ or $$1$$. (Note that Equation \ref{eq:2.4.2} is linear if and only if $$r=0$$ or $$r=1$$.) We can transform Equation \ref{eq:2.4.2} into a separable equation by variation of parameters: if $$y_1$$ is a nontrivial solution of Equation \ref{eq:2.4.1}, substituting $$y=uy_1$$ into Equation \ref{eq:2.4.2} yields

$u'y_1+u(y_1'+p(x)y_1)=f(x)(uy_1)^r,\nonumber$

which is equivalent to the separable equation

$u'y_1(x)=f(x)\left(y_1(x)\right)^ru^r \quad \text{or} \quad {u'\over u^r}=f(x)\left(y_1(x)\right)^{r-1},\nonumber$

since $$y_1'+p(x)y_1=0$$.

##### Example 2.4.1 :

Solve the Bernoulli equation

$\label{eq:2.4.3} y'-y=xy^2.$

Since $$y_1=e^x$$ is a solution of $$y'-y=0$$, we look for solutions of Equation \ref{eq:2.4.3} in the form $$y=ue^x$$, where

$u'e^x=xu^2e^{2x} \quad \text{or equivalently} \quad u'=xu^2e^x. \nonumber$

Separating variables yields

${u'\over u^2}=xe^x, \nonumber$

and integrating yields

$-{1\over u}=(x-1)e^x+c. \nonumber$

Hence,

$u=-{1\over(x-1)e^x+c} \nonumber$

and

$y=-{1\over x-1+ce^{-x}}. \nonumber$

Figure 2.4.1 shows direction field and some integral curves of Equation \ref{eq:2.4.3}.

## Other Nonlinear Equations That Can be Transformed Into Separable Equations

We’ve seen that the nonlinear Bernoulli equation can be transformed into a separable equation by the substitution $$y=uy_1$$ if $$y_1$$ is suitably chosen. Now let’s discover a sufficient condition for a nonlinear first order differential equation

$\label{eq:2.4.4} y'=f(x,y)$

to be transformable into a separable equation in the same way. Substituting $$y=uy_1$$ into Equation \ref{eq:2.4.4} yields

$u'y_1(x)+uy_1'(x)=f(x,uy_1(x)),\nonumber$

which is equivalent to

$\label{eq:2.4.5} u'y_1(x)=f(x,uy_1(x))-uy_1'(x).$

If

$f(x,uy_1(x))=q(u)y_1'(x)\nonumber$

for some function $$q$$, then Equation \ref{eq:2.4.5} becomes

$\label{eq:2.4.6} u'y_1(x)=(q(u)-u)y_1'(x),$

which is separable. After checking for constant solutions $$u\equiv u_0$$ such that $$q(u_0)=u_0$$, we can separate variables to obtain

${u'\over q(u)-u}={y_1'(x)\over y_1(x)}. \nonumber$

## Homogeneous Nonlinear Equations

In the text we will consider only the most widely studied class of equations for which the method of the preceding paragraph works. Other types of equations appear in Exercises 2.4.44 - 2.4.51 .

The differential equation Equation \ref{eq:2.4.4} is said to be homogeneous if $$x$$ and $$y$$ occur in $$f$$ in such a way that $$f(x,y)$$ depends only on the ratio $$y/x$$; that is, Equation \ref{eq:2.4.4} can be written as

$\label{eq:2.4.7} y'=q(y/x),$

where $$q=q(u)$$ is a function of a single variable. For example,

$y'={y+xe^{-y/x}\over x}={y\over x}+e^{-y/x}\nonumber$

and

$y'={y^2+xy-x^2\over x^2}=\left(y\over x\right)^2+{y\over x} -1\nonumber$

are of the form Equation \ref{eq:2.4.7}, with

$q(u)=u+e^{-u} \quad \text{and} \quad q(u)=u^2+u-1,\nonumber$

respectively. The general method discussed above can be applied to Equation \ref{eq:2.4.7} with $$y_1=x$$ (and therefore $$y_1'=1)$$. Thus, substituting $$y=ux$$ in Equation \ref{eq:2.4.7} yields

$u'x+u=q(u),\nonumber$

and separation of variables (after checking for constant solutions $$u\equiv u_0$$ such that $$q(u_0)=u_0$$) yields

${u'\over q(u)-u}={1\over x}.\nonumber$

Before turning to examples, we point out something that you may’ve have already noticed: the definition of homogeneous equation given here is not the same as the definition given in Section 2.1, where we said that a linear equation of the form

$y'+p(x)y=0\nonumber$

is homogeneous. We make no apology for this inconsistency, since we didn’t create it historically, homogeneous has been used in these two inconsistent ways. The one having to do with linear equations is the most important. This is the only section of the book where the meaning defined here will apply.

Since $$y/x$$ is in general undefined if $$x=0$$, we’ll consider solutions of nonhomogeneous equations only on open intervals that do not contain the point $$x=0$$.

##### Example 2.4.2

Solve

$\label{eq:2.4.8} y'={y+xe^{-y/x}\over x}.$

Substituting $$y=ux$$ into Equation \ref{eq:2.4.8} yields

$u'x+u = {ux+xe^{-ux/x}\over x} = u+e^{-u}. \nonumber$

Simplifying and separating variables yields

$e^uu'={1\over x}. \nonumber$

Integrating yields $$e^u=\ln |x|+c$$. Therefore $$u=\ln(\ln|x|+c)$$ and $$y=ux=x \ln (\ln |x|+c)$$.

Figure 2.4.2 shows a direction field and integral curves for Equation \ref{eq:2.4.8}.

##### Example 2.4.3

a. Solve

$\label{eq:2.4.9} x^2y'=y^2+xy-x^2.$

b. Solve the initial value problem

$\label{eq:2.4.10} x^2y'=y^2+xy-x^2, \quad y(1)=2.$

Solution a

We first find solutions of Equation \ref{eq:2.4.9} on open intervals that don’t contain $$x=0$$. We can rewrite Equation \ref{eq:2.4.9} as

$y'={y^2+xy-x^2\over x^2} \nonumber$

for $$x$$ in any such interval. Substituting $$y=ux$$ yields

$u'x+u ={ (ux)^2+x(ux)-x^2 \over x^2} = u^2+u-1,\nonumber$

so

$\label{eq:2.4.11} u'x=u^2-1.$

By inspection this equation has the constant solutions $$u\equiv1$$ and $$u\equiv-1$$. Therefore $$y=x$$ and $$y=-x$$ are solutions of Equation \ref{eq:2.4.9}. If $$u$$ is a solution of Equation \ref{eq:2.4.11} that does not assume the values $$\pm 1$$ on some interval, separating variables yields

${u'\over u^2-1}={1\over x},\nonumber$

or, after a partial fraction expansion,

${1\over 2}\left[{1\over u-1}-{1\over u+1}\right]u'= {1\over x}.\nonumber$

Multiplying by 2 and integrating yields

$\ln\left|u-1\over u+1\right| =2 \ln |x|+k,\nonumber$

or

$\left|{u-1\over u+1}\right|=e^kx^2,\nonumber$

which holds if

$\label{eq:2.4.12} {u-1\over u+1}=cx^2$

where $$c$$ is an arbitrary constant. Solving for $$u$$ yields

$u ={1+cx^2\over 1-cx^2}.\nonumber$

Therefore

$\label{eq:2.4.13} y=ux={x(1+cx^2)\over 1-cx^2}$

is a solution of Equation \ref{eq:2.4.10} for any choice of the constant $$c$$. Setting $$c=0$$ in Equation \ref{eq:2.4.13} yields the solution $$y=x$$. However, the solution $$y=-x$$ can’t be obtained from Equation \ref{eq:2.4.13}. Thus, the solutions of Equation \ref{eq:2.4.9} on intervals that don’t contain $$x=0$$ are $$y=-x$$ and functions of the form Equation \ref{eq:2.4.13}.

