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High School Mathematics Extensions/Counting and Generating Functions

Before we begin: This chapter assumes knowledge of
 * 1) Ordered selection (permutation) and unordered selection (combination) covered in Basic counting,
 * 2) Method of Partial Fractions and,
 * 3) Competence in manipulating Summation Signs

Some Counting Problems
..more to come

Generating functions
..some motivation to be written

To understand this section you need to see why this is true:

 * $$ \lim_{x \to \infty} \frac{x^2 + x}{x^2} = 1 + \lim_{x \to \infty}\frac{ 1}{x} = 1$$

For a more detailed discussion of the above, head to Infinity and infinite processes.

Generating functions, otherwise known as Formal Power Series, are useful for solving problems like:
 * $$x_1 + x_2 + 2x_3 = m$$

where
 * $$ x_n \ge 0$$; n = 1, 2, 3

how many unique solutions are there if $$m = 55$$?

Before we tackle that problem, let's consider the infinite polynomial:
 * $$S = 1 + x + x^2 + x^3 + ... + x^n + x^{n+1}... $$

We want to obtain a closed form of this infinite polynomial. The closed form is simply a way of expressing the polynomial so that it involves only a finite number of operations.

To find the closed form  we starting with our function:

$$S = 1 + x + x^2 + x^3 + ... $$

We multiply both sides of the function by x to get: $$ xS = x + x^2 + x^3 + ... $$

Next we subtract S-xS to get

S - xS = 1 + x + x^2 + x^3 ... - x - x^2 - x^3 ...

Grouping like terms we get

(1 - x)S = 1 + (x - x) + (x^2 - x^2) + (x^3 - x^3)

Which simplifies to

(1 - x)S = 1

Dividing both sides by $$\frac{1}{1 - x}$$ we get $$ S = \frac{1}{1 - x} $$

So the closed form of
 * 1 + x + x2 + x3 + ...

is

\frac{1}{1 - x} $$

For convenience we can write, although this isn't true for any particular value of x.

1 + x + x^2 + x^3 + ... = \frac{1}{1 - x} \ ; \ -1 < x < 1 $$

info - Infinite sums
 The two expressions are not equal. It's just that for certain values of x (-1 < x < 1), we can approximate the right hand side as closely as possible by adding up a large number of terms on the left hand side. For example, suppose x = 1/2, RHS = 2; we approximate the LHS using only 5 terms we get LHS equals 1 + 1/2 + 1/4 + 1/8 + 1/16 = 1.9375, which is close to 2, as you can imagine by adding more and more terms, we will get closer and closer to 2.

Anyway we really only care about its nice algebraic properties, not its numerical value. From now on we will omit the condition for equality to be true when writing out generating functions.

Consider a more general case:

S = A + ABx + AB^2x^2 + AB^3x^3 + ... $$ where A and B are constants.

We can derive the closed-form as follows:

\begin{matrix} S &=& A + &ABx + AB^2x^2 + AB^3x^3 + ... \\ BxS &=& &ABx + AB^2x^2 + AB^3x^3 + ... \\ \\ (1 - Bx)S &=& A \\ S &=& \frac{A}{1 - Bx} \end{matrix} $$

The following identity as derived above is worth investing time and effort memorising.

A + ABx + AB^2x^2 + AB^3x^3 + ... = \frac{A}{1 - Bx} $$

Exercises
1. Find the closed-form:
 * (a)$$ 1 - z + z^2 - z^3 + z^4 - z^5 + ... $$

 * (b)$$ 1 + 2z + 4z^2 + 8z^3 + 16z^4 + 32z^5 + ... $$

 * (c)$$ z + z^2 + z^3 + z^4 + z^5 + ... $$

 * (d)$$ 3 - 4z + 4z^2 - 4z^3 + 4z^4 - 4z^5 + ... $$

 * (e)$$ 1 - z^2 + z^4 - z^6 + z^8 - z^{10} + ... $$

2. Given the closed-form, find a function f(n) for the coefficients of xn:
 * (a)$$\frac{1}{1 + z} $$ (Hint: note the plus sign in the denominator)

 * (b)$$\frac{1}{1 - 2z} $$ (Hint: substitute A=1 and B=2 into $$\frac{A}{1 - Bz} $$)

 * (c)$$\frac{z}{1 - z} $$ (Hint: multiply all the terms in $$\frac{1}{1 - z} $$ by z)

Method of Substitution
We are given that:
 * 1 + z + z2 + ... = 1/(1 - z)

and we can obtain many other generating functions by substitution. For example: letting z = x2 we have:
 * 1 + x2 + x4 + ... = 1/(1 - x2)

Similarly
 * A + ABx + A(Bx)2 + ... = A/(1 - Bx)

is obtained by letting z = Bx then multiplying the whole expression by A.

Exercises
1. What are the coefficients of the powers of x:
 * 1/(1 - 2x3)

2. What are the coefficients of the powers of x (Hint: take out a factor of 1/2):
 * 1/(2 - x)

Linear Recurrence Relations
The Fibonacci series
 * 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...

where each and every number, except the first two, is the sum of the two preceding numbers. We say the numbers are related if the value a number takes depends on the values that come before it in the sequence. The Fibonacci sequence is an example of a recurrence relation, it is expressed as:

\begin{matrix} x_n &=& x_{n-1} &+& \ x_{n - 2}; \ \mbox{for n} \ge 2\\ x_0 &=& 1\\ x_1 &=& 1\\ \end{matrix} $$ where xn is the (n+ 1)th number in the sequence. Note that the first number in the sequence is denoted x0. Given this recurrence relation, the question we want to ask is "can we find a formula for the (n+1)th number in the sequence?". The answer is yes, but before we come to that, let's look at some examples.

Example 1
The expressions

\begin{matrix} x_n &=& 2x_{n-1}& + &3x_{n-2}; \ \mbox{for n} \ge 2\\ x_0 &=& 1\\ x_1 &=& 1 \end{matrix} $$ define a recurrence relation. The sequence is: 1, 1, 5, 13, 41, 121, 365... Find a formula for the (n+1)th number in the sequence.

Solution Let G(z) be generating function of the sequence, meaning the coefficient of each power (in ascending order) is the corresponding number in the sequence. So the generating functions looks like this

G(z) = 1 + z + 5z^2 + 13z^3 + 41z^4 + 121z^5 + ... $$ Now, by a series of algebraic manipulations, we can find the closed form of the generating function and from that the formula for each coefficient

\begin{matrix} & &G(z)   & = &x_0 + &x_1z  + &x_2z^2  &+ &x_3z^3 + x_4z^4 + x_5z^5 + ...\\ 2z&\times &G(z) & = &      &2x_0z + &2x_1z^2 &+ &...\\ 3z^2&\times &G(z)& = &     &        &3x_0z^2 &+ &...\\ \end{matrix} $$

\begin{matrix} G(z) - 2zG(z) - 3z^2G(z) &=& x_0 + (x_1 - 2x_0)z   & + \\ & & (x_2 - 2x_1 - 3x_0)z^2 & + \\ & & (x_3 - 2x_2 - 3x_1)z^3 & + &... \end{matrix} $$ by definition xn - 2xn - 1 - 3xn - 2 = 0

\begin{matrix} (1 - 2z - 3z^2)& \times & G(z)& =& x_0 + (x_1 - 2x_0)z\\ \\              &        & G(z)& =& \frac {1 - z} {1 - 2z - 3z^2}\\ \\              &        & G(z)& =& \frac {1 - z} {(1 - 3z)(1 + z)} \end{matrix} $$ by the method of partial fractions we get:

G(z) = \frac {1} {2} \times \frac {1} {1 - 3z} + \frac {1} {2} \times \frac {1} {1 + z} $$ each part of the sum is in a recognisable closed-form. We can conclude that:

x_n = \frac {1} {2} \times 3^n + \frac {1} {2} \times (-1)^n $$ the reader can easily check the accuracy of the formula.

Example 2

\begin{matrix} x_n &=& x_{n-1} &+& x_{n-2}& -& x_{n-3}; \ \mbox{for n} \ge 3\\ x_0 &=& 1 \\ x_1 &=& 1 \\ x_2 &=& 1 \end{matrix} $$ Find a non-recurrent formula for xn.

Solution Let G(z) be the generating function of the sequence described above.
 * $$G(z) = x_0 + x_1z + x_2z^2 + ...$$

\begin{matrix} G(z)(1 - z - z^2 + z^3) &=& x_0 &+& (x_1 - x_0)z + (x_2 - x_1 - x_0)z^2\\ G(z)(1 - z - z^2 + z^3) &=& 1 - z^2\\ \\ \end{matrix} $$

\begin{matrix} G(z) &=& \frac{1 - z^2}{1 - z - z^2 + z^3}\\ \\ G(z) &=& \frac{1 - z}{1 - 2z + z^2}\\ \\ G(z) &=& \frac{1} {1 - z} \end{matrix} $$ Therefore xn = 1, for all n.

Example 3
A linear recurrence relation is defined by:

\begin{matrix} x_n &=& x_{n-1}& + & 6x_{n-2} + 1; \ \mbox{for n} \ge 2\\ x_0 &=& 1\\ x_1 &=& 1\\ \end{matrix} $$ Find the general formula for xn.

Solution Let G(z) be the generating function of the recurrence relation.
 * $$G(z)(1 - z - 6z^2) = x_0 +  (x_1 - x_0)z + (x_2 - x_1 - 6x_0)z^2  +...$$
 * $$G(z)(1 - z - 6z^2) = 1   +  z^2  + z^3   + z^4 + ...$$
 * $$G(z)(1 - z - 6z^2) = 1   +  z^2(1 + z  +  z^2 + ...)$$
 * $$G(z)(1 - z - 6z^2) = 1   +  \frac{z^2}{1 - z}$$
 * $$G(z)(1 - z - 6z^2) = \frac{1 - z + z^2}{1 - z}$$

\begin{matrix} G(z) &=& \frac{1 - z + z^2}{(1 - z)(1 + 2z)(1 - 3z)}\\ G(z) &=& -\frac{1}{6(1-z)} + \frac{7}{15(1 + 2z)} + \frac{7}{10(1-3z)} \end{matrix} $$

Therefore
 * $$ x_n = -\frac{1}{6} + \frac{7}{15}(-2)^n + \frac{7}{10}3^n$$

Exercises
1. Derive the formula for the (n+1)th numbers in the sequence defined by the linear recurrence relations:

\begin{matrix} x_n &=& 2x_{n-1}& - &1; \ \mbox{for n} \ge 1\\ x_0 &=& 1 \end{matrix} $$

2. Derive the formula for the (n+1)th numbers in the sequence defined by the linear recurrence relations:

\begin{matrix} 3x_n &=& -4x_{n-1}& + & x_{n-2}; \ \mbox{for n} \ge 2 \\ x_0 &=& 1\\ x_1 &=& 1\\ \end{matrix} $$

3. (Optional) Derive the formula for the (n+1)th Fibonacci numbers.

Further Counting
Consider the equation
 * a + b = n; a, b ≥ 0 are integers

For a fixed positive integer n, how many solutions are there? We can count the number of solutions:
 * 0 + n = n
 * 1 + (n - 1) = n
 * 2 + (n - 2) = n
 * n + 0 = n
 * n + 0 = n

As you can see there are (n + 1) solutions. Another way to solve the problem is to consider the generating function
 * G(z) = 1 + z + z2 + ... + zn

Let H(z) = G(z)G(z), i.e.
 * H(z) = (1 + z + z2 + ... + zn)2

I claim that the coefficient of zn in H(z) is the number of solutions to a + b = n, a, b > 0. The reason why lies in the fact that when multiplying like terms, indices add.

