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Computational Physics/Chapter 1

The Spreadsheet Program
This book will describe how specifically to use Microsoft Excel, as this is a widely used program. If you have access to this and are farmiliar with it is suggested that you use it. If you have another preferred spreadsheet program please feel free to use that, however be aware that you will have to refer to the help section where your spreadsheet program differs from this book.

If you would like a free spreadsheet program that works with nearly any operating system then 'Calc' from OpenOffice.org is highly recommended. Hopefully in the future this book will describe how to use this as well as Microsoft Excel.

Why use a spreadsheet? This chapter aims to show you that you can do things quickly and simply without having to learn and type excessive amounts code. We will see towards the end that using a spreadsheet has its limitations, but hopefully also the power of spreadsheets other than for just plotting graphs of your experimental results!

Using Spreedsheets to calculate the Numerical Solutions of Ordinary Differential Equations
If x is a function of a real variable t and f is a function of both x and t, then the equation

x’(t) = f(x(t), t) 		(1.1)

is called a first order differential equation. Solving such an equation involves more than algebraic manipulation; indeed, although the equation itself involves three quantities, x, f, and t, to find a solution we must identify a function x, defined solely in terms of the independent variable t, which satisfies the relationship of (1.1) for all t in some open interval. For many equations, exact solution is not possible and we have to rely on approximations. We will discuss one technique for finding both approximate and, where possible, exact solutions to differential equations. We have already seen many examples of differential equations we discussed finding the position of an object moving in a straight line given its velocity function and its initial position, when we discussed the motion of a projectile, when we considered the two-body problem. Indeed, in many ways the study of differential equations is at the heart of calculus. To study the interaction of physical bodies in the world is to study the ramifications of physical laws such as the law of gravitation and Newton’s second law of motion, laws which frequently lead questions involving the solution of differential equations. Newton was the first to realize the power of calculus for solving a vast array of physical problems. The mathematicians that followed him enlarged and refined his techniques until they began to believe that the entire future of the universe, as well as its past, could be discerned from the knowledge of the current positions and velocities of all physical bodies and the forces at work between them. Today we know more about the limits to our knowledge, but, nevertheless, the study of differential equations remains a key component to our understanding of the universe. To begin our study, consider the equation

x’(t) = f(x(t), t)		(1.2)

with the initial condition x(t0) = x0. To simplify notation, we will frequently omit the independent variable when referring to x(t) and write simply

x’ = f(x, t)		 (1.3)

Now if f(x, t) depends only on the value of t, that is, if f(x, t) = g(t) for all values of x, where g is a function of t alone, then we may solve (1.3) by integration. That is, integrating both sides (1.3) gives us

x(t) - x(t0) =   integral of [g(t), over [t,to]]	(1.4)

which gives

x(t) = x(t0) + integral of [g(t), over [t,to]]	(1.5) Assuming we can compute the definite integral, which can at least always be done numerically, we have solved the differential equation.

For Example

Consider the equation x’ = 4 sin(3t)

with initial condition x(0) = 5. Then

x(t) = 5 - 4/3 cos3t + 4/3 = 1/3 (19 - 4 cos(3t)).

Knowing the value of t we can calculate the exact value of x at t.

Indeed, knowing the difficulty of evaluating ordinary definite integrals, it is not hard to believe that many, if not most; differential equations may be solved only through numerical approximation. Although the function x in equation (1.3) is unknown, we do have enough information to find its best affine approximation at t0. Namely, the best affine approximation to x at t0 is obtained by the Taylor theorem using subinterval [t0,t1] of an entire interval [t0,t].

x(t1) = x0 + f(x0, t0)(t - t0) + 0((t - t0)2 +...). where t1 = t0+h 	(1.6) Equivalently,

x(t0 + h) = x0 + hf(x0, t0)		(1.7)

Thus for a small value of h,

x1 = x0 + hf(x0, t0)			(1.8)

Where h = t1 – t0/n [in the above case n = 1]

Will provide a good approximation to x(t0 + h). However, we want to do more than this; since x is a function, we want to be able to approximate its values over an entire interval, say [t0, t]. To do so, we choose a small value for h and iterate the process that gave us x1. Specifically, we let t = t0 + nh, n = 0, 1, 2. 3,…, where n is chosen large enough and the generalized form of the Euler method is

xn+1 = xn + hf(xn, tn) 			(1.9)

That is, we compute an approximation for x(tn+1) by applying the best affine approximation to our approximation for x(tn), repeating the process enough times until we have approximated x over the entire interval [t0, t]. This method of approximation is known as Euler’s method.

Euler’s method
To approximate a solution to the equation

x’ = f(x, t)

with initial condition x(t0) = x0 on an interval [t0, t], choose a small value for h > 0 and an integer n such that t0 + nh = t. Letting t = t0 + nh, compute x1, x2,. . ., xn+1 using the difference equation xn+1 = xn + hf(xn, tn). (1.10). Then xn+1 is an approximation for x(t0 + nh).

Consider the differential equation x’ = 0.04x with initial condition x(0) = 50. We know that the solution to this equation is

x(t) = 50e 0.04t.

In particular, x(50) = 50e 2 = 369.45, rounding our answer to the second decimal place. To approximate x on the interval [0, 50] using Euler’s method, we will first take h = 1 and, starting with x0 = 50, compute x1, x2. . . x50, where in this case xn+1 = xt will approximate x(t). Using (1.9) with f(x, t) = 0.04x and rounding to two decimal places, we have

n = 1:x1 = x0 + h(0.04x0) = 50 + (1)(0.04)(50) = 50 + 2 + 52.00

n = 2:x2 = x1 + h(0.04x1) = 52 + (1)(0.04)(52) = 52 + 2.08 = 54.08

n = 3: 	x3 = x2 + h(0.04x2) = 54.08 + (1)(0.04)(54.08) = 54.08 + 2.16 = 56.24

and so on, ending with x50 = 355.33.

Notice that the error in our approximation, that is, the difference between xn and x(n), increases as n increases. For example,

x(5) - x5 = 0.24, whereas

x(50) - x50 = 14.12.

This should be expected since the error of the approximation on each step is compounded by the errors made in each of the preceding steps. The only way we can control the amount of error in our approximations is to decrease the step size. For example, if we reduce the step size to h = 0.1, we obtain

x1 = x0 + h(0.04x0) = 50 + (0.1)(0.04)(50) = 50 + 0.2 = 50.2000, x2 = x1 + h(0.04x1) = 50.2 + (0.1)(0.04)(50.2) = 50.4008, x3 = x2 + h(0.04x2) = 50.4008 + (0.1)(0.04)(50.4008) = 50.6024, ………………………………………………………………….	……………………………………………………………….. and ending with

x500 = 367.98 (This can be easily calculated using computers)

We have to calculate the value of x500 and take it equivalent to x50 because from the definition of the parameter h

h = (t – t0)/n

putting h=0.1 clearly the value of n, which actually is the number of iterations to be made, becomes 500 for t0 = 0 and t = 50. [Alay]

Using Spreadsheets to calculate the Numerical Solutions of Integrals
In calculus we come across very complicated and difficult integrals. They are difficult to solve using analytical methods for finding their numerical solutions. This problem can be solved by using numerical integrations e.g. Rectangular rule, limit of Sum definition, Trapezoidal Rule and Simpson’s Rule.

The Trapezoidal Rule
Consider y = f[x] bounded by x=a to x=b. Than by definition integration of the function of x i.e. y w.r.t x is the area under the curve AB shown in figure

Area = F[x] = integral [f(x), over[a,b]dx

now dividing the trapezoid into a Rectangle and an approximate triangle i.e. number of rectangles n = 1 we can write F[x] = h f[a] + ½ h {f[b] – f [a]}

= h {f[a] +1/2f[b] – 1/2f[a]}

= h/2 {f[a] + f[b]} = h/2 {f[x0] + f[x1]}

for n = 2 we have two rectangles and two triangle such that the approximate area under the curve now become

F[x] = h/2 {f[a] +f[{a+b}/2]}+ h/2 {f[{a+b}/2] + f[b]}

= h/2 {f[a] + 2f[{a+b}/2] + f[b]}

= h/2 {f[x0] + 2f[x1] + f[x2]}

So generalizing for n divisions of the trapezoid we have

F[x] = h/2 {f[x0] + 2f[x1] + 2f[x2] + … + 2f[xn-2] + 2f[xn-a] + f[xn]}

Is an approximation of the integral of y = f[x].