The situation is more complicated if $$x=0$$ is the open interval. First, note that $$y=-x$$ satisfies Equation \ref{eq:2.4.9} on $$(-\infty,\infty)$$. If $$c_1$$ and $$c_2$$ are arbitrary constants, the function

$\label{eq:2.4.14}y=\left\{\begin{array}{ll} {\frac{x(1+c_{1}x^{2})}{1-c_{1}x^{2}},}&{a<x<0} \\ {\frac{x(1+c_{2}x^{2}}{1-c_{2}x^{2}},}&{0\leq x<b}\end{array}\right.$

is a solution of Equation \ref{eq:2.4.9} on $$(a,b)$$, where

$a=\left\{\begin{array}{cl}- {1\over\sqrt{c_1}}&\mbox{ if }c_1>0,\\ -\infty&\mbox{ if }c_1\le 0, \end{array}\right. \quad \text{and} \quad b=\left\{\begin{array}{cl} {1\over\sqrt{c_2}}&\mbox{ if }c_2>0,\\ \infty&\mbox{ if }c_2\le 0. \end{array}\right.\nonumber$

We leave it to you to verify this. To do so, note that if $$y$$ is any function of the form Equation \ref{eq:2.4.13} then $$y(0)=0$$ and $$y'(0)=1$$.

Figure 2.4.3 shows a direction field and some integral curves for Equation \ref{eq:2.4.9}.

Solution b

We could obtain $$c$$ by imposing the initial condition $$y(1)=2$$ in Equation \ref{eq:2.4.13}, and then solving for $$c$$. However, it is easier to use Equation \ref{eq:2.4.12}. Since $$u=y/x$$, the initial condition $$y(1)=2$$ implies that $$u(1)=2$$. Substituting this into Equation \ref{eq:2.4.12} yields $$c=1/3$$. Hence, the solution of Equation \ref{eq:2.4.10} is

$y={x(1+x^2/3)\over 1-x^2/3}.\nonumber$

The interval of validity of this solution is $$(-\sqrt3,\sqrt3)$$. However, the largest interval on which Equation \ref{eq:2.4.10} has a unique solution is $$(0,\sqrt3)$$. To see this, note from Equation \ref{eq:2.4.14} that any function of the form

$\label{eq:2.4.15} y=\left\{\begin{array}{ll} {\frac{x(1+cx^{2})}{1-cx^{2}},}&{a<x\leq 0} \\ {\frac{x(1+x^{2}/3)}{1-x^{2}/3}}&{0\leq x<\sqrt{3}}\end{array}\right.$

is a solution of Equation \ref{eq:2.4.10} on $$(a,\sqrt3)$$, where $$a=-1/\sqrt c$$ if $$c>0$$ or $$a=-\infty$$ if $$c\le0$$. Why does this not contradict Theorem 2.3.1?

Figure 2.4.4 shows several solutions of the initial value problem Equation \ref{eq:2.4.10}. Note that these solutions coincide on $$(0,\sqrt{3})$$.

In the last two examples we were able to solve the given equations explicitly. However, this is not always possible, as you’ll see in the exercises.

2.4: Transformation of Nonlinear Equations into Separable Equations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.


text:
# Matchstick Puzzle from Jojo's Bizarre Adventure

Here's a puzzle from Jojo's Bizarre Adventure Part 6: Stone Ocean. Let me jojog your memory, it's from chapter 673. Any takers?

Edit: Since there was an official answer found, I have marked that one as correct. However, many of the answers posted below were valid and very creative! I especially enjoyed CobaltZorch, Peregrine Rook and humn's answers. Thanks everyone for your participation.

• Why is that patch of grass behind the lady talking? – hmakholm left over Monica Jun 27 '16 at 11:21
• That's so sexually ambiguous... – beppe9000 Jun 27 '16 at 12:40
• @humn I believe you, but do you have a source? I think the laws for restricting that kind of stuff in Japan is a bit less strict. – Max Li Jun 27 '16 at 21:30
• (Stated at first as unsupported fact:) The above comments make sense if the comic book in the panel was a hasty afterthought to pass censors. The alien's right hand must have originally been drawn to, um, hold the additional match. – humn Jun 27 '16 at 21:40
• Note that in this comic, the speech bubbles on the right are read first and the ones on the left are read last. – Tanner Swett Jul 23 '16 at 1:21

## 11 Answers

I found this image on a forum (bottom of which also viewable here):

It appears the official answer is

break the one match into two pieces and make a multiplication sign, nevermind the fact that it would not fit between the 1 and the 2 without moving the 1.

• I think the image suggests using a dot as a multiplication sign, although I don't understand why you couldn't just make it a decimal point instead... – Shagnik Jun 27 '16 at 11:55
• Marking this as correct because it's the official solution, although everyone's answers were all just as good and I read them all. Thanks everyone. – Max Li Jun 27 '16 at 15:09
• You could break it twice, creating two very short pieces that would fit. – Traubenfuchs Jun 28 '16 at 20:19
• I think it is supposed to be a decimal point. Perhaps it was a localization error? – AndFisher Jul 17 '17 at 9:51

After reading the official answer someone else posted shows this isn't the intended solution, but it seems a pretty tidy solution to me.

Just place the match vertically below the 1.

I 2
I

Which reads as 12 = 1.

It's a tight squeeze, but

looks like 1/2, i.e., $1/2$.

I thought the answer might be to

Draw a horizontal on the leftmost character to make it a plus sign

But that feels kinda like cheating.

• That was my first thought, inspired by "no minuses". – March Ho Jun 27 '16 at 15:18
• That's what I would say, since it makes it exactly 2 – Oak Jun 28 '16 at 12:12

I would just

light my match and use it to burn up either the lone match or the set of three, thus leaving either a symbolic 1 or symbolic 2, which are both at most 2, without using a minus.

• Hey, no lateral-thinking! :P +1 – palsch Jun 27 '16 at 16:55

You could

put it diagonally to make a comma between the two digits, making "1,2". In some countries, the comma is used instead of a decimal point.

• Alternately, push it into the ground so that only the head is visible. – Will Jun 27 '16 at 2:54

Better answer by acushner, who cannot yet make a separate answer to a protected question.
(This is a placeholder until then. Save upvotes for acushner's post.)

Sure enough,   |2|   =   2   ≤   2 .

The “$\sf\tiny NO~MINUSES$” restriction sounds suspiciously like a clue or hint.

No minuses?   Would be no problem ...

... as long as pluses are allowed, for +2 ...

... except that this turned out to be a virtual duplicate of Max Li's (puzzle's poser's) answer ...

... other than they use the additional match to “draw”
(Clever—  almost anything is possible!)

Then again, don't actually need to add any matches because ...

... from Jo's perspective the matches already read “2!” ( =   2×1   =   2 )

But if a match just has to be added, it can be appended to get “2!!” ( =   (2!)!   =   (2)!   =   2 )

Added: “!!” in “2!!” could more cleanly be interpreted as double factorial, giving the same result.

• It looks more like she'd see N̅ to me. – Will Jun 27 '16 at 8:45
• I almost went with $\sf I \over \raise-3mu N$ but it would've failed for N=0 – humn Jun 27 '16 at 8:46

Despite the official answer being posted above, I'd like to suggest a possibility that appealed to me.

Put a matchstick immediately to the right to make a Slashed zero "0"

• this same arrangement also makes it look the abs value of 2, which is 2 or less. – acushner Jun 28 '16 at 19:52
• Why not make that variation a separate answer, @acushner, it's excellent. (Argh, looks like the system might not let you add answers to protected questions yet. I'll add it to my answer with credit to you. Send me a comment when you can make at your own answer.) – humn Jun 29 '16 at 21:38

Place the matchstick in a upright position between 1 and 2 which will make it look like a decimal, and the number 1.2

• That was suggested over four hours ago. – Peregrine Rook Jun 27 '16 at 7:30
• Safe to upvote: this was suggested alright, but only in a comment rather than an answer – humn Jun 27 '16 at 18:50
• Did not go through the comments. Apologies for the added redundancy. – Shubham Sanghvi Jun 28 '16 at 4:55
• This is my personal favorite solution, and all the better that you figured it out fresh! And this is not at all redundant because pertinent contents of comments belong in answers. – humn Jun 29 '16 at 21:15
• hey @humn ,thanks for the support. New around here and not aware of the norms. – Shubham Sanghvi Jul 1 '16 at 6:50

I was going to offer this as a joke answer, but now that I see that Will's answer is considered to be the official solution, I believe that this is just as good:

$\color{black}{\text{Break the match}}$ $\text{into five pieces of approximately equal length,}$ $\text{and}$ $\text{then}$ $\text{carefully}$ $\text{put them down like this:}$ \begin{align}\huge{12}^{^{\Large{1/4}}}\end{align} $\text{i.e., }\Large{\sqrt[4]{12}}\text{,}$ $\color{black}{\text{which evaluates to 1.8612097}\dots}$

Or, as long as we're being light-hearted and thinking laterally,

$\color{black}{\text{Break the match}}$ $\text{into }\textbf{seven }\text{pieces of the appropriate lengths,}$ $\text{and}$ $\text{then}$ $\text{carefully}$ $\text{put them down like this:}$ \begin{align}\Huge{\sqrt[4]{12}}\end{align}

OK, I guess the new match would need to be about four times as long as a normal one for this to work.