Consider
 * A(z) = 1 + z + z2 + z3 + ...

Let
 * B(z) = A2(z)

it follows
 * B(z) = (1 + z + z2 + z3 + ...) + z(1 + z + z2 + z3 + ...) + z2(1 + z + z2 + z3 + ...) + z3(1 + z + z2 + z3) + ...
 * B(z) = 1 + 2z + 3z2 + ...

Now the coefficient of zn (for n ≥ 0) is clearly the number of solutions to a + b = n (a, b > 0).

We are ready now to derive a very important result: let tk be the number solutions to a + b = n (a, b > 0). Then the generating function for the sequence tk is
 * T(z) = (1 + z + z2 + ... + zn + ...)(1 + z + z2 + ... + zn + ...)
 * $$T(z) = \frac{1}{(1 - z)^2}$$

i.e.
 * $$\frac{1}{(1 - z)^2} = 1 + 2z + 3z^2 + 4z^3 + ... + (n+1)z^n + ...$$

Counting Solutions to a1 + a2 + ... + am = n
Consider the number of solutions to the following equation:
 * a1 + a2 + ... + am = n

where ai ≥ 0; i = 1, 2, ... m. By applying the method discussed previously. If tk is the number of solutions to the above equation when n = k. The generating function for tk is
 * $$T(z) = \frac{1}{(1-z)^m}$$

but what is tk? Unless you have learnt calculus, it's hard to derive a formula just by looking at the equation of T(z). Without assuming knowledge of calculus, we consider the following counting problem.

"You have three sisters, and you have n (n ≥ 3) lollies. You decide to give each of your sisters at least one lolly. In how many ways can this be done?"

One way to solve the problem is to put all the lollies on the table in a straight line. Since there are n lollies there are (n - 1) gaps between them (just as you have 5 fingers on each hand and 4 gaps between them). Now get 2 dividers, from the (n - 1) gaps available, choose 2 and put a divider in each of the gaps you have chosen! There you have it, you have divided the n lollies into three parts, one for each sister. There are $$ n - 1 \choose 2 $$ ways to do it! If you have 4 sisters, then there are $$ n - 1 \choose 3 $$ ways to do it. If you have m sisters there are $$ n - 1 \choose m - 1$$ ways to do it.

Now consider: "You have three sisters, and you have n lollies. You decide to give each of your sisters some lollies (with no restriction as to how much you give to each sister). In how many ways can this be done?"

Please note that you are just solving:
 * a1 + a2 + a3 = n

where ai ≥ 0; i = 1, 2, 3.

You can solve the problem by putting n + 3 lollies on the table in a straight line. Get two dividers and choose 2 gaps from the n + 2 gaps available. Now that you have divided n + 3 lollies into 3 parts, with each part having 1 or more lollies. Now take back 1 lollies from each part, and you have solved the problem! So the number of solutions is $$ n + 2 \choose 2 $$. More generally, if you have m sisters and n lollies the number of ways to share the lollies is
 * $$ {{n + m - 1} \choose {m - 1}} = {{n + m - 1} \choose {n}} $$.

Now to the important result, as discussed above the number of solutions to
 * a1 + a2 + ... + am = n

where ai ≥ 0; i = 1, 2, 3 ... m is
 * $${{n + m - 1} \choose {n}}$$

i.e.
 * $$\frac{1}{(1 - z)^m} = \sum_{i=0}^\infty {m + i - 1 \choose i}z^i$$

Example 1
The closed form of a generating function T(z) is
 * $$T(z) = \frac{z}{(1-z)^2}$$

and tk in the coefficient of zk is T(z). Find an explicit formula for tk.

Solution

\begin{matrix} \frac{1}{(1-z)^2} &=& \sum_{i=0}^\infty (i+1)z^i\\ \\ \frac{z}{(1-z)^2} &=& z\sum_{i=0}^\infty (i+1)z^i\\ \\                 &=& \sum_{i=0}^\infty (i+1)z^{i+1}\\ \end{matrix} $$ Therefore tk = k

Example 2
Find the number of solutions to:
 * a + b + c + d = n

for all positive integers n (including zero) with the restriction a, b, c ,d ≥ 0.

Solution By the formula

\begin{matrix} \frac{1}{(1-z)^4} &=& \sum_{i=0}^\infty {{n + 3}\choose {3}}z^i\\ \\ \end{matrix} $$ so
 * the number of solutions is $${{n + 3}\choose {3}}$$

More Counting
We turn to a slightly harder problem of the same kind. Suppose we are to count the number of solutions to:
 * 2a + 3b + c = n

for some integer $$n \ge 0$$, with a, b, also c greater than or equal zero. We can write down the closed form straight away, we note the coefficient of xn of:
 * $$(1 + x^2 + x^4 + ...)(1 + x^3 + x^6 + ...)(1 + x + x^2 + ...) = \frac{1}{(1 - x^2)(1 - x^3)(1 - x)}$$

is the required solution. This is due to, again, the fact that when multiplying powers, indices add.

To obtain the number of solutions, we break the expression into recognisable closed-forms by method of partial fraction.

Example 1
Let sk be the number of solutions to the following equation:
 * 2a + 2b = n; a, b ≥ 0

Find the generating function for sk, then find an explicit formula for sn in terms of n.

Solution

Let T(z) be the generating functions of tk
 * T(z) = (1 + z2 + z4 + ... + z2n + ...)2
 * $$T(z) = \frac{1}{(1 - z^2)^2}$$

It's not hard to see that
 * $$s_n = 0 \ \mbox{if n is odd}$$

 * $$s_n = {n/2 + 1\choose n/2} = {n/2 + 1\choose 1} = n/2 + 1 \ \mbox{if n is even}$$

Example 2
Let tk be the number of solutions to the following equation:
 * a + 2b = n; a, b ≥ 0

Find the generating function for tk, then find an explicit formula for tn in terms of n.

Solution

Let T(z) be the generating functions of tk
 * T(z) = (1 + z + z2 + ... + zn + ...)(1 + z2 + z4 + ... + z2n + ...)

 * $$T(z) = \frac{1}{(1 - z)} \times \frac{1}{1 - z^2}$$

 * $$T(z) = \frac{1}{(1 - z)^2} \times \frac{1}{1 + z}$$

 * $$T(z) = \frac{Az + B}{(1 - z)^2} + \frac{C}{1 + z}$$

 * A = -1/4, B = 3/4, C = 1/4

 * $$T(z) = -\frac{1}{4}\sum_{i=0}^\infty (i+1)z^{i+1} + \frac{3}{4}\sum_{i=0}^\infty (i+1)z^i + \frac{1}{4}\sum_{i=0}^\infty (-1)^iz^i$$

 * $$t_k = -\frac{1}{4}k + \frac{3}{4}(k + 1) + \frac{1}{4} (-1)^k$$

Exercises
1. Let
 * $$T(z) = \frac{1}{(1 + z)^2} $$

be the generating functions for tk (k = 0, 1, 2 ...). Find an explicit formula for tk in terms of k.

2. How many solutions are there the following equations if m is a given constant
 * $$a + b + 2c = m$$

where a, b and c ≥ 0

Problem Set
1. A new Company has borrowed $250,000 initial capital. The monthly interest is 3%. The company plans to repay $x before the end of each month. Interest is added to the debt on the last day of the month (compounded monthly).

Let Dn be the remaining debt after n months.

a) Define Dn recursively.

b) Find the minimum values of x.

c) Find out the general formula for Dn.

d) Hence, determine how many months are need to repay the debt if x = 12,000.

2. A partion of n is a sequence of positive integers (λ1,λ1,..,λr) such that λ1 ≥ λ2 ≥ .. ≥ λr and λ1 + λ2 + .. + λr = n. For example, let n = 5, then (5), (4,1), (3,2), (3,1,1), (2,2,1), (2,1,1,1), (1,1,1,1,1) are all the partions of 5. So we say the number of partions of 5 is 7. Derive a formula for the number of partions of a general n.

3. A binary tree is a tree where each node can have up to two child nodes. The figure below is an example of a binary tree.

a) Let cn be the number of unique arrangements of a binary tree with totally n nodes. Let C(z) be a generating function of cn.

 * (i) Define C(z) using recursion.

 * (ii) Hence find the closed form of C(z).

b) Let $$P(x)=\sqrt{1+ax}=p_0 + p_1 x + p_2 x^2 + p_3 x^3 ...$$ be a power series.

 * (i) By considering the n-th derivative of P(x), find a formula for pn.

 * (ii) Using results from a) and b)(i), or otherwise, derive a formula for cn.

Hint: Instead of doing recursion of finding the change in cn when adding nodes at the bottom, try to think in the opposite way, and direction.(And no, not deleting nodes)

Project - Exponential generating function
This project assumes knowledge of differentiation.

(Optional)0.
 * (a)

 * (i) Differentiate log x by first principle.

 * (ii)*** Show that the remaining limit in last part that can't be evaluated indeed converges. Hence finish the differentiation by assigning this number as a constant.

 * (b) Hence differentiate $$a^x$$.

1. Consider $$E(x) = e^x$$
 * (a) Find out the n-th derivative of E(x).

 * (b) By considering the value of the n-th derivative of E(x) at x = 0, express E(x) in power series/infinite polynomial form.

(Optional)2.
 * (a) Find out the condition for the geometric progression(that is the ordinary generating function introduced at the beginning of this chapter) to converges. (Hint: Find out the partial sum)

 * (b) Hence show that E(x) in the last question converges for all real values of x. (Hint: For any fixed x, the numerator of the general term is exponential, while the denominator of the general term is factorial. Then what?)

3. The function E(x) is the most fundamental and important exponential generating function, it is similar to the ordinary generating function, but with some difference, most obviously having a fractorial fraction attached to each term.

 * (a) Similar to ordinary generating function, each term of the polynomial expansion of E(x) can have number attached to it as coefficient. Now consider $$A(x) = a_1 + a_2 \frac{x}{1!} + a_3 \frac{x^2}{2!} + a_4 \frac{x^3}{3!} + ...$$

 * Find $$A'(x)$$ and compare it with A(x). What do you discover?

 * (b) Substitute nz, where n is a real number and z is a free variable, into E(x), i.e. E(nz). What have you found?

4. Apart from A(x) defined in question 2, let $$B(x) = b_1 + b_2 \frac{x}{1!} + b_3 \frac{x^2}{2!} + b_4 \frac{x^3}{3!} + ...$$

 * (a) What is A(x) multiplied by B(x)? Compare this with ordinary generating function, what is the difference?

 * (b) What if we blindly multiply A(x) with x(or xn in general)? Will it shift coefficient like what happened in ordinary generating function?

Notes: Question with *** are difficult questions, although you're not expected to be able to answer those, feel free to try your best.


text:
Dutch/Lesson 4

Fill in the blank 4-1-F

 * Peter zegt dat hij die stoelen en tafels echt heeft.
 * Peter wil Elly voor een etentje.
 * Zijn vader zegt: zul je bedoelen!
 * Zijn zoon aan de universiteit.