Where h = b – a/n

n can be even or odd. And approximation can be improved using large value of n. [Alay]

The Random Walk

This is a book for physicists, so let's get right in and describe the physics we want to investigate. The random walk is a problem that occurs in various physics areas such as Brownian motion, which describes how molecules move about in a fluid (a fluid can either be a liquid or a gas). It also plays a role in quantum field theory, but this is much more complicated. Outside of physics the random walk can be applied to share prices. It is interesting that by investigating one problem we find solutions to entirely different ones, that have no other relationship to each other. This is something that occurs throughout physics, such as the way in which waves can be used to model traffic flows.

So what is a random walk? Imagine a drunk guy (yourself after a good night out if you like) who has just left their local bar. He is standing outside the bar and, being rather drunk, he randomly takes one step North up the street or one step South down the street. The probability of him going North or South is equal and all his steps are of the same length. If he does this for 100 steps how far on average has he got away from the bar? You may think on average he ends up very close to the bar, but actually this is not so! (Remember, distance means how many meters is he from the bar, and doesn't take account of where he is. If we found his average position treating North as positive, and South as negative, then by the symmetry of the situation we know that he would indeed end up outside the bar on average.)

We will investigate this problem and find how far he actually ends up away. This is a known as a one-dimensional random walk, as the drunkard walks only North or South. The same ideas can be applied in two dimensions, say a drunk in a field who staggers North, South, East or West. Three dimensions can similarly be investigated, and even four or more, although this is a bit of an abstract idea!

Can you see now how this links to Brownian motion (where molecules move randomly) or to share prices which have approximately equal chances of rising or falling? [John]

Making a one dimensional random walk
On a blank worksheet in your spreadsheet double click on cell B4 and type =RAND and press enter. In cell B4 you should now see a decimal number between 0 and 1. Try hitting the F9 key a few times, which recalculates the entire spreadsheet. You should see numbers being randomly generated in cell B4. By the way cell B4 is specified so there can be some consistency when referring you to specific cells. It will also allow the spreadsheet to be neatly formatted later if you're so inclined! On the subject of formatting the whole random number may not be showing currently, if you click on cell B4 so that it is highlighted and then go to the format menu, select column and then autofit selection the entire contents of the cell is shown. This can be done for the entire selecting the whole sheet, you can do this by clicking the grey box at the top left of the sheet,

So, why did we type what we typed? You may well know if you're familiar with spreadsheets, = specifies that we're typing a mathematical function and not just text. RAND is a function that the spreadsheet recognises as a request to generate a random number. If this isn't clear right now it will become clear as we work through the rest of the chapter.

How do we transform this into a random walk? We need to transform this random number into an event with a 50:50 probability. What we need to do is generate a step South if the number is less than 0.5 and a step North if the number is more than 0.5. This is where we see how a simple spreadsheet can be transformed into a program.

Double click on cell C4 and type: =IF(B4<0.5,-1,1). This was a bit more of a complicated function than the previous! What the IF function does is to set the value of a cell to one value if the condition specified is true, or another value if the condition is false. The structure of this function goes like this: IF(Condition, Value if True, Value if False). B4<0.5 is the condition, this means 'if cell B4 is less than 0.5'.

You should now be able to work out what values of the random number will make it take a step South (generating a -1) or a step North (generating a 1). If not try it and find out! It should work the same way as it was said it would a bit earlier.

Taking one step isn't particularly exciting or impressive (or if you think it is you're going to absolutely love the rest of the book!) so we'll see how we can expand it to 100 steps. It's fairly easy actually, simply click on cell B4 so it is selected and the move the mouse over the little square in the bottom right corner. The mouse cursor will change and you can then drag down to fill the cells below with the same formulae as in this cell. Drag down to cell B103 so that we will be taking 100 steps. It is even easier to fill the cells in column C, all you need do now is double click on the little black square in the bottom corner of cell C4, this will fill it down to the same row as column B is filled down to. Notice that in cell C4 we have a function that relies on the value of the function in B4. But when we fill the other cells they automatically change so that the function depends on the value of the cell in the left. This is usually very handy, but we'll see in a bit that sometimes we need to override it.

Now we can hit F9 and each time generate a random walk! But it's still not very interesting to look at, or a very informative, we can't see where the drunkard is unless we add up all the cells at the moment, nor can we see where he is it any particular position. This can be easily corrected, enter into D4 the function SUM(C$4:C4). Hopefully you can guess what this does, it adds up (sums) all the cells between C4 and, well, um C4. Not too useful as it is you might think, and why is that $ sign there? Well, if you fill down into the cells below you'll see that each cell adds up the total of all the steps taken, and gives the current position of the drunkard after this step. If you look at cell D20 you should get an idea of what's happening, this cell sums between C4 and C20. The $ sign acts to fix a value when you fill cells below.

Now you can use the spreadsheet to make a graph of the random walk, if you don't know how to do this check the help section of your spreadsheet. As a couple of hints to make the graph look good put a '0' in cell D3 so that the drunkard can start outside the bar and select a line graph. Press F9 to generate new graphs.

Do the graphs you've created look like what you would expect, do they actually look random? You'll probably see some strange patterns in some graphs, but they are honestly random! This will hopefully make you think about what a random pattern truly looks like. [John]

How far does the drunk get?
Try and work this out while this section is written!

title: Computational Physics/Chapter 1


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What's up?

 * The full Table of contents was started. (It's recommended to start writing new pages from here.) 10 July 2005
 * A featured article system is ready. This is to make popular and useful articles easier to find for first time users of the book.29 Jun 2005
 * We have added a configuration category, and placed all contents of the installation chapter there.29 Jun 2005
 * Some categorizing has been done: We added the category |MOS to the templates, as well as categorizing the front page of the book a bit. 28 Jun 2005

[ add/edit] Archive

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Mambo Open Source/News archive for MOS NEWS

News archive for the Mambo Open Source wikibook.

 * This news section has just been added, and will contain news about this wikibook. 26 jun 2005
 * Many thanks to all the people who have fixed typos in the book! Please carry on with the good work, and don't hesitate if you want to write your own article. 26 jun 2005


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Talk:Electronics/Amplifiers

 * For more images, see commons:Category:Electrical_diagrams. Everything I have ever made (including the common base, etc. diagrams) is up there now.  You can make a link to them and they will be transwikied automatically. For example:

- Omegatron 20:06, 26 Jun 2005 (UTC)

Spice net lists
Suggestion: Include net-lists for each of the configurations listed here, or at least links to them. --Jcarroll 03:59, 19 March 2006 (UTC)

Other
Would it be appropriate to include the efficiency limitations of each class of amplifiers? 128.101.33.133 16:04, 25 June 2007 (UTC)

Sampling Theorem
The sampling theorem is misstated in the section about class D amps. It should read that the sampling frequency has to be more than double (not half) of the maximum signal frequency.

title: Talk:Electronics/Amplifiers


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How To Tie A Tie

How To Tie A Tie A tie or necktie is a garment, conventionally worn around a man's neck with a collared shirt. There are several ways that the knot can be tied. A common and easy knot is the Four-In-Hand. However this is not the simplest knot.

Contents

 * 1) /The Knots/
 * 2) /Tying the Knots/
 * 3) /General Tips/

Reference

 * Main methods easily explained (with pictures) - How to Tie a Tie
 * Easy Steps - How to Tie a Tie
 * How to Tie a Tie: How to tie a tie
 * Easy way to learn how to tie a necktie, 9 famous knots: How to tie a tie
 * For some pictures: How to tie a tie
 * For a method shown on video: https://www.youtube.com/watch?v=-9u1VK1eTTc
 * And: Theory of Tie Knots
 * And: Tie-a-Tie.net
 * For the mathematical basis of tie knots: Tie knots and random walks
 * The history of the necktie: Necktie Down the Ages
 * With slideshow and instructions: How to Tie a Tie
 * Animated slideshow with clear pictures: - How to tie a tie

title: How To Tie A Tie


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User talk:Whyte~enwikibooks

I finally got around to registering!
 * Whyte 21:52, 26 Jun 2005 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Whyte. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Whyte~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:51, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:38, 19 April 2015 (UTC)

title: User talk:Whyte~enwikibooks


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User:Whyte~enwikibooks

About me

My name is Dave.