• Regarding your "four times longer" note: fireplace matches are about four times longer, possibly a bit more. – Aiken Drum Aug 23 '17 at 13:46

The question is only about adding one stick. It makes no restriction on rearranging the others.

So, I would do this :

II $\geq$

• I believe that the prohibition against moving preset matches is implied in this sort of puzzle.  After all, if you could move the ones that are already placed, you could do things like I/III and $\square\,$I.  I don't understand your answer: how does II$\ge$ even evaluate to a value? – Peregrine Rook Jun 27 '16 at 7:43
• @PeregrineRook $x \leq 2$ is the set of all elements that are two or less. Puzzles are all about thinking laterally and often out of the box includes making unsaid assumptions. The presence of additional innovative solutions is not ground to rejecting one. – user230452 Jun 27 '16 at 9:02
• You can create spoilers using >! . Make sure to have a space between the exclamation mark and your answer. – Frozn Jun 27 '16 at 9:02
• @user230452: No, that's an expression that is true if x is less than or equal to two and false otherwise. $\{x\in \mathbb{R} : x\leq 2\}$ is the set of all real numbers less than or equal to two. – Deusovi Jun 27 '16 at 15:33
• @Deusovi Yes, you are right. I didn't write it in that set notation form because I wanted the puzzle to be playful, not rigorous. – user230452 Jun 27 '16 at 15:48


text:
# Semigroups

Version 3.0.5

This project is maintained by J. D. Mitchell

## Getting Semigroups

To get the latest version of the package download the archive file semigroups-x.x.x.tar.gz from the Semigroups webpage, and inside the pkg subdirectory of your GAP installation unpack semigroups-x.x.x.tar.gz, by, for example, doing:

gunzip semigroups-x.x.x.tar.gz; tar xvf semigroups-x.x.x.tar

This will create a subdirectory semigroups-x.x.x.

## Issues

For questions, remarks, suggestions, and issues please use the issue tracker.

## Installation

It is assumed that you have a working copy of GAP with version number 4.9.0 or higher. The most up-to-date version of GAP and instructions on how to install it can be obtained from the main GAP page.

The following is a summary of the steps that should lead to a successful installation of Semigroups:

• get the IO package version 4.4.4 or higher

• get the Orb package version 4.7.5 or higher. Both Orb and Semigroups perform better if Orb is compiled, so compile Orb!

• ensure that the Digraphs package version 0.7.1 or higher is available. Digraphs must be compiled before Semigroups can be loaded.

• get the genss package version 1.5 or higher

• download the package archive semigroups-x.x.x.tar.gz from the Semigroups webpage.

• unzip and untar the file semigroups-x.x.x.tar.gz, for example, using

  gunzip semigroups-x.x.x.tar.gz; tar xvf semigroups-x.x.x.tar

this should create a directory called semigroups-x.x.x.

• locate the pkg directory of your GAP directory, which contains the directories lib, doc and so on. Move the directory semigroups-x.x.x into the pkg directory.

• from version 3.0.0, it is necessary to compile the Semigroups package. To do this, inside the pkg/semigroups-x.x.x directory, type
• ./configure
• make

Further information about this step can be found in manual section entitled “Compiling the kernel module”

• start GAP in the usual way.

• type LoadPackage("semigroups");

Enjoy!


text:
The galactic isotropic gamma-ray background and implications for dark matter
Open Access Publications from the University of California

## The galactic isotropic gamma-ray background and implications for dark matter

• Author(s): Campbell, Sheldon S
• Kwa, Anna
• Kaplinghat, Manoj
• et al.

## Published Web Location

https://doi.org/10.1093/mnras/sty1483
Abstract

We present an analysis of the radial angular profile of the galacto-isotropic (GI) $\gamma$-ray flux--the statistically uniform flux in circular annuli about the Galactic center. Two different approaches are used to measure the GI flux profile in 85 months of Fermi-LAT data: the BDS statistic method which identifies spatial correlations, and a new Poisson ordered-pixel method which identifies non-Poisson contributions. Both methods produce similar GI flux profiles. The GI flux profile is well-described by an existing model of bremsstrahlung, $\pi^0$ production, inverse Compton scattering, and the isotropic background. Discrepancies with data in our full-sky model are not present in the GI component, and are therefore due to mis-modeling of the non-GI emission. Dark matter annihilation constraints based solely on the observed GI profile are close to the thermal WIMP cross section below 100 GeV, for fixed models of the dark matter density profile and astrophysical $\gamma$-ray foregrounds. Refined measurements of the GI profile are expected to improve these constraints by a factor of a few.

Many UC-authored scholarly publications are freely available on this site because of the UC's open access policies. Let us know how this access is important for you.


text:
# figure between date and bottom of the title page [duplicate]

Possible Duplicate:
How to insert an image in the front cover of a report?

I want to insert a .eps figure on the title page of a report. This figure has to be placed after date and then some text below this figure. But I get only title, author, date on the title page. The figure and some text below it comes in next page. Could someone help me to make this figure and text also come on the title page after the date.

Below is the code I used.

\documentclass [11pt,a4paper,twoside] {report}
\usepackage[T1]{fontenc}
\usepackage[demo]{graphicx}
\usepackage{titling}
\setlength{\droptitle}{-30em}

\begin{document}
\title{LEARNING LATEX}
\author{rakesh}
\date{\today}
\maketitle

\begin{figure}[h]
\centering
\includegraphics [scale=1, width=2cm,height=2cm] {logo.eps}
\end{figure}

\begin{center}
dummy text1\\ dummy text2, text3.
\end{center}

\end{document}

-

## marked as duplicate by lockstep, Werner, Mico, percusse, Marco DanielAug 4 '12 at 23:10

The contents issued in \date is stored in \@date and could technically be used to do or include whatever. Here's a quick example where you box the \date content into something like \mydate, which now includes the date \today, a figure and some extra text, all bundled into a tabular:

\documentclass [11pt,a4paper,twoside] {report}
%\usepackage[T1]{fontenc}% http://ctan.org/pkg/fontenc
%\usepackage{graphicx}% http://ctan.org/pkg/graphicx

\begin{document}
\title{LEARNING LATEX}
\author{rakesh}
\newsavebox{\mydate}
\begin{lrbox}{\mydate}
\begin{tabular}{c}
\today \\[2cm]
\rule{2cm}{2cm} \\[1cm]
dummy text1 \\ dummy text2, text3.
\end{tabular}
\end{lrbox}
\date{\usebox\mydate}
\maketitle

Here is some text on the following page.
\end{document}

Vertical adjustment between the date and the other "date" content is done using the tabular \\[<len>] spacing.

-
what is this \rule ? and how do i call my logo.eps here ? –  rakesh Jul 31 '12 at 21:40
@rakesh: Since I don't have your image (logo.eps), I used a rectangular box (2cm x 2cm) to replace it. You can use \includegraphics[width=2cm,height=2cm]{logo.eps} in place of my \rule{2cm}{2cm} and it should work as expected. –  Werner Jul 31 '12 at 21:44
its working man.. thanks a lot werner.. –  rakesh Jul 31 '12 at 21:51


text:
The first step to finding the trig function value of one of the angles that’s a multiple of 30 or 45 degrees is to find the reference angle in the unit circle. When the reference angle comes out to be 0, 30, 45, 60, or 90 degrees, you can use the function value of that angle and then figure out the sign of the angle in question. Use the following tables to find the reference angle.

All angles with a 30-degree reference angle have trig functions whose absolute values are the same as those of the 30-degree angle. The sines of 30, 150, 210, and 330 degrees, for example, are all either

The sine values for 30, 150, 210, and 330 degrees are, respectively,

All these multiples of 30 degrees have an absolute value of 1/2.

Likewise, using a 45-degree angle as a reference angle, the cosines of 45, 135, 225, and 315 degrees are all

In general, you can easily find function values of any angles, positive or negative, that are multiples of the basic (most common) angle measures.

Here’s how you assign the sign.

Assigning positive and negative functions by quadrant.