Grammatica 4-1 ~ The indefinite articles een en geen

 * ...chairs and a table ... stoelen en een tafel
 * ...has no money... heeft geen geld

In the previous lesson you were introduced to the definite articles—'the' in English and het or de in Dutch. Indefinite articles precede nouns in the same way that definite articles do, but convey a general or indefinite sense. These are 'a' or 'an' in English. Thus, 'the book' or het boek refers to a definite or specific book, whereas 'a book' or een boek is indefinite about which book is referred to. If your mother tongue does not have articles (such as Russian, Polish etc.) try this.

Dutch indefinite articles only come in one form (een), so they don't display gender.

The use of definite and indefinite articles is virtually the same as in English. The few deviations are best learned when listening to the language or speaking it.

Please note (see also previous lesson) that the indefinite article has the same form as the numeral one (één). One could argue that one is a clitic form of the other. To denote the difference, one could place accents on the numeral. Also, there is a difference in pronunciation. The numeral één (one) is pronounced /e:n/, the article een (a) with a much weaker /ən/. Occasionally Dutch has one and English the other:

 * op een middag - one afternoon

Notice that one is used here in the meaning of a certain, not say in contrast to two or three.

There is an inflected form ene that is used independently:

 * Occasionally Dutch has one, English the other.
 * Soms heeft Nederlands het ene, Engels het andere

Negation
In English a negative of an indefinite article is simply formed by adding not:

 * this is a car
 * this is not a car

Alternatively one can drop the article and say:

 * this is no car.

In Dutch there is a special negative of een:

 * dit is een auto
 * dit is geen auto.

Geen is used both with singular and with plural indefinite nouns
 * dat zijn eenden  -- those are ducks
 * dat zijn geen eenden -- those are not ducks

Relation to niet
The combination niet + een is only used in contrasting things:

 * dit is niet een fuut maar een eend.
 * this is not a grebe but a duck.

With definite nouns negation involves adding niet, usually at end:
 * dat is de auto
 * dat is de auto niet

 * Notice that geen is an article . It negates the noun.
 * By contrast niet is an adverb. It negates the verb.

Notice also that Dutch does not use the auxiliary verb to do for negations:

 * Hij zag de auto niet - he did not see the car
 * Hij zag geen eenden - he did not see any ducks

Relation to wel
When a negative is negated Dutch uses the adverb wel to express that. English has to use a construction involving the verb to do:

 * Heb je geen geld? -- Jawel, ik heb wel geld.
 * Don't you have money --  Yes, I do have money

This adverb is also used for contrasting:

 * Hij heeft geen auto, maar wel een motor.
 * He has no car, but he does have a motorbike.

Notice that the second part of the Dutch sentence does not even have a verb. Dutch is quite an 'adverbial' language. If the adverb expresses the meaning sufficiently, why bother with superfluous things like verbs?

A game of "Yes, you do" - "No, I don't" sounds like "Welles!" - "Nietes!" in Dutch

Much like in French, there are two words for yes. A simple confirmation is ja (French: oui); a negation of a negative is jawel (French: si)

Possessives
Recall the following from Gesprek 3-1:
 * Ja. En daarna breng je me op je motor naar huis.

Which translates as:
 * 'Yes. And after that you take me home on your motorcycle'.

The sentence demonstrates one of the possessive pronouns. In the singular these are
 * 'my', 'your', and 'his/her/its' in English
 * mijn, jouw and zijn/haar/(zijn) in Dutch.

The neuter zijn for its is not used very much in Dutch, as we shall see in lesson 8 it often gets replaced by "ervan".

The above pronouns like jouw often turn into a weak (clitic) form je that is used when the emphasis is on something else, such as the motorcycle in this case. In the spoken language this holds for all the ones shown in the table, but in the written language je is the most generally accepted clitic. When written they are:

 * m'n, je and z'n/d'r/(z'n)

Dutch does not have a possessive case as English does. In English one could say this house of mine, where mine (and yours, hers, his, ours, yours, theirs) is possessive case. Dutch uses objective case for this: dit huis van mij as if 'van' (of) is a preposition.

However, for a sentence like is this yours or his? Dutch would use nominalized pronouns (pronouns turned into nouns) with an inflection -e usually accompanied by a definite pronoun de or het:

 * Is dit het jouwe of het zijne? - Is this yours or his?

The possessive for the polite 2nd person pronoun u is uw and in the south this same word is used to refer to the gij pronoun (but no formal usage is implied). Uw does not have a clitic form and the same can be said about the possessives of the plural. Uw does have an inflected form uwe used for nominalization.

 * Dit is uw huis - this is your house
 * Dit is het uwe! - this is yours

For the third person plural the possessive is hun and it follows the same pattern:

 * Dat is hun auto - this is their house
 * Dat is het hunne! - this is theirs

For the first person plural Dutch has ons. This is the only one that follows the rules for inflection of the adjectives. I.e. its inflected form is used for de-words but not for het-words:

 * het huis - ons huis
 * de huizen - onze huizen.
 * de auto - onze auto

The inflected form is once again also used for nominalization:

 * Dat is het onze! - that is ours!

For the second person plural the possessive is jullie which cannot be inflected and cannot be used as a noun, which necessitates a construction involving van:

 * Dat is jullie huis
 * Dat is dat van jullie
 * Dat is jullie auto
 * Dat is die van jullie!

See Dutch/Appendix 3 for a table of the possessive pronouns.

Quizlet
Practice the possessive pronouns at Quizlet (24 term, includes demonstrative and reciprocal pronouns)

Demonstratives
In English, this is used as demonstrative pronoun to indicate something in proximity. That indicates greater distance. In Dutch a similar distinction exists, but gender plays a role:
 * de trein → deze trein - this train
 * het huis → dit huis - this house

So, one replaces 'de' by deze and 'het' by dit.

At a greater distance:

 * de trein → die trein
 * het huis → dat huis

Notice that often when English has th, Dutch will have d:

 * the - de
 * that - dat
 * think - denk

A third, even more distant pronoun exists (gene, gindse), but it is about as common as its English equivalent yon, yonder.

Again, the two languages betray their kinship. In some words, a g in Dutch corresponds to a y in English.. Compare:

 * gisteren - yesterday
 * de gist - the yeast
 * geel - yellow

Using demonstrative pronouns instead of personal pronouns
Recall:
 * Die heb ik echt nodig, hoor!

As we have seen Dutch is on its way to a two-gender system. For inanimate nouns, this makes demonstrative pronouns a more attractive choice to refer things by than personal pronouns. Compare:

As you see demonstratives do not distinguish whether a word is feminine or masculine and follow the same common-neuter pattern as the articles. Compare:

 * Ik zie Jan. Hij is sterk - I see John. He is strong.
 * Ik zie zijn auto. Die is duur. - I see his car. It is expensive.

Note: because de auto is not neuter, it is not correct to say: Het is duur. But saying hij is duur or zij is duur makes the word specifically masculine or feminine. Using die avoids the issue, because die follows the common gender pattern of the definite article.

Increasingly, personal pronouns are reserved for reference to persons (natural gender as in English). To refer to things people resort to substituting the demonstratives.

Reflexive and Reciprocal pronouns
In English reflexive pronouns always carry the ending -self -selves: myself, themselves etc. In Dutch that is not always so. In fact, for a verb that is always reflexive, like zich vergissen (to be mistaken) the ending cannot be used:

 * Ik vergis me
 * Jij vergist je, u vergist u/zich
 * Hij/zij vergist zich
 * Wij vergissen ons
 * Jullie vergissen je
 * Zij vergissen zich

In other words the reflexive pronoun is identical to the clitic object form of the personal pronoun, except in the third person where it is zich. The pronoun u was originally a third person (It stems from U.E. uwe edele, something like: your nobility, your honor) which explains the zich for this pronoun.

Vergissen can only be used with zich, but some verbs can be used with or without a reflexive pronoun. In that case -zelf may be added:

Ik was me/mij/mezelf/mijzelf. - I wash myself.

This topic is revisited in Lesson 16

The most important reciprocal pronoun is elkaar - each other
 * Zij ontmoeten elkaar - They met each other.

Grammatica 4-3 Plural of nouns
We already have seen some things about the plural above:
 * 1) the plural definite article is always de (for all genders),
 * 2) there is no indefinite article, but the negative geen can be used.
 * 3) the demonstrative pronouns are deze and die
 * 4) the personal pronoun is zij or its weak form ze.

Forming the plural of the noun itself is a bit more complicated. In English it is basically always done with -s, but in Dutch that is different

Recall: ...tafels en stoelen...

With few exceptions like ox - oxen pretty much all words simply get an -s in English. Dutch however has two main ways to form a plural: by adding -s and by adding -en. The latter is pronounced /-ən/, /-ə/ or even as a syllabic /-n/ depending on the region.

Which plural applies is best learned case by case as gender is, although we can attempt a general rule:

The ones in -a, -o, -i and -y get an apostrophe before the -s

 * baby - baby's

Unfortunately there are lots of exceptions. Many recent (latinate) loans from English or French and all diminutives get a -s.

 * de tafel - de tafels
 * de familie - de families
 * het meisje - de meisjes

Words in -te and -aar usually get -s:

 * de hoogte - de hoogtes
 * de kandelaar - de kandelaars

Amongst the many words that get -en are the ones in -ing:

 * de helling - de hellingen

Vowel changes
Most monosyllabic words have -en in the plural:

 * de stoel - de stoelen
 * het raam - de ramen

In the latter case, notice that one of the a's is dropped in the spelling of the plural. This difficulty is related to the fact that most Dutch vowels occur in two varieties, a closed one and an open one. Dutch spelling has a rather ingenious and systematic way of denoting which one is intended. It involves the doubling of either vowels or consonants. Compare:

 * het bot /bɔt/ (the bone) has an open vowel /ɔ/ like British pot (or American paw)
 * de boot /bot/ (the boat) sounds much like British boat.

In this case the vowels remain the same in the plural, but notice the doubling:

 * het bot - de botten ['bɔtə(n)] (bot-ten)
 * de boot - de boten ['botə(n)] (bo-ten)

It is customary to call the first sound [ɔ] a 'short o' and the second [o] a 'long o', but this terminology can be rather confusing. There are languages like Czech or Gàidhlig where vowels are indeed distinguished purely on their length. In Dutch, however, the difference in length (quantity) is actually pretty negligible, but the difference in vowel sound (quality) is not. This presents a problem for speakers of the many languages with a five-vowel system, like Italian, Russian, Arabic or isiXhosa whose ears are not accustomed to this kind of difference. Anglophones usually do quite well.

The following five vowels possess open ('long') and covered ('short') varieties:

Vowel changes
For non-native speakers a complication arises in those cases where the actual vowel changes ('lengthens') in the plural, compare:

 * dat pad (/pɑt/) - die paden (/'padən/ - vowel changes) (that path - those paths)
 * die pad (/pɑt/) - die padden (/'pɑdən/ - no vowel change) (that toad - those toads)

The vowel /ɑ/ in pad and padden is approximately as in father. Paden has a vowel /a/ like in broad American 'Oh, my God' (In Dutch the spelling would be: Gaad). Also, notice the gender difference of the two words.

Vowel change is systematic in the plural of the past of certain strong verbs (class 4 and 5; see 6).