Scholastic Interests

I'm currently studying Criminal Justice topics. My concentration is on understanding and mitigating social conflict. I also have a growing interest in international affairs and global social justice issues.

My focus has largely been understanding social phenomenon at an experiential level. Said another way, I am interested in understanding how one's relative level of knowledge about a phenomenon affects the way in which one relates to and with one's social and physical environment. Acedemicly this had involved the study of event level decision making, particular in instances of socially defined criminal behavior.

Scholastic Activity

I'm currently working to finish a Bachelor's of Science in Criminal Justice at Wichita State University. Upon graduation I fully intend to begin work on a Master's of Art in Criminal Justice also at Wichita State University.

Published Works

 * I have not published any works at this time.

Unpublished Works

 * Dynamics in Group Offending: Neo-Nazis and Mujahideen - Undergraduate.

title: User:Whyte~enwikibooks


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Ada Programming/Attributes/'Bit Order

Description
R'Bit_Order is a representation attribute used to specify the bit numbering of a record representation clause (for a record type). The bit ordering is of type .Bit_Order, and may be either: assuming the bit sequence is interpreted as an integer value. The constant .Default_Bit_Order represents the native bit numbering of the platform.
 * High_Order_First when the bit 0 is the most significant bit
 * Low_Order_First when the bit 0 is the least significant bit

It is worth noting that the bit ordering only affects the bit numbering for a record representation clause, and not the byte order of its (multibyte) fields.

Introduction
Storage units are ordered by their addresses. Let's look at an integer occupying several bytes (let's assume a byte being 8 bits for this discussion).

 * On a big endian (BE) machine, the integer's most significant byte is stored at the least address.
 * On a little endian (LE) machine, its least significant byte is stored at the least address.

So in order to be able to count bits consecutively across byte boundaries, Ada counts bits within a byte according to the endianness from most significant bit (MSB) 0 to least significant bit (LSB) on BE and the other way round on LE.

This is how it looks (conveniently writing left to right on BE, right to left on LE):

BE Byte   0 1 2 …  (counting bytes, i.e. addresses, left to right; higher addresses to the right) LE Byte … 2 1 0    (counting bytes right to left; higher addresses to the left) MSB        LSB BE Bit  0 1 2 3 4 5 6 7  (counting bits within a byte left to right) LE Bit  7 6 5 4 3 2 1 0  (counting bits right to left)

We're used to writing left to right, but for LE, as you can see, it's convenient to write addresses from right to left. So on BE, a sequence of bytes and bits is counted like this:

Byte 00                     01                      02 Bit 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 … 00 01 02 03 04 05 06 07 08 09 10 11 12 13 … (we can equally begin to count at byte 01)

In order to be able to write and count consecutively on LE, we have to write right to left like for example in the Arabian or Hebrew scripts (the MSB is always on the left; this seems to be a global trait in all modern scripts):

Byte                 02                      01                      00 Bit … 21 20 19 18 17 16 15 14 13 12 11 10 09 08 07 06 05 04 03 02 01 00 … 13 12 11 10 09 08 07 06 05 04 03 02 01 00

In a record representation like the following (of course only one of the two lines specifying X may be present)

for T use record X at 0 range 13 .. 17; -- these two lines… X at 1 range 5 .. 9; -- … are equivalent end record;

the component X occupies the bits shown in boldface. The byte number (0 resp. is 1) is called Position, the bounds are called First_Bit (13 resp. 5) and Last_Bit (17 resp. 9); there are corresponding attributes 'Position, 'First_Bit and 'Last_Bit. (The meaning of the component X is irrelevant here, only its size is relevant.) As you can see, X is a crossborder component. Thus the result of applying the attribute 'Bit_Order to a record with crossborder components depends on the Ada generation.

Ada 95
Ada 95 defines the result of the attribute for the non-native bit order only when applied within a storage unit or when an item completely fills a sequence of storage units. By applying the attribute to a record, we force the compiler to count bits in the indicated manner, independent of the machine architecture.

Rec A 0  0 .. 5;  B  0  6 .. 7; ;  Rec'Bit_Order  High_Order_First;

The component B occupies the bits shown in boldface:

Byte 0 Bit 0 1 2 3 4 5 6 7

The layout of this record will be the same on BE and on LE machines, i.e. the representation is endian-independent.

We could equally well have defined this record like so and arrived at the same endian-independent layout:

Rec A 0  2 .. 7;  B  0  0 .. 1; ;  Rec'Bit_Order  Low_Order_First;

Byte              0 Bit 7 6 5 4 3 2 1 0

However a record like the following, where B is crossborder, is only valid on a BE machine, i.e. in the native bit order.

Rec A 0  0 .. 5;  B  0  6 .. 9; ;  Rec'Bit_Order  High_Order_First;

Byte 0                       1 Bit  0  1  2  3  4  5  6  7  0  1  2  3  4  5  6  7
 * 1)    00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15

Compiled on a LE machine, the compiler will complain that B occupies non-contiguous storage and reject the code, since the byte order is not affected by the attribute (remember, the next byte with bit numbers 8 to 15 has to be put on the left):

Byte                        1                       0 Bit   00 01 02 03 04 05 06 07 00 01 02 03 04 05 06 07
 * 1)      08 09 10 11 12 13 14 15 00 01 02 03 04 05 06 07

As an example of a record where items fill a range of complete storage units take:

Rec A 0  0 .. 7;  B  1  0 .. 15; ;  Rec'Bit_Order  High_Order_First;

This is valid on both architectures leading to the layout:

BE   0                       1                       2 Bit 00 01 02 03 04 05 06 07 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15

LE                        2                       1                       0 Bit 08 09 10 11 12 13 14 15 00 01 02 03 04 05 06 07 00 01 02 03 04 05 06 07

where A fills the bits in normal face, B those in bold face. As you can see, it doesn't matter that bits are counted in this strange way on LE since component B fills its two bytes completely.

Ada 2005
For the native bit order, there is no change. Record representation specifications have worked since Ada 83. Only the attribute 'Bit_Order has been introduced in Ada 95 with the restrictions as shown above.

In order to improve the situation for non-native bit orders, Ada 2005 introduces the notion of machine scalars. A machine scalar is a conceptual hardware based unsigned integer within which bits are counted and the components positioned as requested.

We need a more elaborated record example now (the values returned by the corresponding attributes 'Position, 'First_ and 'Last_Bit (in native bit order) are given as comments; note that the bit numbers returned are always counted starting with the byte in which the component begins): T    I  0   0 .. 15; -- at 0 range 0 .. 15  A  2   0 .. 12; -- at 2 range 0 .. 12  B  2  13 .. 17; -- at 3 range 5 .. 9  C  2  18 .. 23; -- at 4 range 2 .. 7  D  5   0 .. 7; -- at 5 range 0 .. 7 ;

Let's assume I is a 16 bit (signed or unsigned) integer, the other components are of some unspecified types with the given size requirements. The complete record's size is 48. This is how it looks like on BE and LE machines:

Byte 0      1       2       3       4       5 BE 012345678901234501234567890123456789012301234567 IIIIIIIIIIIIIIIIAAAAAAAAAAAAABBBBBCCCCCCDDDDDDDD DDDDDDDDCCCCCCBBBBBAAAAAAAAAAAAAIIIIIIIIIIIIIIII LE 765432103210987654321098765432105432109876543210 Byte       5       4       3       2       1       0

In the following, we're going to show how far we can go to reach endian-independence of the representation. The result will be that we only have to swap bytes after transfer from one architecture to the other when the new Ada 2005 features are used correctly.

For the following, we'll assume that we are on a big-endian machine, so that we append the corresponding attribute to the representation:

T    I  0   0 .. 15;  A  2   0 .. 12;  B  2  13 .. 17;  C  2  18 .. 23;  D  5   0 .. 7; ;  T'Bit_Order  High_Order_First;

When this is compiled on a little-endian machine, all components with the same Position are taken together and put in a matching machine scalar. The machine scalar is positioned at the given Position, but inside the bits are counted from the opposite side.