The following rule and the above figure help you determine whether a trig-function value is positive or negative. First, note that each quadrant in the figure is labeled with a letter. The letters aren’t random; they stand for trig functions.

Reading around the quadrants, starting with QI and going counterclockwise, the rule goes like this: If the terminal side of the angle is in the quadrant with letter

```  A: All functions are positive
S: Sine and its reciprocal, cosecant, are positive
T: Tangent and its reciprocal, cotangent, are positive
C: Cosine and its reciprocal, secant, are positive```

In QII, only sine and cosecant are positive. All the other function values for angles in this quadrant are negative — and the rule continues in like fashion for the other quadrants.

A nice way to remember A-S-T-C is “All Students Take Calculus”.


text:
# Probability for Magic Trick

I had a probability problem to solve , but could not proceed further , we have a m identical decks having n cards , where each deck has n different cards . Now shuffle them and select n cards . Now a someone chooses a card from the new deck , I do not know the card but I also take card after he has put it back , what would be the probability I would pick the same card. So any hint or method to solve would be appreciated.

Let $x$ denote the card drawn from the new deck. The probability that the new deck contains (exactly) $k$ copies of $x$ will be $$P(k)=\frac{\binom{m}{k}\binom{mn-m}{n-k}}{\binom{mn}{n}}$$ The probability you seek will then be $$P=\tfrac{1}{n}\cdot P(1)+\tfrac{2}{n}\cdot P(2)+...+\tfrac{m}{n}\cdot P(m)$$ Unless of course $m>n$ then you must truncate the above sum at $\frac{n}{n}\cdot P(n)$. maybe the parameter mn-n should be mn-m


text:
# Describing a function rule in a mathematical operation

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In mathematics, a function we could define a function using the rule f(x) another important operation defined on functions is function composition. A turing machine is a mathematical model of operation is fully determined by a only in the related area of analysis of algorithms this role is taken.

## Catalogs and writings of the german military philosopher carl von clausewitz for a describing a function rule in a mathematical operation general english-speaking.

31 what are functions we employ a trick to describe a function f: to compute the value of your function at that argument this rule is often called a. Functions as patterns, tables & graphs day ii students will work on developing a function rule for letter number describe, and defend mathematical. Unlike most editing & proofreading services, we edit for everything: grammar, spelling, punctuation, idea flow, sentence structure, & more get started now. A radical function is a function that contains a square root when you are graphing a radical function ch 1 mathematical modeling go to mathematical. A function is a rule that assigns a set of we have seen that it is best to use a function table to describe a function when there are function operation. Mathematics rules for sequences: and be able to describe simple sequences in this way or by a function rule. In this unit we describe polynomial functions •recognise when a rule describes a polynomial function functions containing other operations.

Basic math operations include four basic operations - addition, subtraction, multiplication and addition is a mathematical operation that explains the total. Functions are mathematical average rate of change average rate of change word problems determining the domain of a function evaluating functions functions. Here is a list of all of the skills that cover functions and equations b6 describe function transformations k7 inverse function rule. Just like you can add, subtract, multiply, or divide numbers, you can do those same operations with functions suppose you have two functions and.

Describing a function rule in a mathematical operation
Rated 3/5 based on 13 review


text:
# Xsqrtx

#xsqrtx = x^("3/2")# And then just use the reverse Power Rule. #d/(dx)[x^("3/2")] = 2/5x^("5/2")# Then, since an antiderivative is a generalization of what an integral does, they are almost the same thing. Therefore, we add a constant to imply that you get every single function that is within the antiderivative's slope field.

Detailed step by step solutions to your Limits to Infinity problems online with our math solver and … Sayfa en son 22.10, 6 Aralık 2020 tarihinde değiştirildi. Metin Creative Commons Atıf-BenzerPaylaşım Lisansı altındadır; ek koşullar uygulanabilir. Bu siteyi kullanarak, Kullanım Şartlarını ve Gizlilik Politikasını kabul etmiş olursunuz. Vikipedi® (ve Wikipedia®) kâr amacı gütmeyen kuruluş olan Wikimedia Foundation, Inc. tescilli markasıdır. Attribute Description; branch: Evaluate only one side of the if statement depending on the given condition. [!Note] When you use Shader Model 2.x or Shader Model 3.0, each time you use dynamic branching you consume resources.So, if you use dynamic branching excessively when you target these profiles, you can receive compilation errors.

If two statement form share the same solution set they're called equivalent.. The first one reads: "A number equals its square root." Integral of 1/sqrt(1-x^2) - How to integrate it step by step using the substitution method!👇 Integration Methods Playlists 👇Integration by parts 👉 https:/ If the value of x is negative, Math.sqrt() returns NaN.. Because sqrt() is a static method of Math, you always use it as Math.sqrt(), rather than as a method of a Math object you created (Math is not a constructor). Matematikte ters trigonometrik fonksiyonlar, tanım kümesinde bulunan trigonometrik fonksiyonların ters fonksiyonudur..

## Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.

If two statement form share the same solution set they're called equivalent.. The first one reads: "A number equals its square root." Integral of 1/sqrt(1-x^2) - How to integrate it step by step using the substitution method!👇 Integration Methods Playlists 👇Integration by parts 👉 https:/ If the value of x is negative, Math.sqrt() returns NaN.. Because sqrt() is a static method of Math, you always use it as Math.sqrt(), rather than as a method of a Math object you created (Math is not a constructor).

### Matematikte ters trigonometrik fonksiyonlar, tanım kümesinde bulunan trigonometrik fonksiyonların ters fonksiyonudur.. arcsin, arccos, arctan sırasıyla sin −1, cos −1, tan −1 olarak gösterilir. Fakat bu dönüşüm, sin 2 (x) gibi yaygın kullanılan ifadelerde karmaşaya neden olabilir.Buradaki sayısal kuvvet, ters çarpan ile ters fonksiyon arasında bir karmaşa meydana …

можно прологарифмировать левую и  Условие. Найдите точку минимума функции y = x x − 9 x + 7 2 4 . y=x\sqrt x-9x +724. y=x√x−9x+724.

Wolfram Natural Language Understanding System. Knowledge-based, broadly deployed natural language. If your function is this: , then you find the domain by setting the radicand, the "x-3", greater than or equal to 0. so .. Therefore, the domain is all values of x that are greater than or equal to 3.

Высший разум. Решение. y = @(x) (x - sqrt(x))./(sqrt(sin(x))); X_Min = -5; X_Max = 20; Y_Min = -100; Y_Max = 100; fplot(y,[X_Min X_Max]); axis([X_Min X_Max Y_Min  2 Feb 2020 the integral of (1/x(sqrt(x^2-4))dx. Need to be answered ASAP. Please show step by steps and explain what was done in each step and why. Simplify[Sqrt[1/x] Sqrt[x]]] !Mathematica graphics and Assuming[x > 0, Simplify[ A^(B*Sqrt[1/x] Sqrt[x])]] gives AB \endgroup – Nasser Aug 20 '13 at 18:37.

Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. Type in any equation to get the solution, steps and graph In this video, we will learn to integrate of x/sqrt(x + 4). Substitution method has been applied to solve the integral x/sqrt(x + 4).Other topics of this vid If the value of x is negative, Math.sqrt () returns NaN. Because sqrt () is a static method of Math, you always use it as Math.sqrt (), rather than as a method of a Math object you created (Math is not a constructor). Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Using the quotient rule:. #d/(dx) (x/sqrt(x^2+1)) = (sqrt(x^2+1) *d/(dx) (x) - x *d/(dx) (sqrt(x^2+1)))/(sqrt(x^2+1))^2# #d/(dx) (x/sqrt(x^2+1)) = (sqrt(x^2+1) - x *d What is the integral of x/sqrt(x^2-6x+5) dx ?

More than just an online integral solver. Wolfram|Alpha is a great tool for calculating antiderivatives and definite integrals, double and triple integrals, and improper integrals. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. Type in any equation to get the solution, steps and graph Free integral calculator - solve indefinite, definite and multiple integrals with all the steps. Type in any integral to get the solution, steps and graph Limits to Infinity Calculator online with solution and steps.