 * ik zat (/zɑt/) - wij zaten (/zatən/) (I sat - we sat)

A few words show vowel changes other than between the open and closed variety of the same vowel:

 * de stad - de steden (city).
 * het schip - de schepen (ship)

Words ending in -heid get -heden:

 * beleefdheid - beleefdheden

-eren
There are about a dozen plurals in Dutch that end in -eren:

 * het kind - de kinderen (child - children)
 * het lam - de lammeren (lamb)

The ending -eren is essentially a double plural. It derives from a plural in -er and in some compounds that is still visible:

 * de kinderkamer - the children's room
 * de lammergier - a species of vulture

-ën
Some words in -ie have an -en plural that requires a diaeresis (trema in Dutch). The spelling depends on where the stress falls:

 * de kolónie - de kolóniën
 * de dynastíé - de dynastíéën

Notice that in Dutch orthography the stress of a word cán be indicated with an acute accent, but this is only permitted if otherwise ambiguity might arise.

A trema (diaeresis) is also used after -ee:

 * de zee - de zeeën
 * de diatomee - de diatomeeën

Changes of consonants
If the root of the word ends in a -z of -v this is both written and pronounced as -s and -f in the singular, but the voiced consonant returns in a plural on -en:

 * de baas - de bazen
 * de raaf - de raven
 * de duif - de duiven

Such words typically have a 'long' vowel (aa in this case) or diphthong.

With short vowel the consonant typically remains voiceless and is doubled in the plural

 * de plas - de plassen
 * de klos - de klossen

Lieden, lui
Some compounds of -man have a plural involving lieden or lui (people)
 * de timmerman - de timmerlieden, de timmerlui

Classical plurals
Occasionally a Latin or Greek plural is preserved in Dutch:

 * het museum - de musea
 * de chemicus - de chemici
 * de crisis - de crises
 * het dogma - de dogmata, dogma's

Quizlet
The vocabulary or this lesson can be practiced at Quizlet (28 terms)

Progress made
Cumulative count

Pronunciation Guide>>

 * Advanced Lesson 4>>

Quizlet
The vocabulary or this lesson can be practiced at Quizlet (28 terms)

Progress made
If you have thoroughly studied the above lesson you should
 * 1) have more knowledge of pronouns (poss. dem, reflex, and reciprocal)
 * 2) know how plurals are formed
 * 3) know the Dutch spelling rule that governs the doubling of vowels and consonants

Cumulative vocabulary count

 * Lesson 1+ : 226
 * Lesson 2+ : 161
 * Lesson 3+ : 89
 * Lesson 4: 52

 * Grand total: 528 terms

Pronunciation Guide>>

 * Advanced Lesson 4>>

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

title: Dutch/Lesson 4


text:
Talk:Cryptography/Introduction

Ist usage of the word 'Cryptography' occured in 1658 in Sir Thomas Browne's 'The Garden of Cyrus. ..quote ....'WE might abate the strange CRYPTOGRAPHY of Gaffarel in his Starrie Booke of heaven'.

Cryptanalysis Classes
Four classes of cryptanalysis;

A) Ciphertext-only. Only the ciphertext is known and the key and/or message must be found from this.

B) Known plaintext. The ciphertext and part of the message is known.  The key and/or the rest of the message must be found from this.

C) Chosen plaintext. The analysist can chose the plaintext to encrypt.  The key must be found from this.

D) Chosen ciphertext. Public key ciphers only, the analysist can chose the ciphertext to decrypt.  The key must be found from this.

Backup of some prior segments
While this could be left to page history, for now I'll leave some bits here so that people can see what major chunks were lifted out and replaced. wikipedia:User:PolyGnome

Prior intro, part 1
Cryptography is generally understood to be the study of the principles and techniques by which information is converted into an encrypted version that is difficult (ideally, impossible) for any unauthorized person to convert to the original information, while still allowing the intended reader to do so. In fact, cryptography covers rather more than merely encryption and decryption. It is, in practice, a specialized branch of information theory with substantial additions from other branches of mathematics. And perspective and approaches, because of the inherent conflict between those authorized, and those not, to see the information / interfere in the exchange / pretend to be other than they are / etc, from such sources as Machiavelli, Sun Tzu, and Karl von Clausewitz.

Unsurprisingly, the study of hiding the meaning of messages from others by encrypting them has been accompanied by the study of how to read such messages when one is not the intended receiver; this area of study is called cryptanalysis. People involved in such work, and with cryptography in general, are known as cryptographers (or for some of those in the other lexigraphic school, cryptologists).

The original unencrypted information being sent from one person (or organization) to another is usually called the plaintext. Encryption is the process of converting the plaintext into some unreadable form (ideally one undistinguisable from random gibberish), and decryption converts this back to the plaintext. Encryption includes two major classes of technique: encoding (yielding codetext) and enciphering (yielding ciphertext). The exact operation of the encryption and decryption, for all schemes with any pretense to security, is controlled by one or more keys.

My intent with the introduction is to try to cover the basic understanding and texture of cryptography without bogging down the reader. Although it will obviously have to be refined after the other modules have been completed. wikipedia:User:PolyGnome

Just a note
Would it be possible to cite a couple books on cryptography? There are several good ones, and I think it's a good idea to refer to something a little more concrete.

Interesting Typos
I discovered some interesting typos, such as "Bprivately" and the like, I don't know why they were there, but it reads better now, I also changed "cypher" in the body to "cipher", to match the introduction

Tyler.szabo 01:54, 6 October 2005 (UTC)

title: Talk:Cryptography/Introduction


text:
Talk:Cryptography/History

Ist usage of the word 'Cryptography' occured in 1658 in Sir Thomas Browne's 'The Garden of Cyrus. ..quote ....'WE might abate the strange CRYPTOGRAPHY of Gaffarel in his Starrie Booke of heaven'.

munitions
What is the best way to copy-and-paste the paper -- "Cryptographic techniques as munitions: US export controls and EU approaches" by TJ Atkins -- into this Cryptography/History? My understanding is that Atkins has already given us permission to do this by releasing with a "Creative Commons 0 License". --68.0.124.33 (talk) 01:18, 5 October 2010 (UTC)

title: Talk:Cryptography/History


text:
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title: User:Karl Wick/Main Page


text:
Talk:German/Appendices/Grammar II

Untitled
There is a tendency for contributors to want to rush ahead and give lots of information in tabular form. While I believe there is a place for this, the text of a lesson book should follow a gradual program of introducing simpler concepts first and then building on these. This page is now a good example. Verb conjugation needs to be spread out over many lessons, although the information and tables collected here make a good reference for the student to fall back on. The layout of this book (IMHO) should allow both approaches. Therefore, I am moving this page to an Appendix to serve as a resource for developing future lessons and as a permanent part of the book intact as a reference for the student. Hope I do not step on too many toes; both approaches are valuable, but different - Marsh 03:08, 10 Feb 2004 (UTC)~

I agree completely that such a clump of tabular information needs to be saved for an appendix, and the work already done is excellent for an introduction. My only concern is that the tabular format for most conjugations seems incompletely labeled; without labelings for each part of speech, I'm afraid that newcomers to the language may find these tables difficult to decipher.

I'm new wiki formatting, but I hope that the 'sein' conjugation chart I modified gives a sense for what I'm talking about. It's certainly not perfect: perhaps someone with more table skills could edit the other tables to include a better set of top-line identifiers. - Saxo, 12 Feb 2004

 * Sorry. I just picked up on your note. When I move pages around I do not always mark them to be watched! You are right.  I can handle the table formatting, but you did good - Marsh 02:59, 22 Feb 2004 (UTC)

title: Talk:German/Appendices/Grammar II


text:
German/Lesson 6

<< Lektion5 | Lektion 6 | Lektion 7 >>

Die Wohnung ~ The Apartment

Gespräch 6-1 ~ Ein Bruder besucht Markus

 * Markus studiert Biologie an der Universität. Er besucht die Vorlesungen und dann geht er nach Hause. Er wohnt nicht bei seinen Eltern; er mietet sich eine kleine Wohnung. Sie hat nur drei Zimmer. Gegen Abend zeigt er sie seinem Bruder.
 * Markus: Karl. Herein!
 * Karl:   Tag, Markus! Mutti grüßt dich.
 * Karl sieht sich um.
 * Karl:    Mir gefällt deine Wohnung.
 * Markus: Danke. Sie hat drei Zimmer. Es gibt eine Küche, ein Wohnzimmer, und ein Schlafzimmer.
 * Karl:   Ich habe sie gern!
 * Markus:

This incomplete story and conversation introduces terms for items around the house (or apartment).

Vokabeln 6-1
der Bruder               brother die Eltern               parents die Küche                kitchen das Schlafzimmer         bedroom die Vorlesung            class, instruction            (at a university) die Wohnung              apartment, flat das Wohnzimmer           living room das Zimmer, die Zimmer   room(s)

es gibt                  there is gegen Abend               towards evening gern haben               like                          (i.e., "to  gladly have") Herein! Come in! sich umsehen             look around

zeigen                   show besuchen                 visit, attend                 (classes) grüßen                   greet mieten                   rent

sein                     his                           (a possessive adjective)

Grammatik 6.1 ~ Introduction to verb conjugations
In German, every grammatical person has, or potentially has, its own unique verb form. Describing the various verb forms is called verb conjugation. This variation in verb form is certainly one of the things that makes German grammar somewhat difficult for English speakers to learn. In English, only the 3rd person singular might differ from the verb form used with all of the other persons (see Grammatik 1-3) and that difference is made by adding an ending of 's' or 'es'. For example: I/you/we/they 'go', but he/she/it 'goes'.

Let us have a closer look at German verbs. Usually, the infinitive form of a verb in German ends with -en—for examples, consider these verbs you have already learned: gehen ('go'), haben ('have'), and studieren ('study'). In order to "build" the different verb forms (that is, conjugate a verb), first cut off the '-en' ending from the infinitive. Then append a new ending according to the grammatical person. For regular verbs it works essentially as follows:

As you see in this example using the verb gehen, the singular 1st person ends with -e, the 2nd person with -st and 3rd person (no matter what gender) ends with -t. As for the plural forms, note that 1st and 3rd person in plural number (see Grammatik 1-3) are built the same way as the infinitive. Again note that, in English, only the verb form for the 3rd person singular is "unique". An easy way to remember the regular verb endings is the following mnemonic "Elephants standing together enjoy trumpeting endlessly".

Seems simple enough. However, realize we are discussing here only the regular verb forms in the present tense (Präsens). You will learn quite soon that, unfortunately, there are many exceptions from these simple rules. An important one is the irregular verb sein ('to be') which is irregular in English as well (I am, you are, he is...).

At least 1st and 3rd person plural are the same. Another important verb is haben ('to have'):

You see, it's not too irregular—only the 2nd and 3rd person singular constitute a small exception since the 'b' has vanished. English is somewhat curious in this respect as well: 'I have', but 'he has'. Future lessons will introduce you to the many irregular verbs in German. But you should now recognize what is happening to the verbs in German sentences. They are reflecting the person and number of their nominative case subjects. Recall these sentences from past lessons (verbs underlined here):

Danke, es geht mir gut                Thanks, it goes well with me (verb is gehen) Ich habe viel Arbeit                  I have much work (verb is haben)  Ist er zu Besuch?                     Is he visiting? (verb is sein) Du bist ein Schwein!                  You are a pig! (verb is sein) Wie heißen Sie?                       What are you called? (verb is heißen, and pronoun is formal) Wir spielen eine Stunde lang!         We play for one hour! (verb is spielen) Sie liegt am Ausfluss des Zürichsees. It lies at the outlet of Lake Zurich (verb is liegen)

Grammatik 6.2 ~ Case in German nouns
Through our discussions on the personal pronouns, you have learned how pronouns have case. Nouns also have case—and in German, noun case can be expressed by the definite article (der). Recall this table from Lektion 3:

These der-words reflect noun gender in the nominative case—appropriate whenever a noun is used as the subject of a sentence. For other cases, the der words change. Expanding the table to present nominative (NOM.), accusative (ACC.), dative (DAT.), and genitive (GEN.) cases:

Note, there are also der-word forms to be used for plural nouns. Fortunately, these are the same, no matter what the gender of the singular noun. For future reference, you can find the der-words summarized in Anhänge Drei.