Let's take the first component, I. It uses 16 bits, so a 16 bit machine scalar will do. It is positioned at byte 0, but inside the compiler will count from the high order bit side. This is how it looks (NNBO - non-native bit order):

IIIIIIIIIIIIIIII NNBO                                0123456789012345 Byte       5       4       3       2       1       0

Now to the next Position 2. The respective components A, B, C use together three bytes, so a 32 bit machine scalar is needed. It is positioned at byte 2, and inside the count will again start from the opposite side. This is how it looks:

IIIIIIIIIIIIIIII NNBO 012345678901234567890123456789010123456789012345 Byte       5       4       3       2       1       0

The bit count starts at the high order bit of byte 5 and continues down to the low order bit of byte 2. The components A, B, C are positioned inside this scalar according to the respective ranges. Thus we arrive at this layout:

AAAAAAAAAAAAABBBBBCCCCCC       IIIIIIIIIIIIIIII NNBO 012345678901234567890123456789010123456789012345 Byte       5       4       3       2       1       0

We immediately see the conflict with component D, whose range is already occupied by A, and the compiler will of course complain and reject the code. The solution is simple: Just instead of locating D at Position 5, we change this to the (on BE) equivalent line like so:

T    I  0   0 .. 15;  A  2   0 .. 12;  B  2  13 .. 17;  C  2  18 .. 23; --D at 5 range 0 .. 7;  D  2  24 .. 31; ;  T'Bit_Order  High_Order_First;

And, drum roll, on LE, we now have (for comparison, the native layout is also given):

AAAAAAAAAAAAABBBBBCCCCCCDDDDDDDDIIIIIIIIIIIIIIII NNBO 012345678901234567890123456789010123456789012345 Byte       5       4       3       2       1       0 IIIIIIIIIIIIIIIIAAAAAAAAAAAAABBBBBCCCCCCDDDDDDDD BE 012345678901234501234567890123456789012345678901 Byte 0      1       2       3       4       5

In the non-native bit order, the values returned by the corresponding attributes 'Position, 'First_ and 'Last_Bit are exactly those given in the record specification.

As an additional service, GNAT's compilation output will give you the values as counted in the native bit order within the machine scalar:

range "I"  0 .. 15 "A" 19 .. 31 "B" 14 .. 18 "C"  8 .. 13 "D"  0 .. 7     AAAAAAAAAAAAABBBBBCCCCCCDDDDDDDDIIIIIIIIIIIIIIII NNBO 012345678901234567890123456789010123456789012345 LE 109876543210987654321098765432105432109876543210 Byte       5       4       3       2       1       0

This is as far as we can get with the current Ada standard.

Data Transfer
Let us transfer a value of this record from the native big-endian machine to a little-endian machine. For demonstration purposes, the high-order parts of the crossborder items are shown in upper case, the low-order parts in lower case.

IIIIIIIIiiiiiiiiAAAAAAAAaaaaaBBBbbCCCCCCDDDDDDDD BE 012345678901234501234567890123456789012345678901 Byte 0      1       2       3       4       5

The bytes will be transferred in the given order. Since the bit order attribute does not reorder the bytes after transfer, on the target machine we will receive the data in scrambled order:

DDDDDDDDbbCCCCCCaaaaaBBBAAAAAAAAiiiiiiiiIIIIIIII NNBO 012345678901234567890123456789010123456789012345 Byte       5       4       3       2       1       0

All we have to do to arrive at the desired end, is to swap bytes 0↔1, 2↔5, 3↔4:

AAAAAAAAaaaaaBBBbbCCCCCCDDDDDDDDIIIIIIIIiiiiiiii NNBO 012345678901234567890123456789010123456789012345 Byte       5       4       3       2       1       0

Example
The following two sets of representation clauses specify the same register layout in any machine / compiler (Ada 2005 and later):

Device_Register Ready : Status_Flag; Error : Error_Flag; Data : Unsigned_16; ;  Device_Register Ready 0   0 .. 0;       Error  0   1 .. 1;       Data   0  16 .. 31;     ;  Device_Register'Size       32;  Device_Register'Bit_Order System.Low_Order_First; (Device_Register);

If the bit order is modified, the bit numbering of all the elements of the record representation clause must be reversed:

Device_Register Ready : Status_Flag; Error : Error_Flag; Data : Unsigned_16; ;  Device_Register Ready 0  31 .. 31;          Error  0  30 .. 30;          Data   0   0 .. 15;        ;  Device_Register'Size       32;  Device_Register'Bit_Order System.High_Order_First; (Device_Register);

Both can be interchangeably used in the same machine. But note that in two machines with different endianness the Data field will have the native byte order regardless of the bit order specified in the representation clauses.

Interfacing
Record representation clauses are a unique feature of the Ada language as far as the authors know, so there is no need to have a feature similar to attribute 'Bit_Order in other programming languages. It is common practice in other programming languages to use masks and bit operations explicitly, thus the native bit numbering must always be used.

Wikibook

 * Ada Programming
 * Ada Programming/Attributes
 * Ada Programming/Attributes/'Position
 * 'First_Bit
 * 'Last_Bit

Ada Rationale

 * Ada 95:
 * Ada 2005: 9.2.1 Incompatibilities with original Ada 95

title: Ada Programming/Attributes/'Bit Order


text:
User talk:136.159.123.211

Hi, please do not just remove the product mentions on Cookbook:Vegan_Substitutions. Please engage in the conversation about them on the talk page (Talk:Cookbook:Vegan_Substitutions) or on the main cookbook talk page (Talk:Cookbook). Kellen 00:43, 27 Jun 2005 (UTC)

title: User talk:136.159.123.211


text:
Talk:How To Tie A Tie

Hey, Welcome to my How-To-Tie-A-Tie discussion page.

Basically if you have any advice, changes, tips or anything else just leave a message.

Also, this is where I'll keep credits for now for all the contributors.

UPDATE: T-Marsh is currently working on the illustrations. Since we all have a life outside of Wiki, this is a SLOW process but we'll post them as soon as they become available. The Four-in-Hand should be up soon. Thank you for your patience. --Shmeedogg 00:32, 15 July 2005 (UTC)

Comments Section
Yea, i came to this page to learn how to tie a tie, but was just told what different kinds of knots there are. Could you maybe include a step by step procedure for actually making all of the different knots? That would be a GREAT help!

 * Working on it buddy-- I'm working on the line drawings and illustrations. If anyone wants to collaborate and help that would be much appreciated. --Shmeedogg 20:57, 27 Jun 2005 (UTC)

There are drawings of "how to" tie knots in the wikimedia commons and on wikipedia. Bowtie and. Kellen 04:32, 28 Jun 2005 (UTC)

 * Thanks for the tip Kellen. The problem, though, is that those drawings are craptastic! No offense, but we need some quality illustrations to help tie-newbies.

Shmeedogg: Begging your pardon if this is not the right protocol/format but I'm a newbie to the wiki thing. However, I am an experienced professional at technical illustrating. If you can send me some kind of pencil sketch that outlines how the ends of the tie are positioned/move maybe I can help. All else failing, I can cheat with screen shots from from the following website if you are willing to suggest changes that will improve things: www.tieanecktie.com. FROM:tmmarshall2@comcast.net (please put your screenname/WIKI in the subject heading so that I accept your e-mail).

 * That is awesome T-Marsh, I'll get in contact w/ you ASAP. I have some ideas (pictures I've been working on) that might inspire you. --Shmeedogg 19:18, 29 Jun 2005 (UTC)

btw, if you use a colon (":") for each level of indentation in comments, it'll make things more readable. e.g:
 * This is a 2nd level comment using ":"
 * and this is the response using "::"
 * and another response. using ":::" Kellen 02:07, 30 Jun 2005 (UTC)

 * Point taken. BTW, this is a WIKI so feel free to improve the document on your own. Editing is always welcome. --Shmeedogg 04:07, 30 Jun 2005 (UTC)

Your book will be awesome, i might add a few things, cant wait till the pics are up, i need help... lol. --wrestlenovi 07:16, 08 Jul 2005 (UTC)

Please hurry. --209.161.213.26 09:55, 14 July 2005 (UTC)

Credits:
Rick H. --Shmeedogg 01:15, 27 Jun 2005 (UTC) Keith S Brown Ksbrowntalk 21:31, 30 December 2006 (UTC)

An idea for your next book
How To Wiki A Wiki.