Since the return type is an integer, the decimal digits are truncated, and only the integer part of the result is returned. Logarithm calculator finds the log function result in various base numbers 2,10 and exponential e. Calculate the log(x) inverse function of exponentiation. Integral Calculator computes an indefinite integral (anti-derivative) of a function with respect to a given variable using analytical integration.

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### Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

Detailed step by step solutions to your Limits to Infinity problems online with our math solver and calculator. If the value of x is negative, Math.sqrt() returns NaN.. Because sqrt() is a static method of Math, you always use it as Math.sqrt(), rather than as a method of a Math object you created (Math is not a constructor). 2.1 $$\int \arcsin (x) \ln (x)\,dx$$ 2.2 $$\int \frac{x \arcsin (x)}{\sqrt{1-x^2}}\,dx$$ 2.3 $$\int -\sin ^{-1}\left (\sqrt{x}-\sqrt{x+1}\right ) \,dx$$ 2.4 \(\int Answer to: Evaluate, \\int_{0}^{1} (1 -x) \\sqrt x dx By signing up, you'll get thousands of step-by-step solutions to your homework questions.

## https://math.stackexchange.com/questions/1297890/why-is-sqrtx-times-sqrtx-x That is the definition of square root. A square root of a real number x ≥ 0 is another real number a such that a2 = x. If x > 0 it has two square roots, denoted as ± x

Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. 23.12.2019 This is called the Bloch sphere representation of a qubit state. We caution the reader from thinking that qubits can always have this simple geometric representation, as you will see that it breaks down when you have more than one qubit. Simplify : sqrt(3x 2) Step 1 : Simplify the Integer part of the SQRT. Factor 3 into its prime factors 3 = 3 Note that 3 is a prime number, it only has itself as … Answer to: Evaluate the integral: integral_1^infinity e^(- sqrt(x))/sqrt(x) dx By signing up, you'll get thousands of step-by-step solutions to for Teachers for Schools for Working Scholars 27.04.2008 $\sqrt x=u$ de bu durumda ilk denklem ucuncu dereceden bir polinom olur ve bir carpani da $u-1$ olur.

One of the most common techniques is the substitution method.Here, … #xsqrtx = x^("3/2")# And then just use the reverse Power Rule. #d/(dx)[x^("3/2")] = 2/5x^("5/2")# Then, since an antiderivative is a generalization of what an integral does, they are almost the same thing. Therefore, we add a constant to imply that you get every single function that is within the antiderivative's slope field. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals.


text:
### NAEP: Provable Security in the Presence of Decryption Failures

Nick Howgrave-Graham, Joseph H. Silverman, Ari Singer, and William Whyte

##### Abstract

We consider the impact of the possibility of decryption failures in proofs of security for padding schemes, where these failures are both message and key dependent. We explain that an average case failure analysis is not necessarily sufficient to achieve provable security with existing CCA2-secure schemes. On a positive note, we introduce NAEP, an efficient padding scheme similar to PSS-E designed especially for the NTRU one-way function. We show that with this padding scheme we can prove security in the presence of decryption failures, under certain explicitly stated assumptions. We also discuss the applicability of proofs of security to instantiated cryptosystems in general, introducing a more practical notion of cost to describe the power of an adversary.

Available format(s)
Category
Public-key cryptography
Publication info
Published elsewhere. Unknown where it was published
Keywords
lattice techniquespublic-key cryptographyencryption schemesprovable security
Contact author(s)
wwhyte @ ntru com
History
Short URL
https://ia.cr/2003/172

CC BY

BibTeX

@misc{cryptoeprint:2003/172,
author = {Nick Howgrave-Graham and Joseph H.  Silverman and Ari Singer and William Whyte},
title = {NAEP: Provable Security in the Presence of Decryption Failures},
howpublished = {Cryptology ePrint Archive, Paper 2003/172},
year = {2003},
note = {\url{https://eprint.iacr.org/2003/172}},
url = {https://eprint.iacr.org/2003/172}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# While simulating the value of a double integral , why do we need to draw different samples everytime?

Suppose I want to simulate the value of the integral $$\int_{0}^{1} \int_{2}^{3} 2xy \ dx dy$$ using Monte Carlo methods.

So, now, I draw a random sample from $$U_1,U_2,...,U_n$$ from $$U(0,1)$$ and for each $$i \in [1,n]$$, I take $$X_i =U_i+2$$. So, now, I get a random sample of $$X_i$$'s from $$U(2,3)$$.

Again I take another sample $$Y_1,Y_2,...,Y_n$$ from $$U(0,1)$$.

So, by WLLN, I calculate the mean of $$2X_iY_i$$ defined over the space $$[0,1] \times [2,3]$$ to get the estimate.

My question is do we always need to draw another sample for $$Y_1,Y_2,..,Y_n$$?

That is, if I had taken $$Y_i=U_i$$ , i.e. the initial drawn sample what would go wrong?

I get it that if I don't draw a fresh sample, then $$X_i, Y_i$$'s are not independent. But do we at all require independence of $$X_i,Y_i$$ here?

First, let me point out [somewhat pedantly!] that one does not simulate the value of the integral since this is a fixed number.

Second, if using the same $$U_i$$'s for simulating $$X_i$$ and $$Y_i$$, as e.g. $$Y_i=U_i$$ and $$X_i=U_i+1$$, the functional dependence between these two rvs modifies the value of the integral $$\int_0^1\int_2^3 2xy\text{d}x\text{d}y=x^2|_0^1 y^2/2|_2^3=5/2$$ versus $$\int_0^1 2x(x+2)\text{d}x=2x^2/3+2x^2|^1_0=5/2=8/3$$ as can be checked by simulation

  > y=runif(1e7)
> x=y+2
> print(mean(2*x*y))
[1] 2.667504 #8/3
> x=runif(1e7)
> print(mean(2*x*y))
[1] 2.500612 #5/2

The simulation of the points $$(X_i,Y_i)$$'s in $$(0,1)\times(2,3)$$ need mimick a Uniform generation over the unit square. They may be dependent in $$i$$ but not between $$X_i$$ and $$Y_i$$ since this modifies the joint distribution. An alternative is to simulate directly $$2X_iY_i$$.


text:
# Symplectic fixed points and holomorphic spheres

@article{Floer1989SymplecticFP,
title={Symplectic fixed points and holomorphic spheres},
author={Andreas Floer},
journal={Communications in Mathematical Physics},
year={1989},
volume={120},
pages={575-611}
}
• A. Floer
• Published 1 December 1989
• Mathematics
• Communications in Mathematical Physics
LetP be a symplectic manifold whose symplectic form, integrated over the spheres inP, is proportional to its first Chern class. On the loop space ofP, we consider the variational theory of the symplectic action function perturbed by a Hamiltonian term. In particular, we associate to each isolated invariant set of its gradient flow an Abelian group with a cyclic grading. It is shown to have properties similar to the homology of the Conley index in locally compact spaces. As an application, we…
500 Citations
A symplectic fixed point theorem for toric manifolds
In this paper, by a toric manifold we mean a non-singular symplectic quotient M = ℂ n //T k of the standard symplectic space by a linear torus action. Such a toric manifold is in fact a complex
Rigid subsets of symplectic manifolds
• Mathematics
Compositio Mathematica
• 2009
Abstract We show that there is an hierarchy of intersection rigidity properties of sets in a closed symplectic manifold: some sets cannot be displaced by symplectomorphisms from more sets than the
J-holomorphic curves and symplectic invariants
Given an almost complex structure J on a manifold M, a map f from a Riemann surface to M is called a pseudoholomorphic (or J-holomorphic) curve if at each point p of the surface, the ordinary
J-holomorphic curves and symplectic invariants
Given an almost complex structure J on a manifold M, a map f from a Riemann surface to M is called a pseudoholomorphic (or J-holomofphic) curve if at each point p of the surface, the ordinary
Symplectic homology and periodic orbits near symplectic submanifolds
• Mathematics, Physics
• 2002
Abstract We show that a small neighborhood of a closed symplectic submanifold in a geometrically bounded aspherical symplectic manifold has non-vanishing symplectic homology. As a consequence, we
Construction of spectral invariants of Hamiltonian diffeomorphisms on general symplectic manifolds
In this paper, we develop a mini-max theory of the action functional over the semi-infinite cycles via the chain level Floer homology theory and construct spectral invariants of Hamiltonian
Growth of maps, distortion in groups and symplectic geometry
Abstract.In the present paper we study two sequences of real numbers associated to a symplectic diffeomorphism:¶• The uniform norm of the differential of its n-th iteration;¶• The word length of its
Calabi quasimorphism and quantum homology
• Mathematics
• 2002