The following examples demonstrate the use of the definitive article in various parts of speech:

Du hast die Wurst und den Käse.         You have the sausage and the cheese. (accusative case)  Die Geschäftsleute verstehen die Arbeit The business associates understand the work. (nominative and accusative cases) Sie liegt am Ausfluss des Zürichsees.   It lies at the outlet of ( the ) Lake Zurich. (genitive case) Zürich ist die größte Stadt der Schweiz. Zurich is the largest city in (of the ) Switzerland. (nominative and genitive cases)

In the last example, remember that in both English and German, the noun (or pronoun) that follows the verb 'to be' is a predicate noun, for which the correct case is the nominative. That is why, in English, 'It is I' is grammatically correct and 'It is me' is incorrect. So consider the following (and note that case of each definite article is the same as in the last example above):

Zürich ist der Kanton der gleichnamigen Stadt. Zurich is the canton of the same named city.

Grammatik 6.3 ~ Commands
Ruf sie an, bitte!        Call her, please. or Ruf sie bitte an! Gehen Sie nach Hause!     Go home (formal). Kommt mit!                Come with (plural)! Gib es mir!               Give me it!

Notice that in these sentences there are no subjects (except for #2). In German, as in English, there is a commandative form, a way to demand something using an understood you. In English, there is only one you-form and one command form. In German, since there are three you's, there are three ways to command.

If the subject is singular (du), then the verb has no ending. If it is irregular, it takes the du-form, such as in essen (Iss!) or lesen (Lies!). If there is a plural subject (ihr), then the verb takes the ihr-form. Nothing else is changed. Most of the time, ihr-commands are used with children, but that is not always the case. In both of these sentences, the du or ihr is omitted.

Formal is normal. The Sie stays (after the verb) and the verb is in its formal form. Although it is worded like a question, in written or spoken form, it is easy to tell the difference.

title: German/Lesson 6


text:
Talk:German/Lesson 6

Untitled
This Lesson 5 is a real conglomeration of pieces contributed by several of the contributors. Although it may (from the Page History) appear that Marshman did most of the work, this was definitely not the case. I think this is one of the best lessons to date (February 2004) and is actually mostly the work of Hansm and SamE, with Marshman structuring their contributions. Unfortunately, that structuring (moving stuff aroud) gives me more credit than I'm due. Thanks guys - Marsh 04:03, 24 Feb 2004 (UTC)

Okay, a minor local one (but a killer), a major local (text) and a major general (vocab):

1. "Gib mich es!" is shocking because it contradicts the rules for word order and indirect object from the previous lesson. Correct: "Gib es mir!"

2. I think we do have a problem with the vocabulary lists. All of them! They are absolutely correct. But only for a very specific context. For exampel: You can translate 'office' with German 'Sitz' - possibly. The banks have their main offices in Zurich and therefore it is their 'Sitz' (site of most important building). Now imagine a student on his fifth lesson in German doing his first free translation. He has to check up the German word for 'office' (in the sense of -white collar workspace-) to form a new sentence and turns to his true and trusted Wikibook vocabulary list. Let's say the sentence is "I will go to the office later today." -> "Ich werde später zum !Sitz! gehen." I don't think so. "Na, wie war es im !Sitz! heute?" No! The word he is looking for is 'Büro'.

 * That is the whole point. Each Vokabeln is NOT a partial German dictionary, but reflects exactly how the word or term is used in that lesson (in the material above the Vokabeln) with minor "the same" meanings. If the word is used later with a different meaning (or a different word is introduced with a similar meaning), then it is repeated in the new Vokabeln It is worth thinking about to have an appendix that does what you are suggesting. Each word in the lessons given as in a dictionary, wirth some more detail such as you give. - marsh 19:10, 22 Dec 2004 (UTC)

I don't think we should burden expressions with highly specific meanings before their standard interpretations have been established.

3. I wonder if the introductory text is too difficult. There is a lot of grammar in it that has not been introduced yet (passive, irregular verbs, complex side clauses). And lot of fillers muddy the waters.

 * I do agree to the extent that as much as possible the introductory passages should be simplified or (if to difficult) moved to later in the book. This is a slow, on-going process and would be a messy one if whole passages were just thrown out.  In the beginning, it is impossible not to have grammar that hass yet to be introduced. To do so would require very preschool, then first grade, then 2nd grade, etc. apperoach to including passages.  The point uis whether the student can understand a passage given the vocabulary presented; not whether he can understand the grammar behind each sentence. Much of the latter is picked up in the review process: sentence grammar in Lesson 1 is suddenly obvious after the stuident returns from having done Lessons 2-4. - marsh 19:10, 22 Dec 2004 (UTC)

I don't know what of these issues has been discussed before and what the consensus is. Help me out there, please!

 * Floflei6

Okay I will try to list some more doubts about the vocabulary

Take this one, No 5:

der Ausfluss           outlet, effluence (of a stream, river, or lake)  minor: only with lakes, with rivers it's 'Mündung'
 * I will correct

das Ende               end  now, is it 'to end' or 'an end'?
 * It is in the "noun" section so it must be a noun.

das Haus               house verb or noun?
 * It is in the the noun section so it has to be a noun

der Sitz               office see above

nach Hause             to home ''' 'to home' is an infinitive for an English verb. The phrase 'nach Hause' is idiomatic: "Wir gehen nach Hause" or "Ich will nach Hause". The correct English translation would be "We go home" and "I want to go home". It is not "We go to home" or "Let's go to home" or "Do you fly to home tonight?" and so on. Therefore, word for word translation in the lists is not very feasible. In those cases we need phrases to demonstarte proper usage!'''
 * This should be done like this: nach Hause = (to) home

anrufen                call, telephone '''Is 'telephone here an equivalent of anrufen or a modifier for 'to call'? Obviously the latter. So, why not "jemanden anrufen     ->     to call someone on the telephone, to call"'''

 * it is in the "verb" section, so it must be the verb to telephone.
 * You should read the Introduction to the book as it explains this set-up of the vocabularies. Also, Lesson 1 emphasizes this again.  - marsh 19:21, 22 Dec 2004 (UTC)

kommen                 come ''' 'come' could be commandative form. I think infinitives need to be marked as such. '''

liegen (lag, gelegen)  lie (lay, lain)

am (an dem)            at the ''' How are people to know that prepositions plus articles can be contracted because that's what they need to know to get the point here. '''

bei                    in ''' Nope, 'bei' means 'near' or 'close to'. Then again it can mean a lot of other things, none of them is close to English 'in'. '''

beiden                 two ''' What happened to 'both'? Can you reverse it? 'Two' does not mean 'beide'. '''

gleichnamigen          same named ''' I am not an English native-speaker but 'same named' sounds a bit weird. '''

klein                  small ''' I love that case, because we had 'klein' already in an earlier unit and it meant 'short'. What now? '''

neben                  besides ''' Can you realy use 'besides' spatially? Shouldn't it be 'next to'? '''

nördlichen             northern schweiz                Swiss  'die nördliche Schweiz'

I think the word for word vocabulary lists don't answer. We need to provide aditional information about function and use of expressions. Infinitives need to be marked as such. Nouns likewise. Words have to stripped of aspect, form and tense (not Dativ/Akkusativ 'beiden' but Nominaitiv 'beide'). If possible, translations should be mutually entailing and reversable - work both ways.

I think that's it for the moment.

 * Again, most of the problems you indicate come from your not reading the introduction and Lesson 1 examples of how the Vokabeln are set up. While everything you say is true, it would apply only if the Vokabeln were dictionary entries, which they are not. They were set up to be simple and not confuse the learning stuident with all of the other meanings, exceptions, etc.  As they are now is very straight-forward, simple, and useful for the lesson in which they appear. You are essentially taking them out of context, and then saying: "Look, there is no context here" The context is there.  Irt is just not obvious when you jump into Lesson 4 from having not read all that went before - marsh 19:21, 22 Dec 2004 (UTC)

 * Floflei6

"Mutti sagt Hallo." is nonsens! -> deleted --217.231.147.46 17:56, 24 Mar 2005 (UTC)
 * Such actions are not the least bit helpful. You can correct the statement to "good" German, but your calling it "nonsense" is nonsensical - marsh 04:28, 25 Mar 2005 (UTC)
 * sorry, but there is realy NO SENSE in this sentence. Do you want to say somthing like: "Ich soll dich von Mutti grüßen."?--217.231.132.9 06:58, 25 Mar 2005 (UTC)
 * Yes, but is there an easier way to say that (mine was just a direct English formulation, obviously) since at this point in the text modals are complications. - marsh 17:26, 25 Mar 2005 (UTC)
 * OK, what's about: "Mutti grüßt dich."? I wouldn't say this, but it's correct --217.231.137.143 23:25, 25 Mar 2005 (UTC)
 * Danke. We can use that until something better comes along - marsh

title: Talk:German/Lesson 6


text:
General Chemistry/Measurement

Moles
The mole is a unit used to measure the number of particles in a substance (where particles can be atoms, molecules, electrons, ions etc.). One mole of substance contains 6.02205 × 1023 particles. This rather large quantity is a constant called Avogadro's Number, which is abbreviated as NA. It simply refers to the number of particles in one mole.

$$N_A = 6.02214 \times 10^{23}\hbox{ mol}^{-1}$$

For example, one mole of helium gas contains 6.022×1023 atoms; 2 moles of oxygen gas contains 1.204×1024 molecules.

Moles and Mass
It would be useful to be able to convert between the number of moles and the mass of a substance. The molar mass is the mass of one mole of a substance, measured in grams per mole or $$\hbox{g}\cdot\hbox{mol}^{-1}$$. Not by coincidence, the molar mass is the same as the atomic weight of a substance (this is how the concept of moles was originally defined).

For example, the molar mass (and atomic weight) of carbon-12 is 12 g·mol−1, so 1 mole of carbon-12 weighs 12 grams. Similarly, chlorine has a molar mass of 35.45 g·mol−1, so one mole of chlorine atoms weighs 35.45 g. There are two chlorine atoms in one molecule of Cl2 gas, so one mole of Cl2 gas weighs 70.90 g.

Measurement and Quantities
See the Units Appendix for tables of SI fundamental and derived units.