The Pratt is not a poor cousin...
I prefer the Pratt because it is symmetrical without the bulk of the Windsor. I think it is more elegant and streamlined. Some ties with thicker linings are almost untieable with the Windsor, but look quite good with a Pratt. Many of my friends wonder why I have to untie my tie after I take it off, I explain, and most of them have adopted it themselves. Please give the Pratt its due. It was only discovered in the 1980's, so it is not mainstream, but it certainly is not analogous with a VW. Try it before you dismiss it as "commie" garb. The Four-in-Hand would be at the bottom of my list just behind the Half-Windsor.
 * Peasant. I suppose you also think that knots the size of a man's fist are the current trend, eh??  WRONG.––193.118.203.3 08:39, 18 July 2006 (UTC)
 * Section know copied to the page Talk:How To Tie A Tie/The Knots - Ksbrowntalk 21:38, 30 December 2006 (UTC)

How to tie a tie
Yeah, so this says absolutely nothing about how to actually tie a tie. It is quite a humorous read though. — 208.180.131.161 a.k.a.: w:User:ACupOfCoffee 22:50, 16 September 2006 (UTC)

How to tie a tie???
This page has been up for over a year and there are still no instructions on how to tie a tie. Time to add these or delete the entry.--146.145.140.155 19:12, 20 October 2006 (UTC)

Four-In-Hand (feedback) +Oriental
I have just put a step by step guide up from tying my own tie several times. I have checked it from starting either way round and find the narrow to my (biological/medical) right to be easiest starting position, but this is because I always tie it that way. They both give the same knot.

I am unsure if it is affected by handedness, I am right handed.

If anyone who doesn't know how to tie a tie would like to post a message if they cannot follow the instructions let me know so I can improve the instructions. Please let me know where it all started to go wrong if you have an idea. If you know exactly where the instructions make you tie it wrong, change them if you wish. This is to make the instrucitions better so please do not send me messages of panic, as I am unable to be able to answer them fast enough and will feel bad at not being able to help you in time. Ksbrown 21:30, 29 December 2006 (UTC)

 * I have also added the oriental instructions for people to try. Ksbrowntalk 10:29, 30 December 2006 (UTC)

Reorg suggestion
I'd suggest that, rather than having "The Knots" and "How to tie them", that these instead be divided into pages on a per-knot basis, ie. one page for "Windsor", one for "Pratt", and the like

title: Talk:How To Tie A Tie


text:
User talk:Tilly~enwikibooks

Biodiesel
Continued from discussion with Tilly as 61.9.214.113: I'm sorry if you felt that I was editing your work unfairly. But, because you were contributing anonymously, I had no way of knowing what your intentions were (also, you didn't fill out your edit summaries, as you should). I'm also concerned that you have replaced the current process with your own proprietary process, that can be seen as advertising, and is not allowed on Wikibooks. I see that you have toned down your claims for remuneration, that's good. But, you do understand that you lose any clam to the process once you have submitted it to Wikibooks (as you have licensed it under the GFDL by the act of submitting it)? It will be modified without mercy. Additionally, you don't need to sign your work unless it's on a talk page. (Donovan|Geocachernemesis|Interact) 02:24, 27 Jun 2005 (UTC)

Hello Geocachernemesis, I just did not understand how my edits were changing back so fast.

I am struggling with the operations of this forum, computers are not my strong point. I did not see any edit summeries to fill out or anything else and I do not even know how to reply to you here properly, but I hope this reaches you. I finally found the registration feature and am now logged in. Believe it or not I got to this page by accident when I was "Googling" something else altogether different and I have not got a clue exactly "Where I am" now. I could never re-locate the page where I wrote to you the first time.

The Work I was editing is an early version of a Biodiesel production procedure which I developed. It is probably the single most popular method of making mini-batches of biodiesel in the English speaking Biodiesel World and is known throughout the biodiesel world as the "World Famous Dr Pepper Technique (Pat Pend)" I did not post it to this forum just like I have not posted it to many other forums across the internet where it appears. It makes me feel good to see that something I have developed is spreading all over the internet helping people.

Because this forum is editable I was updating the posting to include the latest changes to the procedure. The (Pat Pend) and reminder of royalties due after 53,459 litres production payable through your "Favourite Swiss Bank account" are probably tongue in cheek, but if some one wants to send me money... While I am flattered that people think so highly of my work that it has been posted here without asking my permission, I do think that as it is my work it should have my "Name" on it.

Tilly

I'm sorry if my reaction seemed harsh. It's nothing personal, but I'm simply trying to uphold Wikibooks policy. Unfortunately, your biodiesel recipe is original research and will probably not be allowed by Wikibooks, see What Wikibooks is not.

If you would like to learn more about Wikibooks, then here are some tips to help you get started:

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Good luck!

(Donovan|Geocachernemesis|Interact) 22:13, 28 Jun 2005 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Tilly. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Tilly~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:50, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:35, 19 April 2015 (UTC)

title: User talk:Tilly~enwikibooks


text:
User talk:68.112.230.211

Hello and welcome to Wikibooks!

Here are some tips to help you get started:

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(Donovan|Geocachernemesis|Interact) 03:13, 27 Jun 2005 (UTC)

title: User talk:68.112.230.211


text:
User talk:80.56.38.110

Pages too short
I see that you have been creating numerous empty pages containing only links to a corporate website. It is Wikibooks policy to immediately delete such pages. If their content is expanded dramatically, then they may stay. Please remember to complete you edit summaries for every edit that you make. (Donovan|Geocachernemesis|Interact) 03:34, 27 Jun 2005 (UTC)

title: User talk:80.56.38.110


text:
User talk:64.203.53.206

Hello and welcome to Wikibooks!

Here are some tips to help you get started:

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(Donovan|Geocachernemesis|Interact) 03:36, 27 Jun 2005 (UTC)

title: User talk:64.203.53.206


text:
GRE

The GRE (Graduate Record Exam) is required for entry into graduate programs at most institutions in the United States. As with most standardized tests, individuals who study and practice the test in advance can improve their scores by more than enough to justify the effort. This book will attempt to show the basic format of the GRE, how it is taken, how it is scored, and how to do well.

Table of Contents

 * /About the Test/
 * /Quantitative/
 * /Verbal/
 * /Analytical Writing/
 * /Subject Tests/
 * /Resources/
 * /How to start preparing for GRE/
 * /Frequently asked questions about GRE Preparation/
 * /What should be GRE goals/
 * /How to approach professor for funding/
 * /Preparing for GRE while you are working/
 * /Frequently asked questions on VISA process/
 * /Some questions about Credit Card/
 * /GRE Aspirations with a 3 year Degree/
 * /Taking GRE the Second time/
 * /Rescheduling your GRE Exam/
 * /All about education in USA/

title: GRE


text:
Cookbook:Buttermilk Fried Chicken with Pan Gravy

| Southern cuisine (USA) | Fried Chicken

Fried chicken is a Southern main meal in the USA, served often at "Sunday dinner" ("dinner" being the midday meal for many Southerners) or church "suppers" ("supper" being the term for the evening meal). Recipes are legion; every family has one handed down.