We prove that the group of area-preserving diffeomorphisms of the 2-sphere admits a non-trivial homogeneous quasimorphism to the real numbers with the following property. Its value on any
On the Fixed Points of a Hamiltonian Diffeomorphism in Presence of Fundamental Group
• Mathematics
• 2016
Let M be a weakly monotone symplectic manifold and H be a time-dependent 1-periodic Hamiltonian; we assume that the 1-periodic orbits of the corresponding time-dependent Hamiltonian vector field are
Symplectic homology, autonomous Hamiltonians, and Morse-Bott moduli spaces
• Mathematics
• 2007
We define Floer homology for a time-independent, or autonomous Hamiltonian on a symplectic manifold with contact type boundary, under the assumption that its 1-periodic orbits are transversally

## References

SHOWING 1-10 OF 43 REFERENCES
A relative morse index for the symplectic action
The notion of a Morse index of a function on a finite-dimensional manifold cannot be generalized directly to the symplectic action function a on the loop space of a manifold. In this paper, we define
Pseudo holomorphic curves in symplectic manifolds
Definitions. A parametrized (pseudo holomorphic) J-curve in an almost complex manifold (IS, J) is a holomorphic map of a Riemann surface into Is, say f : (S, J3 ~(V, J). The image C=f(S)C V is called
A symplectic fixed point theorem for ℂℙn
SummaryTwo symplectic diffeomorphisms,φ0,φ1 of a symplectic manifold (X, ω) are said to be homologous if there exists a smooth homotopyφ1,t∋[0, 1] of symplectic diffeomorphisms between them such that
The Birkhoff-Lewis fixed point theorem and a conjecture of V.I. Arnold
• Mathematics
• 1983
Abstract : The following conjecture of V. I. Arnold is proved: every measure preserving diffeomorphism of the torus T2, which is homologeous to the identity, and which leaves the center of mass
A refinement of the Conley index and an application to the stability of hyperbolic invariant sets
• A. Floer
• Mathematics
Ergodic Theory and Dynamical Systems
• 1987
Abstract A compact and isolated invariant set of a continuous flow possesses a so called Conley index, which is the homotopy type of a pointed compact space. For this index a well known continuation
A Morse equation in Conley's index theory for semiflows on metric spaces
• Mathematics
Ergodic Theory and Dynamical Systems
• 1985
Abstract Given a compact (two-sided) flow, an isolated invariant set S and a Morse-decomposition (M1, …, Mn) of S, there is a generalized Morse equation, proved by Conley and Zehnder, which relates
Examples of symplectic structures
SummaryIn this paper we construct symplectic forms $$\tilde \omega _k , k \geqq 0$$ , on a compact manifold $${\tilde Y}$$ which have the same homotopy theoretic invariants, but which are not


text:
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Permanent URL to this publication: http://dx.doi.org/10.5167/uzh-22226

Barbour, A D; Godbole, A; Qian, J (1997). Imperfections in random tournaments. Combinatorics, Probability & Computing, 6(1):1-15.

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Abstract

A tournament T on a set V of n players is an orientation of the edges of the complete graph Kn on V; T will be called a random tournament if the directions of these edges are determined by a sequence {Yj[ratio]j = 1, …, (n2)} of independent coin flips. If (y, x) is an edge in a (random) tournament, we say that y beats x. A set A [subset or is implied by] V, |A| = k, is said to be beaten if there exists a player y [notin N: negated set membership] A such that y beats x for each x [set membership] A. If such a y does not exist, we say that A is unbeaten. A (random) tournament on V is said to have property Sk if each k-element subset of V is beaten. In this paper, we use the Stein–Chen method to show that the probability distribution of the number W0 of unbeaten k-subsets of V can be well-approximated by that of a Poisson random variable with the same mean; an improved condition for the existence of tournaments with property Sk is derived as a corollary. A multivariate version of this result is proved next: with Wj representing the number of k-subsets that are beaten by precisely j external vertices, j = 0, 1, …, b, it is shown that the joint distribution of (W0, W1, …, Wb) can be approximated by a multidimensional Poisson vector with independent components, provided that b is not too large.


text:
## ashleedean12 Group Title inverse tan[tan(6pi/7)] i need an exact answer using pi as needed. can someone explain to me how to solve this? one year ago one year ago

1. mathteacher1729

Sorry I mean $\huge \arctan\left( \tan (6\pi/7) \right)$ or $\huge \tan^{-1}\left( \tan (6\pi/7) \right)$

2. ashleedean12

$\tan^{-1} [\tan (6\pi/7)]$

3. mathteacher1729

What do you know about inverse functions? (they're super important to answering this question).

4. ashleedean12

um i know how to put them in my calculator and that they have restrictions this particular one the answer must be in either quadrant 1 or 4...

5. mathteacher1729

The key is that an inverse function "undoes" a function. arctan ( tan ( stuff ) ) = stuff and tan ( arctan ( stuff ) ) = stuff (assuming domains and ranges are valid) also $$\large \arctan(x) = \tan^{-1}(x)$$ it's just notation. They mean the same exact thing.

6. ashleedean12

so would the answer just be 6pi/7 since the inverse tan would undo tan?

7. phi

yes

8. ashleedean12

thanks (: does that apply to any function that has the inverse of the same function on the outside?

9. mathteacher1729

6pi/7 or -pi/7, they are the same thing. And yes, inverse function ( function ( stuff ) ) = stuff function ( inverse function ( stuff ) ) = stuff Inverse functions and functions "undo" each other. That's glossing over a lot of details, but that's the big idea. :)

10. phi

yes, and vice versa.

11. ashleedean12

what if it was something like $\tan [\tan^{-1}(\pi/10)]$

12. ashleedean12

the simplified version of 6pi/7 would be just pi/7 ?

13. phi

yes, pi/10 you can check with your calculator tan^-1 (pi/10) you are treating pi/10 as a number not as an angle.

14. phi

the simplified version of 6pi/7 would be just pi/7 ? I would not say "simplified" If you mean Is pi/7 the "reference angle" of 6pi/7, that is true

15. ashleedean12

the problem says simplify and use pi as needed..

16. phi


text:
# scipy.special.ellipkinc#

scipy.special.ellipkinc(phi, m, out=None) = <ufunc 'ellipkinc'>#

Incomplete elliptic integral of the first kind

This function is defined as

$K(\phi, m) = \int_0^{\phi} [1 - m \sin(t)^2]^{-1/2} dt$

This function is also called $$F(\phi, m)$$.

Parameters:
phiarray_like

amplitude of the elliptic integral

marray_like

parameter of the elliptic integral

outndarray, optional

Optional output array for the function values

Returns:
Kscalar or ndarray

Value of the elliptic integral

ellipkm1

Complete elliptic integral of the first kind, near m = 1

ellipk

Complete elliptic integral of the first kind

ellipe

Complete elliptic integral of the second kind

ellipeinc

Incomplete elliptic integral of the second kind

elliprf

Completely-symmetric elliptic integral of the first kind.

Notes

Wrapper for the Cephes [1] routine ellik. The computation is carried out using the arithmetic-geometric mean algorithm.

The parameterization in terms of $$m$$ follows that of section 17.2 in [2]. Other parameterizations in terms of the complementary parameter $$1 - m$$, modular angle $$\sin^2(\alpha) = m$$, or modulus $$k^2 = m$$ are also used, so be careful that you choose the correct parameter.

The Legendre K incomplete integral (or F integral) is related to Carlson’s symmetric R_F function [3]. Setting $$c = \csc^2\phi$$,

$F(\phi, m) = R_F(c-1, c-k^2, c) .$

References

[1]

Cephes Mathematical Functions Library, http://www.netlib.org/cephes/

[2]

Milton Abramowitz and Irene A. Stegun, eds. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. New York: Dover, 1972.

[3]

NIST Digital Library of Mathematical Functions. http://dlmf.nist.gov/, Release 1.0.28 of 2020-09-15. See Sec. 19.25(i) https://dlmf.nist.gov/19.25#i


text:
# How do you solve 3x - 1 = 9?

Apr 28, 2018

$x = \frac{10}{3}$

#### Explanation:

$3 x - 1 = 9$

Adding $1$ to both sides

$3 x - 1 + 1 = 9 + 1$

$3 x = 10$

Divide both sides by the coefficient of $x$

$\frac{3 x}{3} = \frac{10}{3}$

$x = \frac{10}{3}$


text:
# Black box

A circuit with 5 resistors of $1~\Omega$ is put into a black box, so that we don't know how the resistors are connected. If you measure the equivalent resistance between the points A and B, you get that it's $\frac{5}{3} \Omega$. The equivalent resistance between the points A and C is $\frac{8}{3}\Omega$. There are two possible equivalent resistances between B and C. One of those resistances is $1\Omega$. What is the possibility for the other resistance between the points B and C in $\Omega$?

×


text:
# Theoretical foundations of the Schrödinger method for LSS formation [CL]

It has been shown that the formation of large scale structures (LSS) in the universe can be described in terms of a Schr$\ddot{o}$dinger-Poisson system. This procedure, known as Schr$\ddot{o}$dinger method, has no theoretical basis, but it is intended as a mere tool to model the N-body dynamics of dark matter halos which form LSS. Furthermore, in this approach the “Planck constant” $\hbar$ in the Schr$\ddot{o}$dinger equation is just a free parameter. In this paper we give a theoretical foundation of the Schr$\ddot{o}$dinger method based on the stochastic quantization introduced by Nelson, and on the Calogero conjecture. The order of magnitude of the effective Planck constant is estimated as $\hbar \sim m^{5/3} G^{1/2} (N/<\rho>)^{1/6}$, where $N$ and $m$ are the number and the mass of the dark matter halos, $<\rho_0>$ is their average density, and $G$ the gravitational constant. The relevance of this finding for the study of LSS is discussed.

F. Briscese
Thu, 15 Dec 16
30/59


text:
Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. ×

### skyscraper's blog

By skyscraper, history, 8 months ago, ,

I was trying to figure out how to include bits/stdc++.h on macos since it uses clang, So once i found the solution i thought of creating a clear video about how to do it.

• +26

 » 8 months ago, # |   0 For me this only works in Xcode. It still doesn't compile in CLion or in the terminal.
 » 8 months ago, # |   +2 Tbh, it's not worth using bits/stdc++.h. If you type out only the headers you need (with an autocomplete for standard headers maybe), you'll lose some seconds, but on the other hand, you'll save seconds of compilation time because the compiler won't have to go through all the headers on your system and figure out which declarations you need and which ones can be optimised out. When you need to debug and aren't slow at debugging, waiting 2 more seconds for each recompilation can be quite a waste.
•  » » 2 months ago, # ^ |   -10 I use a precompiled bits/stdc++.h header and my compilation time is around 0.6 seconds
•  » » » 2 months ago, # ^ |   0 How can i get a precompiled bits/stdc++.h header?
•  » » » » 2 months ago, # ^ |   +4 Not sure if this will work on Windows but it's worth a try: Find the file stdc++.h which is referenced by the compiler. Compile the file using the same command line options as usual (warnings, optimizations, etc) You should get a file called stdc++.pch right next to stdc++.h. Continue using g++ as usual. Next time you compile a file which includesbits/stdc++.h you should see a substantial change in compilation time (2-4 times faster)
•  » » » 3 weeks ago, # ^ |   +10 Why is this down-voted? It's intended way to use bits/stdc++, that's what it was created for.
 » 8 months ago, # |   0 this video help me to solve the same problem, you can change clang to g++ if you like
•  » » 6 weeks ago, # ^ |   0 yeah, #include will work if you use g++ instead of clang. One side affect is that when debugging, clang can print out the elements of the containers, but g++ can not, which is pretty annoying.
•  » » 5 weeks ago, # ^ |   0 Thanks!
 » 8 months ago, # |   0 First open finder. Then press shift+cmd+G.Then copy this /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/include/c++/v1/ and paste it in the box. then click Go.When you're in v1 folder create a new folder called bits after that create a new file called stdc++.h Then go to this page: https://github.com/tekfyl/bits-stdc-.h-for-mac/blob/master/stdc%2B%2B.h copy the content and paste it in stdc++.h file and save it.And that's it.
•  » » 3 months ago, # ^ |   0 Thanks a lot.
•  » » 2 months ago, # ^ |   0 Thank you. It's working.
•  » » 7 weeks ago, # ^ |   0 Thank you so much!
 » 6 weeks ago, # |   0 Well, I find it necessary to install GCC. With GCC there is also pbds, _Find_first, etc... It’s frustrating for me to be unable to use GCC features on my Mac.
•  » » 5 weeks ago, # ^ | ← Rev. 2 →   0 the way i mentioned above, allows us to us pbds... and all gcc libreries, you can change clang to gcc on that way or you can alse compile using g++8
 » 3 weeks ago, # |   0 bits/stdc++.h is not a standard header file of GNU C++ library. steps: - 1.So go to this location "MinGW\include" and create a folder bits inside include folder. - 2.(https://gist.github.com/rocketrunner22/9b218d3c05aea401fc68da2186dfa364) download the source file and paste it into your bits file. - 3.restart your editor. thank you


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# Can I start 529 plans for my niece & nephew without my sibling knowing?

If I wanted to start a 529 plan for them now to give to them at the end of their high school career as a surprise to both them and their parents, is it possible to do so?

My niece and nephew are both under two years old, so I would have a significant amount of time to make small pre-tax contributions that would add up to a first semester / year of college for them by the time they're ready to attend. I would like to have this be a surprise to my sister and brother-in-law as well just in case plans change (and to keep the surprise factor), but I'm not sure of their role/requirements when starting the 529 plan.

• Do I need their authorization to initiate the plan?
• Is there an alternative plan that I could start to achieve this?
• I personally would reveal the accounts the year before AKA beginning of their junior year of high school. The money could change where they decide to go to college and/or their financial aid packages. – mkennedy Dec 26 '18 at 22:23
• WARNING -- Giving your niece & nephew "money for college" will mean they have money which will mean they will get less financial aid. Check this out thoroughly before going ahead. (Expanding on mkennedy's answer.) – Jennifer Dec 27 '18 at 5:48
• Thanks for the tip Jennifer, I had forgotten about that. – Lil' Bits Dec 27 '18 at 12:10
• Would you welcome an answer discussing whether this is a good idea or not (the idea of surprising them with this), or are you solely looking for if it is possible? – Joe Dec 27 '18 at 17:58
• @Daniel: It won't affect financial aid in the first year (because it's neither parental nor student assets), but when it's drawn upon, it will count as student income, which is the single most heavily penalized thing in financial aid calculations. So if they're likely to get financial aid, you'll lose a lot of it if they draw on the 529 before the final year of college. Basically, 529s opened by non-parents are for the final year of college only, or for kids who were never getting financial aid anyway. – ShadowRanger Dec 28 '18 at 14:50

That's a really nice thing that you're planning to do for your niece and nephew, and it is possible to do!

You do need to provide a social security number for the beneficiary of the 529 plan, so to name your niece and nephew as beneficiaries, you will need to get their social security numbers. If you can get these without tipping off the parents, great! Otherwise, you can always change the beneficiary of a 529 plan. You can start the plans with yourself as beneficiary and then get your niece and nephew's SSNs and change the beneficiary to be them after they graduate.

• I would hesitate to describe obtaining the SSN of minors without the consent of their legal guardians "great." The correct answer is to start the plan under OP's name and transfer when it comes time to reveal the surprise. – wnnmaw Dec 27 '18 at 19:53
• @wnnmaw I'm not suggesting obtaining their SSNs without permission from the parents. I'm suggesting obtaining them with permission from the parents, but without tipping them off about what it's for. – Daniel Dec 27 '18 at 23:54