Commonly used units in Chemistry
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Conversions (etc)
amount of substance mole, mol 1 mol = 6.022×1023 particles concentration (solutions) mole per litre, mol L-1 gram per litre, g L-1 (mol L-1)(g mol-1) = g L-1 (g L-1)/(g mol-1) = mol L-1 electric charge coulomb, C electric current ampere, A electromotive force volt, V energy joule, J kilojoule, kJ 1000 J = 1 kJ density gram per litre, g L-1 mass gram, g kilogram, kg ton 1 tonne = 1000 kg = 1×106 g molar volume litre per mole, L mol-1 22.41 L (S.T.P) 22.47 L (25 °C, 101.325 kPa) pressure kilopascal, kPa atmosphere, atm millimetre of mercury, mmHg 1 atm = 101.324 kPa = 760.0 mmHg temperature kelvin, K degree celsius, °C TK = TC + 273.15 time second, s hour, h 1 h = 3600 s volume litre, L millilitre, mL cubic centimetre, cm3 cubic decimetre, dm3 1 L = 1000 mL = 1 dm3 = 1000 cm3
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tera T 1012 giga G 109 mega M 106 kilo k 103 milli m 10−3 micro µ 10−6 nano n 10−9 pico p 10−12
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title: General Chemistry/Measurement


text:
Template:New

Newest Books category=Freshly started books namespace=Main suppresserrors=true order=ascending ordermethod=categorysortkey start a book • suggest a book

title: Template:New


text:
LPI Linux Certification/Use Debian Package Management

Detailed Objective
(LPIC-1 Version 5.0

Weight: 3

Description: Candidates should be able to perform package management using the Debian package tools.

Key knowledge areas:
 * Install, upgrade and uninstall Debian binary packages.
 * Find packages containing specific files or libraries which may or may not be installed.
 * Obtain package information like version, content, dependencies, package integrity and installation status (whether or not the package is installed).
 * Awareness of.

The following is a partial list of the used files, terms and utilities:

Package Structure
In order to understand how to use Debian's package management system, it would be useful to first have an understanding of how a Debian package is named. For example, the package ncftp_3.1.3-1_i386.deb has 5 major parts: Note that there is special significance to the use of underscores(_) and hyphens(-); an underscore shall separate the name of the program and its version, a hyphen shall separate a version number and the revision number, and an underscore shall separate the revision number and the architecture.
 * ncftp - the name of the program/application/library
 * 3.1.3 - the version of the program/application/library assigned by the original (upstream) author(s)
 * 1 - the revision number of the package assigned by the person(s) who packaged the program for a Debian system
 * i386 - the architecture the packaged program is designed to run on
 * .deb - signifies this is a Debian package

dpkg
dpkg is the "grandaddy" or "back-end" of the Debian Package Management System. Features present in the more advanced tools are not available to dpkg but it is nevertheless a useful tool.

Some notes:

 * dpkg keeps its record of available packages in /var/lib/dpkg/available.

Some of the more common functions used by administrators by dpkg are:

Adding, Removing, and Configuring Packages
 * dpkg {-i|--install}  will install the specified package
 * dpkg {-r|--remove}  will remove the specified package (but leave the configuration files intact)
 * <tt>dpkg {-P|--purge} <package-name></tt> will remove the specified package and the corresponding configuration files
 * <tt>dpkg --root /target -i </tt> will install a package into a unbootable system by specifying the system root.
 * <tt>dpkg --unpack <package-name></tt> will unpack (but do not configure) a Debian archive into the file system of the hard disk
 * <tt>dpkg --configure <package-name></tt> will configure a package that already has been unpacked

<tt>Querying Package Information</tt>
 * <tt>dpkg --info <package-name></tt> will print out the control file (and other information) for a specified package
 * <tt>dpkg {-l|--list}</tt> this will give you a list of installed packages.
 * <tt>dpkg {-a|--pending}</tt> is given instead of a package name, then all packages unpacked, but marked to be removed or purged in file /var/lib/dpkg/status, are removed or purged, respectively.
 * <tt>dpkg -s (--status)</tt> <package-name> will give you a description of installed package

<tt>Updating Package Information</tt>
 * <tt>dpkg --update-avail <package-name></tt> will replace old information with with new information from package.
 * <tt>dpkg --merge-avail <package-name></tt> will combine old information with new information from package.

dpkg-reconfigure
<tt>dpkg-reconfigure</tt> reconfigures packages after they have already been installed.

 * <tt>dpkg-reconfigure <package-name></tt> to reconfigure the initial installation settings
 * <tt>dpkg-reconfigure --priority=medium package [...]</tt> will set the minimum priority of question that will be displayed
 * <tt>dpkg-reconfigure --all</tt> will reconfigure all packages
 * <tt>dpkg-reconfigure locales</tt> will generate any extra locales
 * <tt>dpkg-reconfigure --p=low</tt> xserver-xfree86 will reconfigure X server

Dselect
The utility that allows you on debian to easely add/remove packages is dselect.
 * Choose the access method to use.
 * Update list of available packages, if possible.
 * Request which packages you want on your system.
 * Install and upgrade wanted packages.
 * Configure any packages that are unconfigured.
 * Remove unwanted software.

There is on dselect an interactive menu that will allow you to install/remove packages. Care must be taken with this utility. You can damage your system.

Dselect menu example: Debian `dselect' package handling frontend.

0. [A]ccess   Choose the access method to use. 1. [U]pdate   Update list of available packages, if possible. 2. [S]elect   Request which packages you want on your system. 3. [I]nstall  Install and upgrade wanted packages. 4. [C]onfig   Configure any packages that are unconfigured. 5. [R]emove   Remove unwanted software. 6. [Q]uit     Quit dselect.

$ dselect - list of access methods Abbrev. Descriptio cdrom          Install from a CD-ROM. * multi_cd      Install from a CD-ROM set. nfs           Install from an NFS server (not yet mounted). multi_nfs     Install from an NFS server (using the CD-ROM set) (not yet mounted). harddisk      Install from a hard disk partition (not yet mounted). mounted       Install from a filesystem which is already mounted. multi_mount   Install from a mounted partition with changing contents. floppy        Install from a pile of floppy disks. apt           APT Acquisition [file,http,ftp]

apt-get
If you know the name of a package you want to install, use apt-get. You must configure the sources.list file. This same file is used when you choose the apt access method of dselect. The location is <tt>/etc/apt</tt>.

 * <tt>apt-get install <package-name></tt> will search its database for the most recent version of this package and will retrieve and install it from the corresponding archive as specified in sources.list. In the event that this package depends on another APT will check the dependencies and install the needed packages.
 * <tt>apt-get install <package-name>=<version></tt> will install a package at the version specified
 * <tt>apt-get install <package-name> -o DPkg::options::="--force-overwrite"</tt> installs a package ignoring "error processing ..., which is also in package ..." errors.
 * <tt>apt-get remove <package-name></tt> will remove the specified package but keep its configuration files.
 * <tt>apt-get --purge remove <package-name></tt> will remove the specified package and its configuration files.
 * <tt>apt-get -u install <package-name></tt> will upgrade and install a specific package.
 * <tt>apt-get -u upgrade</tt> will upgrade packages within the same distribution packages except those which have been kept back because of broken dependencies or new dependencies.
 * <tt>apt-get -u dist-upgrade</tt> will upgrade an entire Debian system at once.

apt-file

 * <tt>apt-file search <file-name></tt> will search for a package which includes the specified file.
 * <tt>apt-file list <package-name></tt> will list the contents of a package matching the pattern. This action is very close to the dpkg -S command except the package does not need to be installed or fetched.

Apt-cache
To find the name of a package that you want to install use apt-cache. apt-cache main options are :

 * add - Add a package file to the source cache
 * showpkg - Show some general information for a single package
 * stats - Show some basic statistics
 * search - Search the package list for a regex pattern
 * show - Show a readable record for the package
 * depends – Show raw dependency information for a package

user@host:~$ apt-cache search gimp babygimp - An icon editor in Perl-Tk blackbook - GTK+ Address Book Applet cupsys-driver-gimpprint - Gimp-Print printer drivers for CUPS escputil - A maintenance utility for Epson Stylus printers filmgimp - A motion picture editing and retouching tool

Resources
APT HOWTO http://www.debian.org/doc/manuals/apt-howto/index.en.html dselect Documentation for Beginners http://www.debian.org/doc/manuals/dselect-beginner/

Exercises

 * Exercises results

 * 1) Install a system with Debian.
 * 2) Get familiar with dselect and remove the tcpdump utility.
 * 3) Install back with apt-get the package that contains the tcpdump utility.
 * 4) Try kpackage to install ethereal


text:
Trigonometry/Law of Sines

For any triangle with vertices $$A,B,C$$ corresponding angles $$A,B,C$$ and corresponding opposite side lengths $$a,b,c$$, the Law of Sines states that
 * $$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$$

Each of these expressions is also equal to the diameter of the triangle's circumcircle (the circle that passes through the points $$A,B,C$$). The law can also be written in terms of the reciprocals:
 * $$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$$

Proof

Dropping a perpendicular $$OC$$ from vertex $$C$$ to intersect $$AB$$ (or $$AB$$ extended) at $$O$$ splits this triangle into two right-angled triangles $$AOC$$ and $$BOC$$. We can calculate the length $$h$$ of the altitude $$OC$$ in two different ways:

 * Using the triangle AOC gives
 * $$h=b\sin(A)$$ ;

 * and using the triangle BOC gives
 * $$h=a\sin(B)$$.

 * Eliminate $$h$$ from these two equations:
 * $$a\sin(B)=b\sin(A)$$.

 * Rearrange to obtain
 * $$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}$$

By using the other two perpendiculars the full law of sines can be proved. QED.

Application
This formula can be used to find the other two sides of a triangle when one side and the three angles are known. (If two angles are known, the third is easily found since the sum of the angles is $$180^\circ$$ .) See Solving Triangles Given ASA. It can also be used to find an angle when two sides and the angle opposite one side are known.

Area of a triangle
The area of a triangle may be found in various ways. If all three sides are known, use Heron's theorem.

If two sides and the included angle are known, consider the second diagram above. Let the sides $$b$$ and $$c$$, and the angle between them $$\alpha$$ be known. The terms /alpha and /gamma are variables represented by Greek alphabet letters, and these are commonly used interchangeably in trigonometry just like English variables x, y, z, a, b, c, etc. From triangle $$ACO$$, the altitude $$h=CO$$ is $$b\sin(\alpha)$$ so the area is $$\frac{bc\sin(\alpha)}{2}$$.

If two angles and the included side are known, again consider the second diagram above. Let the side $$c$$ and the angles $$\alpha$$ and $$\gamma$$ be known. Let $$AO=x$$. Then
 * $$\frac{x}{h}=\tan(\alpha)\text{ ; }\frac{c-x}{h}=\tan(\gamma)\text{ ; adding these, }\frac{c}{h}=\tan(\alpha)+\tan(\gamma)$$

Thus
 * $$h=\frac{c}{\tan(\alpha)+\tan(\gamma)}\text{ so area }=\frac{c^2}{2\bigl(\tan(\alpha)+\tan(\gamma)\bigr)}$$.

Matemática elementar/Trigonometria/Lei dos senos e dos cossenos

title: Trigonometry/Law of Sines


text:
Trigonometry/Law of Cosines

Law of Cosines

The Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:
 * $$a^2+b^2-2ab\cos(\theta)=c^2$$

where $$\theta$$ is the angle between sides $$a$$ and $$b$$.

Does the formula make sense?
This formula had better agree with the Pythagorean Theorem when $$\theta=90^\circ$$.

So try it...

When $$\theta=90^\circ$$, $$\cos(\theta)=\cos(90^\circ)=0$$

The $$-2ab\cos(\theta)=0$$ and the formula reduces to the usual Pythagorean theorem.