Chicken

 * 1 whole chicken
 * 2 cups buttermilk
 * 1 tsp mustard powder
 * 1 tsp cayenne powder
 * 2 tsp cracked black peppercorns
 * 1 tsp cinnamon
 * 1 tablespoon oregano
 * 1 tablespoon parsley
 * 1 tsp paprika
 * Flour seasoned with herbs and spices (e.g. pepper, cayenne, salt, ground coriander, cinnamon, thyme, oregano)
 * Oil (e.g. peanut or corn oil)

Gravy

 * Oil
 * 1 small onion (or shallots), chopped
 * 1 clove of garlic
 * 3–4 tablespoons flour
 * 3 cups chicken stock (or 2 cups stock + 1 cup cream)
 * Salt
 * Pepper

Chicken

 * 1) Cut the chicken into 8–10 pieces.
 * 2) Combine the buttermilk and spices, and marinate the chicken pieces for 4 hours in the mixture.
 * 3) Take each chicken piece out of the marinade and allow it to drip dry for a few seconds. Then, dredge it in seasoned flour. An easy way to dredge involves putting flour and seasonings in a small plastic bag, dropping the piece in, and shaking.
 * 4) Let the pieces rest (separately; do not stack) for about 10 minutes. Then dredge them again. The first dredging absorbs the buttermilk marinade. The second dredging gives a seasoned flour outer coat.
 * 5) Heat the oil in a deep, wide pan. When oil is smoking hot, drop chicken pieces into the pan, and be careful to not overcrowd (pieces should not touch; fry in multiple batches). After 3 or 4 minutes, turn the chicken, and reduce the heat. Let the chicken fry for 20–25 minutes, turning the pieces every few minutes. When the chicken is a deep golden-brown, and the juices run clear, it's done. The chicken will continue to cook after being removed from the pan, so err on the side of "just about done".
 * 6) Drain the chicken on absorbent paper or cloth. If you are not serving the chicken immediately, place it in a lightly-warmed oven, uncovered, separately (not stacked).

Gravy

 * 1) Pour off the accumulated oil from the pan, reserving 3–4 tablespoons and all the flour and chicken residue. Keeping the pan on medium flame, add the chopped onion and garlic. Sauté for a few minutes. Whisk in 3–4 tablespoons of flour, and keep stirring until the mixture becomes light to medium brown. Do not burn.
 * 2) Whisk in 3 cups of chicken stock. Simmer over low heat, covered, for 15 minutes. Salt and pepper to taste.
 * 3) Serve chicken with rice and some green vegetables. Spoon gravy over rice and chicken.

Notes, tips, and variations

 * The spice mix isn't fixed. You can use more or less as desired.


text:
User:Muhgcee~enwikibooks

my blog my pictures

Contact information:
 * email stu@fourmajor.com
 * Jabber muhgcee@jabber.org.uk
 * Skype fourmajor
 * AIM muhgcee at y
 * Secondary AIM muhgcee
 * Yahoo muhgcee
 * ICQ 3133801

title: User:Muhgcee~enwikibooks


text:
User talk:Muhgcee~enwikibooks

I don't want the link to this page to be red any more. . . so here we go. weeee

Abc123: Cheers for updating my bad grammar. Must be my Kiwi English coming through...

Abc123 Since when was my announcer not announcing or my book not being knowledge? I don't really see how Marcel Prouse and a map of 13th Century Poland enhance. I go back to what I originally said, if you wish to remove something then go ahead - but enhance rather than rip down to bare text.
 * Since when was my announcer not announcing or my book not being knowledge? I don't really see how Marcel Prouse and a map of 13th Century Poland enhance. What?  Muhgcee 05:21, 19 July 2005 (UTC)

Class Content Update
A course you are taking part in has new content available. Please visit the course page for more information.

Your account will be renamed
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The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Muhgcee. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Muhgcee~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

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Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:17, 19 April 2015 (UTC)

title: User talk:Muhgcee~enwikibooks


text:
Cookbook:Fried Okra

| United States | Southern U.S. cuisine

Fried okra means summer in the southern states of the USA. A quick mid-day meal; a side-dish at the evening meal—easy to make and always tasty.

Ingredients

 * Fresh okra
 * Cultured buttermilk or plain milk
 * Flour
 * Spices as desired (e.g. salt, pepper, cayenne, thyme, etc.)
 * Corn meal

Procedure

 * 1) Wash and drain a couple of handfuls of fresh okra.
 * 2) Cut each okra into three or four slices, about ¾–1 inch (2–2.5 cm) long. Discard the stem.
 * 3) Soak the okra in buttermilk or plain milk, drain, then dredge in seasoned flour and corn meal. A quick way to do this is to put ½ cup flour and 1 cup of corn meal in a plastic bag, add the spices you like, put in the okra, and shake.
 * 4) Let the okra rest a few minutes. If any pieces look soggy, dredge them again in a flour/corn meal mix.
 * 5) Deep fry the okra in hot oil until a light to medium brown. They're done in 2–3 minutes.
 * 6) Drain, and enjoy!

title: Cookbook:Fried Okra


text:
Cookbook:Garlic and Mustard Vinaigrette

| Salad Recipes

Vinaigrette (from French vinaigre, vinegar) is an emulsion of acid and fat, usually with other flavorings. It is one of the most basic salad dressings.

Ingredients

 * 1 clove of garlic, smashed
 * 2 tablespoons vinegar
 * 1 teaspoon prepared mustard
 * 6 tablespoons oil (vegetable, extra-virgin olive, almond, hazelnut, etc.)
 * Salt to taste
 * Freshly-ground pepper to taste
 * Herbs to taste

Procedure

 * 1) Mix garlic, salt, mustard, and vinegar until smooth.
 * 2) Add oil and mix until smooth.
 * 3) Add pepper and herbs to taste.
 * 4) Let stand; the longer the herbs and mustard soak in the oil, the better-blended the flavor will be. You may need to quickly remix the vinaigrette just before dressing the salad if the oil and vinegar have separated while standing.

Notes, tips, and variations

 * The classical ratio of vinegar to oil is 1:3. Some recipes, however, suggest a ratio as low as 1:5.
 * It is important to mix the mustard and vinegar first, then add the oil. This helps the emulsion hold better, resulting in a smooth vinaigrette that is slow to separate out. Adding the oil in a stream and mixing as it is added helps even more.
 * Use a pleasantly-flavored vinegar such as balsamic vinegar or red wine vinegar.
 * Mustard powder may be used instead of regular prepared mustard.
 * Instead of the garlic, a similar amount of very finely diced shallot may be used.
 * Toss the dressing with a salad just before serving. Greens that have been dressed wilt and become unappetizing within a few minutes.
 * Different oils can be used for different flavors. Some suggest the use of vegetable oil, but this tends to add little in the way of flavor and most would find it unsuitable. Extra-virgin olive oil is always a good choice, but nearly any plant oil can be used, and experimentation is always good.

title: Cookbook:Garlic and Mustard Vinaigrette


text:
User talk:62.252.192.6

Hello and welcome to Wikibooks!

Here are some tips to help you get started:

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Good luck!

(Donovan|Geocachernemesis|Interact) 12:00, 27 Jun 2005 (UTC)

title: User talk:62.252.192.6


text:
User:MyNameIsClare~enwikibooks

Wikipedia page

title: User:MyNameIsClare~enwikibooks


text:
Talk:Puzzles/Statistical puzzles/The Pentagon/Solution

The solution offered is not correct. There is a bigger chance that you will see 2 sides then 3. I SUSPECT the actual chance decreases the further you move away, approaching 50%, but I haven't calculated this (I haven't calculated anything, because the maths on this are pretty complex).

Now I hear you guys/girls shouting: WHERE IS YOUR PROOF? Well, very simple really: While I agree all the triangles have a 36° corner at their point, they do NOT share the same origin. Thus, a cross-section for a green triagle (3 side view) will be smaller then one from a yellow (2 side view) triangle at the same distance from the pentagon center. The author actually admits this: "When the distance from the Pentagon becomes large, the initial difference between the yellow and green sectors becomes negligible, and so the probability of being in a yellow sector (and seeing exactly two walls) is 50%"

The puzzle states that the distance should be larger then the 'only able to see 1 side' distance (it even says: "You are standing near The Pentagon."). To make this puzzle work it should state something like: You are looking at the Pentagon from a great distance. What is the probability that you are only seeing 2 sides.

Also, the last line offering 'proof' is clearly wrong ("If you see two sides of building, you friend at opposite side sees three of them. For each spot where you can see two sides, there is one spot where you can see three sides, and for each spot where you can see three sides, there is one spot where you can see two sides. This gives us exacly 50%."). Explanation: Alice takes position exactly left of the Pentagon (as seen in the diagram on the solution page), at a large distance (at least outside the 'red' part). She is in the yellow triangle. Bob takes position exactly right of the Pentagon (as seen in the diagram on the solution page), at a large distance (at least outside the 'red' part). He is in the yellow triangle. You can draw a straight line from Alice to Bob. Alice and Bob have found a spot where they both see 2 sides and are directly opposite each other.