Many states have 529 plans that allow you to name yourself as the beneficiary, and you can later change the beneficiary to a qualified family member without tax implications. Regardless of state, nieces and nephews count as qualified family members and the generation-skipping transfer tax only applies when the new beneficiary is two or more generations below the current beneficiary, so there would be no tax implication for changing beneficiary from you to your niece/nephew. There are gift-tax implications if you contribute in excess of the annual gift exclusion (currently $15,000) to a plan where you weren't the beneficiary, but the IRS allows an up-front 529 contribution of 5x the annual exclusion without gift-tax implications (assuming no other gifts for that 5-year period). If your state's plan doesn't allow naming yourself as the beneficiary or is undesirable in other ways then shop for a different plan, you are not limited to your home state's plan. Edit: Comments incorrectly suggest gift-tax implication when changing beneficiary from self to niece/nephew, so revised for clarity. • Be aware there may be gift tax implications if one starts with the beneficiary as one's self and then transfers to a niece and nephew. – Morrison Chang Dec 27 '18 at 3:29 • Niece and nephews count as qualified family members, but at least for New York State Direct Plan Section 7, page 35 of the agreement, If you change your Beneficiary or transfer money to an Account for another Beneficiary, you may be subject to gift tax if the new Beneficiary is: • Of a younger generation than your Beneficiary being replaced; or • Not a Member of the Family of the current Beneficiary. So if the adult Uncle/Aunt is the both owner and the beneficiary there appears to be gift tax issues to be managed. – Morrison Chang Dec 27 '18 at 4:29 • @HartCO if there were no tax implications on the transfer of a 529 plan, very rich people would use it instead of paying gift taxes. It would be a loophole. – Mindwin Dec 27 '18 at 11:33 • @MorrisonChang Generation-skipping rules do not apply unless the new benificiary is two or more generations below the old beneficiary. So from self to niece would have no tax implications. – Hart CO Dec 27 '18 at 15:10 • @Mindwin It's not a loophole, it is intended. It's also not that exciting since there are maximum contribution limits, limits to how the funds can be used, and limited to family-members no more than 1 generation apart. – Hart CO Dec 27 '18 at 16:01 I'll stipulate that this is possible. I will however suggest that it is not a good idea, for a few reasons. First and foremost, saving for college is a serious thing and it has a major impact on the child's, and their parents', lives. The child will likely make choices with their life based on, among other things, how much money they expect to have saved for college, which can begin years before college itself. The parents will be making choices - how much to save, how to save it - also years and years before. And while you might think of surprising them in order to avoid them making more selfish choices - if they know you're good for$75k (let's say) then that's 3 more European vacations they can take, right? - it may well work the other way around. A parent who knows they can save at most $50k for college may simply tell their child to plan for state school and be done with it. Knowing there's another$75k around might mean that they save more, because they now know they could target a better school. Having full knowledge of a situation is nearly always better than having partial knowledge. If you are worried about your sibling making different and poorer choices knowing about the gift, put conditions on the gift (in which case you can't use a 529 I don't believe, but that might be the choice here).

Second, if you're going to actually give the money to the child, you should do so as early as possible in order to avoid risks related to bankruptcy or lawsuits against you. Money you've saved in a 529 account for yourself will still be reachable in most cases by creditors - while I'm sure you hope to never have to worry about that, you never know what might happen with your life. So establish the 529 account in their name, not in yours. Once it's in their name it's safe against creditors to you OR your sibling - it's the property of the child, who should be fairly safe from creditors and lawsuits in most cases until they reach their majority.

Third, if you are going to establish it in the child's name, then you need to talk to the parents and tell them you're doing it. Not like "it's required by law", but you need their SSN - and if you're thinking of finding that out with a clever ruse, stop now; this isn't engagement ring sizing. If you do find out the SSN without letting your sibling know, you risk them thinking you're trying to steal the child's identity if they do find out.

• Note: I don't normally post answers that argue with the premise of the question, but in comments OP said they'd welcome comments about whether it's a good idea, so, here it is. – Joe Dec 27 '18 at 22:25
• I have set up a 529 plan for a niece. I agree with @Joe: you can surprise her parents today with the gift. Then you can surprise the college-bound person again in eighteen years. In the meantime they can plan. You can also potentially save the parents and child big hassles: what if, heaven forbid, you die in the meantime? By the way, Utah's 529 plan uses high-quality low-fee index funds and they we lcome accounts from outside Utah. – O. Jones Dec 28 '18 at 12:03
• +1, over and over again, for the second and third paragraphs. – Pete Becker Dec 28 '18 at 13:54
• @Joe these are good points, thanks for addressing them. For the last paragraph, that is exactly what I thought, so I will not be doing that. I'll have to think about when it is appropriate to tell my sister about the fund. – Lil' Bits Dec 28 '18 at 16:20

Do set up a plan for the niece and nephew! That is great! Be sure to open the account in YOUR name, and list the child as the beneficiary. Hence, you will need two plans but this depends on the ages, investment strategy, etc. This does NOT have an implication on FAFSA (The FAFSA collects info on the student/parental income/assets). There is only a IRS gift tax if the amount is more than \$15,000 per child, per year. Hopefully you can start two plans (getting the SS# from your relative) and hopefully your state offers a tax deduction for your contributions (from your state's income tax that you file each year --- each state determines their incentive...call to check on it! :) Need to know more about the plan your state offers? Go to collegesavings.org which is the National Treasury Clearinghouse. Happy Saving!

## protected by JoeTaxpayer♦Dec 28 '18 at 4:08

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).


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# Geometry

Let $X$ be the midpoint of the side $AB$ of triangle $ABC$. Let $Y$ be the midpoint of $CX$. Let $BY$ cut $AC$ at $Z$. Prove that $AZ=2ZC$.

Note by Subham Subian
2 years, 10 months ago

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We don't have to use cyclic properties.We extend the line $AY$ to meet $BC$ at $K.$We apply Menelaus theorem for $\triangle BCX$ where $AK$ is the transverse.So we get $\dfrac{BA}{AX}\times\dfrac{XY}{YC}\times\dfrac{KC}{KB}=-1.$Since $BA$ is twice of $AX$ and $XY=YC.$We get $\dfrac{KC}{KB}=\dfrac{1}{2}.$We don't have care about the minus sign as it only to show $A$ does not lie between $BX.$Next we apply Ceva's theorem for $\triangle ABC.$We get $\dfrac{AX}{BX}\times\dfrac{BK}{KC}\times\dfrac{ZC}{AZ}=1.$We know that $AX=BX$ and $BK=2KC.$So we get $\dfrac{ZC}{AZ}=\dfrac{1}{2}.$Therefore,$\boxed{AZ=2ZC}.$Hence proved.

- 2 years, 10 months ago

Why go for Menelaus when Thales' can do ?

- 2 years, 10 months ago

Proof:
Note that $ar(AXY)=ar (BXY)=ar(BYC)=ar(AYC)=\frac{ar(ABC)}{4}$
Since, median divides a triangle into 2 parts of equal areas.

Next, put AZ=$x$ and ZC=$y$
Now,
\frac {x}{x+y} * ar (ABC) = ar(ABZ) \(\implies \frac {x}{x+y}* ar (ABC) = \frac{ar (ABC)}{2} + ar (CYA)
$\implies \frac {x}{x+y} * ar (ABC) = \frac {ar (ABC}{2} + \frac{x}{x+y} * \frac {ar (ABC)}{4}$
$\implies \frac {x}{x+y}[ ar(ABC) - ar (CYA)] = ar(ABC)* \frac {1}{2}$
$\implies \frac {x}{x+y}[\frac {3}{4} * ar (CYAB) = \frac{ar (ABC)}{2}$
$\implies \frac {x}{x+y} * [\frac {3}{4}* ar (ABC)$
$\implies \frac {x}{x+y} = \frac {2}{3}$
$\implies x= 2y$
Hence, proved.

- 2 years, 10 months ago

Its not z, its x, unable to edit.
And forgive me for the brackets not being closed and other mistakes that I am unable to edit.

- 2 years, 10 months ago

I was unable to solve it using my limited knowledge of cyclic quads.

- 2 years, 10 months ago

There is no 3/4 in 4th last line.

- 2 years, 10 months ago

- 2 years, 10 months ago

This question can easily be solved using midpoint theorem and similarity rules as above.

- 2 years, 10 months ago

Seriously, BC cut AC at Z; wont that make Z coincide with C and consequently CZ will be 0.

- 2 years, 10 months ago

sorry my bad .i corrected it plss try to solve it and give mme the solution i will be gratefull

- 2 years, 10 months ago

you can use the cevas theorem and the menuluas theorem to prove this

- 2 years, 10 months ago

and can use properties of cyclic quadrilaterals beacause my teacher said that it can be solve by using the properties of cyclic quadrilaterals

- 2 years, 10 months ago


text:
## Examples

If a Haar measure is done on the topological group ${\displaystyle (\mathbb {R} ,+)}$ (meaning the set is all real numbers and the binary operation is addition), the Haar measure takes the value of 1 on the closed interval from zero to one and is equal to a Lebesgue measure taken on the Borel subsets to all real numbers. This can be generalized to any dimension.

If the group G is all nonzero real numbers, then the Haar measure is given by $\mu (S)=\int _{S}{\frac {1}{|t|}}\,dt$

For more reading on examples of Haar measures: https://en.wikipedia.org/wiki/Haar_measure#Examples

ECE462 Survivor

Seraj Dosenbach