Permutations
For any triangle with angles $$A,B,C$$ and corresponding opposite side lengths $$a,b,c$$, the Law of Cosines states that
 * $$a^2=b^2+c^2-2bc\cdot\cos(A)$$
 * $$b^2=a^2+c^2-2ac\cdot\cos(B)$$
 * $$c^2=a^2+b^2-2ab\cdot\cos(C)$$

Proof

Dropping a perpendicular $$OC$$ from vertex $$C$$ to intersect $$AB$$ (or $$AB$$ extended) at $$O$$ splits this triangle into two right-angled triangles $$AOC$$ and $$BOC$$, with altitude $$h$$ from side $$c$$.

First we will find the lengths of the other two sides of triangle $$AOC$$ in terms of known quantities, using triangle $$BOC$$.

 * $$h=a\sin(B)$$

Side $$c$$ is split into two segments, with total length $$c$$.
 * $$\overline{OB}$$ has length $$\overline{BC}\cos(B)=a\cos(B)$$
 * $$\overline{AO}=\overline{AB}-\overline{OB}$$ has length $$c-a\cos(B)$$

Now we can use the Pythagorean Theorem to find $$b$$, since $$b^2=\overline{AO}^2+h^2$$.

 * $$b^2$$
 * $$=\bigl(c-a\cos(B)\bigr)^2+a^2\sin^2(B)$$
 * $$=c^2-2ac\cos(B)+a^2\cos^2(B)+a^2\sin^2(B)$$
 * $$=a^2+c^2-2ac\cos(B)$$
 * }
 * $$=a^2+c^2-2ac\cos(B)$$
 * }
 * $$=a^2+c^2-2ac\cos(B)$$
 * }

The corresponding expressions for $$a$$ and $$c$$ can be proved similarly.

The formula can be rearranged:
 * $$\cos(C)=\frac{a^2+b^2-c^2}{2ab}$$

and similarly for $$cos(A)$$ and $$cos(B)$$.

Applications
This formula can be used to find the third side of a triangle if the other two sides and the angle between them are known. The rearranged formula can be used to find the angles of a triangle if all three sides are known. See Solving Triangles Given SAS.

title: Trigonometry/Law of Cosines


text:
Talk:Chinese (Mandarin)/Lesson 7

Untitled
England is translated as 英国 or "country of the brave"? I don't know that! i think it's transliterated from England ['ing-]. it may be an abbreviation for 英格兰 or 英吉利 or something. --Yacht (talk)Q 02:28, 11 Feb 2004 (UTC)

 * Well, I suppose the transliteration serves both the phonetic and semantic purposes, but you bring up a good point. --Taoster

Vocabulary list and pinyin
1. The vocabulary list looks different to lesson 1 and 2, they should the same, i would change it, but we should choose one or the other a standard, please some suggestions. My suggestion would be the style of lesson 1 and 2.

2. Pinyin, sometime with numbers for the tones, sometimes with tone marks, i think the tone marks are better. Suggestions are welcome here as well.

title: Talk:Chinese (Mandarin)/Lesson 7


text:
Computer Programming/Database Programming

Introduction
A database is a repository of information managed by a database engine which ensures integrity of data and fast access to the data.

A very common implementation of a database is a Relational Database Management System (RDBM). To users, the information in a database can be accessed by using Structured Query Language (SQL) a database language common to most databases. However, SQL only provides commands to access and manipulate the data in the database.

For any complex application, there commonly is a need for conditions, branching, and loops. None of these are a part of the SQL language. In order to fill this gap, many common programming languages allow integration with SQL through a common library such as Open Data Base Connectivity (ODBC), Object Linking and Embedding (OLE), and sometimes with application programming interfaces or libraries supplied with the database.

In addition, most databases now have a simple language of their own which allows simple control for applications which do not need the full power of standard languages like C++ and Pascal. These simple languages are used to write stored procedures and are proprietary to each database.

An alternate approach taken by other languages like MUMPS is to have a data model that includes persistent variables that are shared among multiple processes. This approach hides the database operations within the programming language instead of hiding the language within the database.

Stored procedure languages

 * Oracle PL/SQL Programming


text:
LPI Linux Certification/LPI Permission

Date: Fri, 16 Jan 2004 14:33:06 -0500 From: "Wilma Silbermann"  To: "m strawser"  Subject: Re: reprinting of exam objectives

Hi there Mark,

We are pleased that you are creating more training material for the LPI exams and would have no problem approving your request to reprint the objectives as long as you ensure you show that LPI does not approve or endorse your book in any way.

You may use the logo in the same manner - please read the logo license;

https://www.lpi.org/en/logo.html

Please give me a call if you have any further concerns or questions on this or any other matter.

Thanks again for your support of LPI.

Wilma

title: LPI Linux Certification/LPI Permission


text:
User:Thomas~enwikibooks

Ich bin jetzt als Thomas Gehrlein im Wiki unterwegs.

title: User:Thomas~enwikibooks


text:
Talk:Puzzles/Decision puzzles/Monty Hall/Solution

The solution is wrong! Moreover, it's obvious for any mathematician...

If something's wrong, by all means, fix it. Be bold! :) 129.94.6.30 22:19, 11 Feb 2004 (UTC) (User:Dysprosia, logged out)

I concur, the solution is incorrect. This is basic Probability. Consider:
 * goat
 * goat
 * prize

No matter which door you select, a door with a goat will be revealed. The fact that the door opened is one other than you choose is irrelevant. The remaining doors contain:
 * goat
 * prize

While 2 of the 3 original doors remain, the odds of selecting the prize from the two remaining doors is 1:2 NOT 2:3.

However, if the chosen door is revealed you effectively have two chances to select the prize from the three doors -- the odds are 2:3. The odds of selecting the prize from the initial round is 1:3.

--Eibwen 19:30, 25 Sep 2004 (UTC)

The solution is right, you are wrong
Your reasoning is flawed. Read the explanation again. For that matter, did either of you try doing a monte carlo experiment and seeing for yourself?

If this makes it clearer, suppose you choose door A. Then:

Star prize is in   If you stick    If you switch A                  Star prize      Goat B                  Goat            Star prize C                  Goat            Star prize

The same applies whichever door you pick. -- Smjg 11:25, 20 Oct 2004 (UTC)

Elaboration
There are four (and only 4) permutations for a given door. To illustrate, suppose you choose door A:

Prize Door  Revealed Door   Contents   Remaining Doors   If you stay  If you switch   Prize recieved by A            B               Goat       A, C              A - Prize    C - Goat        Staying C              Goat       A, B                           B - Goat        Staying B           C               Goat       A, B              A - Goat     B - Prize       Switching C           B               Goat       A, C              A - Goat     C - Prize       Switching

Independent of whichever door is selected only two doors remain, of which only one contains the prize, 1:2. Moreover, the odds of recieving the prize by either staying (2) or switching (2) relative to the number of permutations (4) is 2:4 or 1:2.

While there are 4 unique scenarios, staying or switching for any given door yield the same result. Consequently, two of the four scenarios are identical and the practical odds are 2:3 in favor of switching.

Both odds are valid, but they reference the outcome from different perspectives. 2:3 is the practical odds and is the answer given the context of the question. However, both the question and the solution could be further clarified to prevent this discrepancy while maintaining the integrity of the puzzle.

--Eibwen 01:58, 2 Nov 2004 (UTC)

 * What matters is not so much which possibilities are the same as their relative probabilities. Though statement 3 below does depend on the symmetry in B and C, and hence the equivalence of the first two....

 * The key statements are:
 * The prize is equally likely to be in A, B or C.
 * Hence once a door is picked, the probability of that door having the prize is 1/3.
 * Monty opening a door cannot alter the fact of statement 2, since his opening a door you didn't choose is unconditional.

 * Maybe the solution can be improved. But how would you improve the question?
 * -- Smjg 18:27, 2 Nov 2004 (UTC)

A demonstration
I've just written this C program for the doubters among you to try for yourself:

int main { const int ATTEMPTS = 10000; int stickWins = 0, switchWins = 0, attempt; srand(time(NULL)); for (attempt = 0; attempt < ATTEMPTS; attempt++) { int prizeIn = rand % 3; // prize is behind this door int choose = rand % 3;  // contestant's first choice of door int open;                 // door opened by Monty int switchTo;             // door that contestant switches to         // now reveal a goat if (prizeIn == choose) { open = rand % 2; if (open == choose) open = 2; } else { for (open = 0; open < 3; open++) { if (open != choose && open != prizeIn) break; }        }         // try sticking if (prizeIn == choose) { stickWins++; }        // try switching for (switchTo = 0; switchTo < 3; switchTo++) { if (switchTo != choose && switchTo != open) break; }        if (prizeIn == switchTo) { switchWins++; }    }     // now report on result printf("Wins by sticking: %5d\n", stickWins); printf("Wins by switching: %5d\n", switchWins); return 0; } -- Smjg 16:26, 21 Oct 2004 (UTC)
 * 1) include
 * 2) include
 * 3) include

Never mind me
OK, I just ran the program, modified it to make it more fair (took out some unneeded code and made the choosing algorithm more transparent. My version keeps using rand untill it finds a valid door to open, instead of falling back to a 'default' 2 if it detects a double. It STILL ends up is 33%/66%. I was suprized... Untill I realized that the clue behind this 'riddle' is that the door that gets opened isn't a random door. If you choose a goat door the door that get's opened is non random. So I'm not sure yet how the math on this works, but because of this you chance goes up. The description on the solution page is still wrong though. However, the solution mentioned here (mapping all possibilities out if you choose door A) does privide a sufficient explanation, if you explain properly that you can see the 2 equal situations as 1 situation (resulting in 1 no switch win and 2 switch wins).

The current solution, as of june 21 2005, is correct.

Scenario A: Select prize; Switching after goat revealed means you lose. Scenario B: Select Goat A; Switching after goat revealed means you win. Scenario C: Select Goat B; Switching after goat revealed means you win.

Scenario A, B and C are equally likely; Switching means you win 2/3 times.

Merged Puzzles:Goats and sports car
... which was a duplicate solution to this problem. --Kernigh 00:47, 14 February 2006 (UTC)


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User talk:212.14.13.203

If something's wrong, by all means, fix it. Be bold! :) 129.94.6.30 22:18, 11 Feb 2004 (UTC) (User:Dysprosia, logged out)

title: User talk:212.14.13.203


text:
Talk:Puzzles/How do you ... ?/Crossing the Bridge/Solution

The solution in my opinion is not correct. Even though during the juggling process, one of the apples is in the air, if you consider the man+ 3 apples as a system, this system is applying an average force of 101kg on the bridge. (actually, while throwing the apple up in the air, you exert a very large impulsive force and then you go back to your man + 2 apples weight)

I believe the solution is incorrect.