ShadowLord 14:18, 27 Jun 2005 (UTC)

 * Oh, solution really is wrong. I just took it from some book, so i didn't really think if it correct or not. Then this thing is probably much more complex, and probably should be removed. --Divinity 19:31, 27 Jun 2005 (UTC)

The solution is somewhat correct. The problem is you're looking at a fraction of an infinite plane. If you start with a disk with as center the center of the pentagon, then indeed as the radius of the circle increases, the two fractions are the same (the seemingly extra bit of the yellow area is a zero-fraction of the plane, just like the pink areas). So the probability would be 50-50.

However, you could also look at an ellipse, instead of a disk. With the pentagon in on of the foci, and the other focus either in a green or red area. Now if you increase the size of the ellipse (along the large axis) and take the limit to infinity, then the probability gets skewed tremendously towards whatever area the second focus is in.

And that's just two possible approaches to find 'the' probability. Infinity is not a concept to mess around with indiscriminitely.

For fun compare this to taking the sum over all integer numbers. You can go 0 + 1 + -1 + 2 + -2 + 3 + -3 ... which has a cauchy limit of 0. But there is no reason why you couldn't do  0 + 1 + 2 + -1 + 3 + 4 + -2 ... which goes to infinity in the limit. Despite that in both cases you add every single number from the set once.

-- towr 17:44, 28 Jun 2005 (UTC)

Yes, the 50% probability is a limit of the probability as we choose random points in increasingly large circles centred at the Pentagon. If you have a better interpretation of "the probability", please share it.

--Borbrav 19:15, Jun 30, 2005 (UTC)
 * I'm not entirely convinced that is the only valid interpretation of 'the probability', although it certainly is the most natural, and probably the only one to give a sensible answer (but then not all good puzzles have answers, let alone sensible ones). If you want the probability for when you're at a given (be it unknown) distance, that I'd agree it's unambiguously 50%. But what the stochastic variable actually is in this case seems ambiguous to me. But maybe I'm just being daft :P -- towr

 * Actually, at any given fixed radius (larger than the red triangles) you have a probability of seeing two sides in excess of 50%, but this difference decreases as the circle's radius increases inversely to the radius. I too agree the puzzle is ambiguous. Borbrav July 6, 2005 04:43 (UTC)

Also, another thing in defense of the proof: look at the diagram from a very large distance above it. All you will see will be a point with five yellow and five green triangles, which will all appear identical.

--Borbrav 19:22, Jun 30, 2005 (UTC)


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Puzzles/Subdivisions/Subdividing a triangle

Consider an equilateral triangle.

 * You intend to subdivide it into 3 identical parts. How can you do so?
 * If the parts are to be triangles again, can you still find a solution?
 * Can you demonstrate all possible solutions and prove that you have found them all?


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Talk:Puzzles/Subdivisions/Subdividing a triangle

Untitled
How can I refer to a picture from wikipedia?

e.g.

I believe the answer is that you can't. You would have to upload it to Wikimedia Commons. When this is done, you can include it as if it's in the same wiki. As I've just done here. -- Smjg 16:58, 1 Jul 2005 (UTC)

Solving for three cannot be done. Solving for four is an easy job. All that takes is finding the midpoints of all the sides and connecting them to form 4 congruent triangles. —Preceding unsigned comment added by 68.255.27.230 (discuss • contribs) 23:02, 10 February 2006
 * It can be done. The simplest solution is to join each vertex of the triangle to the centre. -- Smjg (talk) 16:20, 9 September 2010 (UTC)


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User:Fenice~enwikibooks

hi!

title: User:Fenice~enwikibooks


text:
User:ZeiP

More information about me available in Finnish on my Finnish Wikipedia's user page.

title: User:ZeiP


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User:Daniel575~enwikibooks

'Note: I'm on vacation from July 4th until July 21th and will not be active on Wikibooks during this period.'

Name: Daniel

Location: The Netherlands

Occupation: Currently none

Interested in: Medicine, Judaism

Contact me: Through this page

title: User:Daniel575~enwikibooks


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Cookbook:Rice Wheat Spread

| Vegan Cuisine | Breakfast

A spread that consist of the two base ingredients rice and wheat (optionally oat flakes).

Ingredients

 * 400 g cooked rice
 * 150 g wheat flour
 * Margarine or rapeseed/other vegetable oil
 * Soy sauce (optional)
 * Mustard (optional)
 * Ketchup (optional)
 * Garlic (optional)
 * Pepper (optional)
 * Onions (optional)
 * Thyme (optional)
 * Seasoning mix (corn flour, salt, chili, onion, garlic, paprika; e.g. Mexican taco seasoning mix)

Procedure

 * 1) Mash the rice. Mix in the flour, margarine, and some water until you get a homogeneous mass.
 * 2) Mix in the optional ingredients to taste. You can also try to just add your favorite herbs and no other ingredients.
 * 3) Let rest to absorb the flavor of the herbs.

Kochbuch/ Brotaufstrich aus Reis und Weizen

title: Cookbook:Rice Wheat Spread


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Wikijunior:Solar System/Saturn/Titan

Titan is the largest moon of the planet Saturn. It is also the second largest moon in the solar system.

How big is Titan?
The moon Titan is 5150 km wide. It is the second largest moon in the Solar System. Only Ganymede, a moon of Jupiter, is larger. Titan is bigger than either the planet Mercury or dwarf planet Pluto. It is nearly one and a half times the size of the Earth's Moon, and has nearly twice the mass of the Moon. Titan is the only body besides Pluto, that we know of, that has a colder surface temperature because it has an atmosphere. The surface temperature of Titan is -180 degrees centigrade.

What is its surface like?
Titan is the only moon in the Solar System to have a thick atmosphere. Until the Cassini spacecraft arrived in orbit around Saturn, we did not know what the surface looked like. The ground is very cold on Titan, and all of the water is frozen. The surface looks like it is smooth, with areas of light and dark material. Most of the craters have been filled in.

It was thought that Titan may have seas of liquid methane, but it now looks like there may only be a few lakes. The Huygens probe was dropped into the atmosphere of Titan and slowly fell to the ground. This gave us our first close-up pictures of the surface. These showed what appear to be channels among hills that lead down to flat regions below. These may have been made by flows of liquid that left dark materials behind.

An atmosphere, complex molecules and the presence of methane on Titan mean that many people think there may be simple forms of life on or under Titan's surface.

How long is a day on Titan?
A day on Titan is the same as the time it takes for the moon to orbit around Saturn, about 15 days, 22 hours, and 41 minutes. The same side of Titan always faces toward Saturn, in much the same way as the same face of our Moon always points toward the Earth. This was caused by the gravity of Saturn creating a tidal force on Titan, which changed its rotation.

What is it made of?
The atmosphere of Titan is mostly nitrogen, with small amounts of other gases. Many of these gases are called hydrocarbons because they include hydrogen and carbon. These hydrocarbons are created in the upper atmosphere from Ultraviolet sunlight. Ultraviolet is the same light that causes sunburn on your skin. The hydrocarbons create a thick orange smog that hides the surface.

The moon is made of about half ice and half rocks. Most of this rock may be in a core at the center of the moon. The core may still be hot, as is the core of our own world, and there could be some volcanic activity. The surface is mostly ice.

How much would Titan's gravity pull on me?
The gravitational pull on Titan is about one seventh the Earth's gravitational pull. So if you weighed 100 pounds here you would weigh 14 pounds on Titan.

Who is it named after?
Titan was named after the race of children borne by Gaea and Uranus in Greek mythology.

When was it discovered?
It was discovered in 1655 by the Dutch astronomer Christiaan Huygens.

Wikijunior Sonnensystem/ Saturn/ Titan

title: Wikijunior:Solar System/Saturn/Titan


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User talk:62.252.64.14

Hello and welcome to Wikibooks!

Here are some tips to help you get started:

 * Remember to conduct any editing experiments in the sandbox.
 * Check out our help page if you need additional help.
 * Eventually, you might want to read the Manual of Style and Policies and Guidelines.
 * Remember to maintain a Neutral point of view.
 * Explore, be bold in editing pages, and have fun!

Good luck!

(Donovan|Geocachernemesis|Interact) 15:26, 27 Jun 2005 (UTC)

title: User talk:62.252.64.14


text:
User:Zellin

Zellin's main user page is located at w:User:Zellin.

title: User:Zellin


text:
User talk:Zellin

Zellin's main talk page is located at w:User talk:Zellin.

title: User talk:Zellin


text:
Japanese/Grammar/～ほど

Comparison

 * Only used for negative verbs.
 * Has a different nuance when used with positive verbs.

 * Examples:
 * バンクーバーは東京ほど暑くない.
 * Vancouver is not as hot as Tokyo.

It was ... to the extent that I could/couldn't/did/didn't ...

 * ほど
 * The condition (e.g., heat, noise) was so great, that it caused something else.
 * In "Making Sense of Japanese" by Jay Rubin, he compares it to the old joke of "It was so hot/cold/etc that ."

 * Examples:
 * 勉強ができないほど暑かった.
 * It was so hot that I couldn't study.

title: Japanese/Grammar/～ほど


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Talk:A-level Mathematics/OCR/C1/Indices and Surds

What do you do if the power isn't a rational number? For example, how is x to the power of root two defined?
 * The definition of irrational powers isn't required in C1, in fact I don't recall anywhere in A-Level Mathematics where it is formally taught. I can't remember the formal definition but $$x^{\sqrt{2}}$$ is the limit of $$x^1, x^{1.4}, x^{1.41}, x^{1.414}, x^{1.4142}\ldots$$.-- mattrix 21:30, 5 August 2005 (UTC)
 * Here's one formal definition: $$x^a = e^{\ln x^a} = e^{a\ln x}$$. That might not look any better but $$e^x$$ can be defined for any real $$x$$ without using irrational indices (for example, with the Maclaurin series, see Exponential_function). -- mattrix 21:43, 5 August 2005 (UTC)

rooth rule
The rule $$\sqrt{x\cdot y }=\sqrt{x}\cdot\sqrt{y}$$ only works if both x and y are non-negative, or one can easily "prove" that -1=1. Se http://en.wikipedia.org/wiki/Square_root

question
What if the indice is not 1/x to the power 3, but n/x to the power 3? what does the n do?

$$\frac{n^3}{x^3}$$ This is dictated by rule #6.Keytotime 20:23, 27 November 2006 (UTC)

Old introduction to indices.
I tried writing my own introduction as I found this one a little wordy.

You are probably already familiar with indices, for example $$x^2$$ is just a shorter way of writing $$x \times x$$ and $$x^4$$ is similarly $$x \times x \times x \times x$$. With $$x^5$$, $$x$$ is called the base and $$5$$ is called the power or exponent. $$x^4$$ is pronounced "x to the four", or "x raised to the 4th power" in full. Some powers are so useful that they have special names: $$x^2$$ is referred to as "x squared", $$x^3$$ is "x cubed" and $$x^{-1}$$ (which you will soon learn about if you haven't already encountered it) is called "the reciprocal of x".

Note: The "law of indices" is sometimes also called the "exponent laws" or "power rules". More generally, an index in mathematics is a superscript or subscript to a symbol.

Using this notation you might notice several patterns.

Conrad.Irwin 18:06, 29 November 2006 (UTC)

Division by Zero
Please be careful - x^4 / x is not x^3 if x is zero, as the division has undefined value. Simmilarly, x^2 times x^-1 is not x if x is zero, but rather is undefined for x^-1.

History???
i want to know who have founded these rules of indices??

Indices
Someone should solve this indices 8x-^3×3x^2/6x-^4 Thanks so much (discuss • contribs) 11:54, 31 January 2019 (UTC)


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User:Terrible Tim

I wrote the recipie for Corn Bread

I'm collecting a list of my favorite foods.

 * Grilled Cheese Sandwiches
 * Rigatoni pasta with Ricotta in Tomato Cream Sauce
 * Southwest Pasta
 * Tomato Pasta
 * Lasagne with bean sauce

title: User:Terrible Tim


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Geometry/Chapter 14

The Pythagorean Theorem
The Pythagorean Theorem shows the relationship between the sides (a and b) and the hypotenuse (c) of a right triangle. The right triangle I will be using is shown below. The Pythagorean Theorem states that, in a right triangle,the square of a (a²) plus the square of b (b²) is equal to the square of c (c²). $$a^2+b^2=c^2 $$

Summary: The Pythagorean Theorem is a²+b²=c², or leg² + leg² = hyp². It works only for right triangles.

Proof of the Pythagorean Theorem
Now that we know the Pythagorean Theorem, take a look at the following diagram. Look at the large square. The large square's area can be written as:

(a+b)(a+b)

or as:

(a+b)²

since each side's length is (a+b). Look at the tilted square in the middle. Its area can be written as:

c².

Now, look at each of the triangles at the corners of the large square. Each triangle's area is:

½ab

There are four triangles, so the area of all four of them combined is:

4(½ab)

The area of the large square is equal to the area of the four triangles plus the area of the tilted square. This can be written as:

(a+b)²=c²+4(½ab)

Using Algebra, this can be simplified.

(a+b)²=c²+4(½ab) (a+b)(a+b)=c²+2ab a2+2ab++b2=c²+2ab

-2ab     -2ab

a²+b²=c²

Now we can see why the Pythagorean Theorem works, or, in other words, we can see proof of the Pythagorean Theorem.

''However, this proof is not based on Euclidean Geometry. It is not elementary.''

There are thousands more proofs of the Pythagorean theorem, too.
 * You should be able to explain why the following is proof of the Pythagorean Theorem:

Summary: The Pythagorean Theorem can be proved using diagrams.

title: Geometry/Chapter 14


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Using GNOME/File manager home folder

The home folder contains a list of all the users on the computer with the exception of the root user. Therefore, if your username was, then your files and settings would be at the location. This ensures that each user has separate storage, also each user cannot see the others’ folder by default to prevent unauthorised access on public computers and the like.

title: Using GNOME/File manager home folder


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File:Similar1.PNG

Two similar triangles.

title: File:Similar1.PNG


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Geometry/Chapter 4

Translations, Reflections and Rotations
Before we continue, you need to know what Translations, Reflections, and Rotations are. Let us start with the following image of a triangle. The image above was translated a few units horizontally (along the x axis) and a few units vertically (along the y axis). The image above was flipped over the x axis (a horizontal line). The image above was rotated 90 degrees clockwise from a certain pivot point.
 * Translation means moving the image horizontally (along the x axis) or vertically (along the y axis).
 * Reflection means flipping the image either over the x axis (a horizontal line) or over the y axis (a vertical line).
 * Rotation means moving the image from a pivot point.

Notice that in all of the operations performed on the triangle above, none of the operations changed the angles of the triangle, or the lengths of any of the line segments. In all of the operations shown above, the only things that change are the location of the three points that make up the triangle.

Translations, reflections and rotations do not fundamentally change a shape. Terms for shapes that undergo any of these transformations are covered in the next section.

Congruence and Similarity
Intuitively, congruent shapes are shapes that are exactly the same. Technically speaking, two shapes are congruent if you can translate, rotate and/or reflect one of them in such a way that it coincides exactly with the other shape. Hence, a shape may be translated, reflected or rotated and remain congruent to its counterpart. These two triangles above are congruent, even though they are rotations of each other.

Similar shapes are shapes that, when scaled, are exactly the same. A shape may be translated, reflected or rotated and remain similar to its counterpart. In a sense, similar shapes are scale models of each other, that is, they are proportional. These triangles are similar because when one is scaled down, reflected, and rotated, it becomes congruent with the other.

Vocabulary

 * Congruent Shapes - Shapes that coincide exactly when translated, reflected, and/or rotated.
 * Similar Shapes - Shapes that coincide exactly when translated, reflected, rotated and/or scaled.

Exercises
1) On graph paper, draw a triangle with the points (0, 0), (0, 15) and (15, 0). Draw the triangle reflected over the Y axis (the vertical axis).  Draw the triangle translated up 25 units.

2) On graph paper, draw a square with the points (0, 0), (0, 15), (15, 15) and (15, 0). When you scale this shape by any amount, what happens to the point at (0, 0)?

3)On graph paper, draw a square with the points (0, 0), (0, 6), (6, 6) and (6, 0). When you scale this shape by 2, what happens to the point at (0, 6)?

title: Geometry/Chapter 4