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Talk:High School Mathematics Extensions/Counting and Generating Functions

I think the guessing part is bad mathematics
I think the guessing solution is bad mathematics. I dont know who came up with the idea of guessing. I know that part of the mathematics isn't guessing, it's an observation turned generalisation. Nothing is guesswork, but careful deduction and elimination. I think it would be good to just keep it there for the moment being. But it will need to be heavily edited before the first draft.Xiaodai 22:54, 5 Feb 2004 (UTC)

Uh, smart guessing isn't bad mathematics - if you have a problem and you have a good idea that it may lead to a positive outcome, then there's no real harm in guessing - afterward you can clean up, and justify the guess. I suppose you'd like some justification of the "guess", and I should have done that earlier, so I apologise Dysprosia 23:20, 5 Feb 2004 (UTC)

 * Yeah. Justification would be nice, but I think the idea comes from the derivation method. I dont think we should say it's guesswork, because it's something that follows naturally once you know the work. Xiaodai 00:29, 6 Feb 2004 (UTC)

 * I would argue the opposite, if, say, you were confronting the theory of linear recurrence relations for the first time. You might try polynomials at first and see that they don't work, or try something else - then see that exponentials work, then try and fiddle around with exponentials and then see that they form the general solution, then try and explain why exponentials work. It's much more difficult to come up with some general theory, and then make it more specific. Dysprosia 01:41, 6 Feb 2004 (UTC)

Alternate solution to example 1
We have $$x_n = 2x_{n-1} + 3x_{n-2}$$. Let us suppose that we will have a solution will be in the form rn for some r (Later we will show you that this is just the right form of solution to choose). If we substitute this guess in, we obtain
 * rn = 2rn-1+3rn-2

Divide throughout by rn-2
 * r2 = 2r+3

But this is just a quadratic equation - we can easily solve to get roots
 * α=-1, β=3

So we have two solutions, A(-1)n, and B(3n). We can add these solutions together to get a new solution*, but we need to find out what coefficients A and B are. We've been given initial conditions x0 = 1, x1 = 1. So, let's substitute x0 into xn=A(-1) n +B(3 n).
 * 1 = A(-1)^0+B(3^0)
 * 1 = A+B

Now substitute x1
 * 1 = -A+3B

Solve these two to get A=B=1/2.

So our solution is xn=1/2((-1)n+3n), same as above.

 * We know this because, if an and bn are both solutions, and we substitute into the general recurrence relation
 * xn=Axn-1+Bxn-2

we get, for an and bn, since they are solutions
 * 0=Aan-1+Ban-2-an
 * 0=Abn-1+Bbn-2-bn

Substituting their sum,
 * A(an-1+bn-1)+B(an-2+bn-2) - (an+bn)
 * =Aan-1+Ban-2-an+
 * Abn-1+Bbn-2-bn
 * =0+0=0

So both the solutions and their sum satisfy the linear recurrence.

Now why is rn such a good solution for linear recurrences? For simplicity, we'll prove this for recurrences in the form an=Aan-1+Ban-2, but the proof can be extended easily to general linear recurrences like the ones you see below.

Now r must be a solution of the quadratic equation x2-Ax-B=0 to be a solution. Why is this so? We have the recurrence
 * an=Aan-1+Ban-2

rearranging
 * an-Aan-1-Ban-2=0

Substitute rn (r not zero) into the recurrence to get
 * rn-Arn-1-Brn-2=0

Now factor out by rn-2, the term farthest on the right
 * rn-2(r2-Ar-B)=0

We know that r isn't zero, so rn-2 can never be zero. So r2-Ar-B must be zero, and so rn, with r a solution of r2-Ar-B=0, will be a solution of the linear recurrence.

Why is this removed, now? If you don't like the approach, just say the word, but it's just a good and logical method as the ones featured in the text. Dysprosia 05:34, 12 Feb 2004 (UTC)

 * This removal is temporary. I think what you have done there is good work. What I would like to do with is to modify it a little so that it's a demonstrastion of how this method of guessing is derived, not just an alternate solution to example 1, but something bigger and more general. The work you have done on this section isn't all that convincing, because the justification (derivation) part isn't fully complete. This will need to be modified (by you or me), before incorporating back into the main article. That's why I just copy-and-pasted it, I see value in this, but not just a solution for example 1. Xiaodai 00:35, 13 Feb 2004 (UTC)

Thanks Xiaodai. How's this: moved to Discrete mathematics/Recursion Feel free to merge some stuff across here, if you like :)

Note there are special circumstances when the roots of the characteristic equation are not distinct, and I'll add them later. Dysprosia 04:49, 13 Feb 2004 (UTC)

Partial fraction
Has anyone done a tutorial on how to do partial fractions? Like how to separate into fraction with unique denomiators? Or do I have to write my own?

 * It's not so easy to cover systematically and gently, but I'll have a go if I have some time to spare... Calculus/Integration techniques or something ... Dysprosia 00:04, 18 Feb 2004 (UTC)

Some hints please (1.1 2c)
I am kind of stuck on finding the answer to 1.1 2c (finding f(n) for $$\frac{z^2 - 1}{1 + 3z^3} $$) I would really like it if somebody could give me some hints on how to solve this problem. Mmartin 15:18, 15 Aug 2004 (UTC)

 * Well this is a hard exercise and it's requires a deep understanding of the topic. Try to emulate the derivation of the closed form of other generating functions. Note that
 * $$\frac{1}{1 - z} = 1 + z + z^2 + ... $$

now replace z by 3w^3, what happens? Xiaodai 07:10, 16 Aug 2004 (UTC)

 * If you know Taylor series, you can try and apply that and glean the general form for the coefficients :P Dysprosia 12:48, 30 Oct 2004 (UTC)

Project
I'm considering writing a project on power sum,it may cross this chapter and the mathematical proof chapter. Anyway here is the working draft, I'll place it in the correct location once it has been done to a certain degrees.

Note: Notation is quite a problem here, in question 1, S(z) is a generating function of square number, Sn is the sum of the first n square number, and P2 is a generating function of power sum in degrees 2, i.e. the sum of square number.

Project-Power Sum-First draft 1.a) Find the closed-form of the generating function
 * $$S(z)=1^2+2^{2}z+3^{2}z^{2}+4^{2}z^{3}+...$$
 * (Hint:Reduce it into a linear generating function first)
 * b) Hence find the generating function of
 * $$P_{2}(z)=S_{1}+S_{2}z+S_{3}z^2+S_{4}z^3+...$$
 * Where $$S_n=1^2+2^2+3^2+...+n^2$$
 * And then find out the general formula for Sn

1.a)

$$ \begin{matrix} & & S(z) &=& 1^2 & +2^{2}z & +3^{2}z^{2} & +4^{2}z^{3}+...\\ z & \times & S(z) &=& & z & +2^{2}z^2 & +3^{2}z^3+...\\ (1-z) & \times & S(z) &=& 1 & +3z & +5z^2 & +7z^3+...\\ \end{matrix} $$

$$ \begin{matrix} & & (1-z)S(z) &=& 1 & +3z & +5z^2 & +7z^3+...\\ z & \times & (1-z)S(z) &=& & z & +3z^2 & +5z^3+...\\ (1-z) & \times & (1-z)S(z) &=& 1 & +2z & +2z^2 & +2z^3+...\\ & & (1-z)^{2}S(z) &=& \frac{2}{1-z} -1\\ \end{matrix} $$

$$S(z) = \frac{2}{(1-z)^3} - \frac{1}{(1-z)^2}$$

 * b)

$$ \begin{matrix} & & S(z) &=& 1^2 & +2^{2}z & +3^{2}z^{2} & +4^{2}z^{3}+...\\ z & \times & S(z) &=& & 1^{2}z & +2^{2}z^2 & +3^{2}z^3+...\\ z^2 & \times & S(z) &=& &  & 1^{2}z^2 & +2^{2}z^3+...\\ z^3 & \times & S(z) &=& &  &  & 1^{2}z^3+...\\ ...\\ (1+z+z^2+z^3+...) & \times & S(z) &=& S_{1} & + S_{2}z & + S_{3}z^2 & + S_{4}z^3+ ...\\ \end{matrix} $$

$$\frac{S(z)}{1-z} = P_{2}(z)$$ $$P_{2}(z) = \frac{2}{(1-z)^4} - \frac{1}{(1-z)^3}$$

$$\sum_{i=0}^{\infty} [2{4+i-1 \choose i} z^i] - \sum_{i=0}^{\infty} [{3+i-1 \choose i} z^i] = P_{2}(z)$$ $$\sum_{i=0}^{\infty} [2{3+i \choose i} - {2+i \choose i}] z^i = \sum_{i=0}^{\infty} S_{i+1} z^i$$ $$[2{3+i \choose i} - {2+i \choose i}] z^i = S_{i+1} z^i$$ $$2{3+i \choose i} - {2+i \choose i} = S_{i+1}$$ So n = i+1. $$2{n+2 \choose n-1} - {n+1 \choose n-1} = S_{n}$$ $$S_{n} = 2{n+2 \choose (n+2)-(n-1)} - {n+1 \choose (n+1)-(n-1)}$$ $$S_{n} = 2{n+2 \choose 3} - {n+1 \choose 2}$$ $$S_{n} = \frac{2(n+2)(n+1)n}{6} - \frac{(n+1)n}{2}$$ $$S_{n} = \frac{2(n+2)(n+1)n - 3(n+1)n}{6}$$ $$S_{n} = \frac{[2(n+2) - 3](n+1)n}{6}$$ $$S_{n} = \frac{n(n+1)(2n+1)}{6}$$

2.Let $$p_n = 1^n + 2^n z + 3^n z^2 + 4^n z^3 + ...$$
 * a)Recall that $$\frac{d}{dx}x^n = nx^{n-1}$$
 * Try to define pn recursively.
 * b)Using 2a), or otherwise, find out the closed-form of the generating function of p3, p4, and p5.

 * Ok, cool. I will read the thing carefully now. Xiaodai 23:39, 27 Dec 2004 (UTC)

 * Update:It appears that power sum is a deep and difficult topic, so... I've discovered that the topic of generating function is deep and wide, that we couldn't possibly cover 'em all in those 25 pages. I propose exploring exponential generating function in the project, is it ok? Somehow it is the most common generating function right after ordinary generating function. --Lemontea 10:49, 12 August 2005 (UTC)

 * Sounds good, i can't pretend to have cover everything i know about generating functions in those pages. Feel free to add your own content! It's great. But somehow i dont know how ppl will discover how to diff e^x. Xiaodai 14:14, 12 August 2005 (UTC)

 * Ha! When I learn differentiating, I was told to define e so that we can differentiate log x, then went on to discover e^x is the solution of the differential equation f'(x) = f(x) lol. (Opps, I spoiled what will appear later on!) --Lemontea 14:33, 12 August 2005 (UTC)

Generating functions and the z-transformation
What do the counting and generating functions have in common with the z-transformation. (The z-transformation of $$f(n)$$ I came across is defined by $$\sum_{n=0}^{\infty}f(n)z^n$$) It seems to me that the generating function of some series is the z-transform of the function that gives the coefficients of the polynomial. If that's the case it might be nice to inform the reader about it.

Counting and Generating Functions
In the first section of "Counting and Generating Functions", there is a note about it called "Info - Infinite Sums". It says that the two expressions (the infinite geometric series and the formula for it) are not equal. An example was also given by adding a finite amount of terms of the series and comparing the sum with the result from the formula. I disagree with this statement that the two expressions are not equal. First, infinite sums have to be defined and it is DEFINED to be the limit of the partial sum as the number of terms go to infinity if the limit exists. So when the formula was derived, it IS equal to the geometric series according to the definition. The example is not a valid way of showing that the two things are different since it was a finite amount of terms, so of course the number will be close but not equal. According to the definition the value of the infinite series is the LIMIT defined before, so in the example it is 2 and the formula also gives 2 which are equal. Also, I am not just making up this definition, its in any rigorous calculus book such as by Spivak.