January 23

I noticed the wrinkles in somebody's shirt today and wondered of a description of the exact mechanism and simulation of a computer modeling someone's shirt wrinkle. These were wrinkles like one gets from just being in a shirt, not necessarily anything the slightest bit permanent. X [Mac Davis] (DESK|How's my driving?) 01:56, 23 January 2007 (UTC)[reply]

Catastrophe theory has something to say on the topic. There they examine local behaviour and have models describing local behaviour and deformations which can occur. There are folds and pleets (another word for the cusp catestrophy). --Salix alba (talk) 14:09, 23 January 2007 (UTC)[reply]

In computer graphics, cloth is generally modeled using some form of subdivision surface. A number of control points throughout the piece of cloth have simulated physics (some network of calculations approximating springs or other forces), and then a more detailed curve for the cloth is generated by interpolating between these control points with a Catmull-Rom spline or whatever technique is preferred. Realism can be increased by using a denser net of control points, but the calculation takes longer. - Rainwarrior 16:41, 23 January 2007 (UTC)[reply]

I've noticed wrinkles tend to recur in the same location, indicating a flaw in the material may be instrumental in causing the process, much like a nucleation site is needed to start a crystal. StuRat 07:18, 24 January 2007 (UTC)[reply]

I remember A Wrinkle in Time talking about this kind of stuff to quite a great extent, though can't remember if they went into much of the mathematical details. Mathmo ^Talk 00:30, 29 January 2007 (UTC)[reply]

How do you separate the numbers in Cartesian coordinates if you use a comma for a decimal point? —The preceding unsigned comment was added by 155.43.64.40 (talk) 02:04, 23 January 2007 (UTC).[reply]

The concept of a Cartesian coordinate system is fairly abstract. In the US, most of us use a period (".") withing a number, so we can use the comma {","} to separate numbers. We write a two-vector as (1.2,2.45). to descrobe a point on the cartesian plane. a 3-vector looks like this: (1,5,7) or (1,23,5.67,8.9). When working wiht dollar amounts, US accountants use the comma character to separate powers of 1000. thus, $10,000.01 is ten thousand dollars and one cent. This convention cannot be use in conjunction with vector notation, because it is ambiguous. Thus, in accounting (10,000.01) meand "a debit of ten thousand collard and one cent," but in a mathematical context, (10,000.01) is a point on the plane whose x displacement is 10 and whose y displacement is 0.01." If you use a comma for a decomal point, you might use a colon for the vector separator. —The preceding unsigned comment was added by Arch dude (talk • contribs) 03:40, 23 January 2007 (UTC).[reply]

In practice, you could use a semicolon, or, better, a vertical bar like (0.5|1). – b_jonas 10:41, 23 January 2007 (UTC)[reply]

A third possibility, at least when typing the coordinates on a computer, is to use a whitespace character. TERdON 19:40, 29 January 2007 (UTC)[reply]

You could write points as a column vector (easier if writing by hand). e.g (x)

                                                                           (y)

except the brackets are all one. See I said it was easier by hand!

I'm having trouble remembering how to use the binomial theorem and Pascal's triangle. I read the articles and they have not helped me very well. If I had something like (x³-3)¹², how would I expand that? X [Mac Davis] (DESK|How's my driving?) 03:03, 23 January 2007 (UTC)[reply]

The coefficients of the bionomial to the n power are the nth row of pascals triangle. I ususally write out the terms first, then multiply each by its coefficient. So take the first term, x³ and raise it to the 12th power, and the second term, -3, to the zero power. Add to this x³ to the 11th power times -3 to the first power. Continue on, counting down in the first exponent and counting up in the second. Now multiply each term by the number in the 12th row of pascal's triangle (the row with a 12 as the second term). You should get

${\displaystyle 1*(x^{3})^{12}+12*(x^{3})^{11}*(-3)+66*(x^{3})^{10}(-3)^{2}+...+12*(x^{3})(-3)^{11}+1*(-3)^{12))$

Now just simplify each term (carry out the exponentiation of the -3 and multiply by the coefficient). Hope this helps. --TeaDrinker 03:22, 23 January 2007 (UTC)[reply]

Let's try to build a vivid mental model. For convenience, we'll use (a+b)⁴. Here is the top of Pascal:

1
1	1
1	2	1
1	3	3	1
1	4	6	4	1
1	5	10	10	5	1
1	6	15	20	15	6	1

Here is the product:

(a+b) (a+b) (a+b) (a+b)

Before combining like terms, each term in the result chooses either a or b from one of the four factors. For example, if we choose a from each factor, we get a⁴. If we choose a from the first two factors and b from the last two, we get a²b². (Since we have four factors in this example, the sum of the a and b exponents in a term must be four.) We can only get a⁴ in this one way, but we can get a²b² in other ways, such as by choosing a from the last two factors and b from the first two. This explains why we read a binomial coefficient as "n choose k", and sometimes write it C(n,k).

We can also write (a+b)⁴ in terms of (a+b)³:

(a+b)³ (a+b)

This helps us get from one line of the table to the next. For, suppose we have worked out (a+b)³.

(a+b)³ = a³ + 3a²b + 3ab² + b³

When we look at the product

(a³ + 3a²b + 3ab² + b³) (a+b)

we again see that we must choose one term from each factor. As the triangle is displayed above, choosing a contributes to the term straight down, while choosing b contributes diagonally right down. For example, we can get a²b² as either (ab²)a or as (a²b)b, so C(4,2) = C(3,2)+C(3,1). This is the source of the famous relation C(n+1,k) = C(n,k)+C(n,k−1).

If we view C(n,k) as choosing b from k factors out of n, then it also chooses a from n−k factors out of n. Thus we have the symmetry C(n,k) = C(n,n−k), seen on each line of the triangle: it reads the same left-to-right as it does right-to-left.

This method generalizes easily to multinomial coefficients, such as those for (a+b+c)ⁿ, though we don't seem to need these nearly so often. The general term will be

${\displaystyle {\frac {(k_{1}+\cdots +k_{m})!}{k_{1}!\cdots k_{m}!))x_{1}^{k_{1))\cdots x_{m}^{k_{m)).}$

Here the sum in the numerator will always be n, and counts the number of permutations of n factors. Why does this make sense? Conceptually, we can always choose all x₁ from the first k₁ factors, all x₂ from the next k₂ factors, and so on until we have exhausted all n factors of (x₁+⋯+x_m). We get the different possibilities by rearranging the factors, leaving the choice fixed. For example, here are the possibilities for ab² in the expansion of (a+b)³.

(a+b) (a+b) (a+b)

(a+b) (a+b) (a+b)

(a+b) (a+b) (a+b)

(a+b) (a+b) (a+b)

(a+b) (a+b) (a+b)

(a+b) (a+b) (a+b)

We have three unique choices.

abb + abb + abb

But we see that each choice occurs twice among the six permutations, because we have two (that is, 2!) permutations of the factors from which we choose b. This accounts for the denominator of the multinomial coefficient.

Short question, long answer; hope it helps. --KSmrq^T 16:39, 23 January 2007 (UTC)[reply]

Another(s)

Find term by degree

What would be a quick way of finding the term of a specific degree? Like, if I wanted to know the term whose x was raised to the 400th? Ok, I'm completely making this up too, so tell me if I mess up along the way. If we had the expression

${\displaystyle (x^{a}-y^{b})^{n))$ (probably not standard typographical convention, what is for here?)

and according the binomial theorem we could expand that to

${\displaystyle C(n,0)(x^{a})^{n}+C(n,1)(x^{a})^{n-1}(y^{b})+C(n,2)(x^{a})^{n-2}(y^{b})^{2}\cdots +C(n,n-2)(x^{a})^{2}(y^{b})^{n-2}+C(n,n-1)(x^{a}(y^{b})^{n-1}+C(n,n)(y^{b})^{n))$

So...


*think*
If I'm looking for x⁴⁰⁰ somewhere in there, I would want: 400=az, with z identifying the term by order of left to right. Would that mean z is a function of n, or something similar? Is this logic making sense so far? How far is this logic making sense? :) X [Mac Davis] (How's my driving?) 22:53, 23 January 2007 (UTC)[reply]

Find fifth term

This problem wants me to find the fifth term of the expanded ${\displaystyle (x^{2}+2)^{1}2}$. I've did first:

${\displaystyle (x^{2})^{12}+11(x^{2})^{11}(2)+55(x^{2})^{10}(2)^{2}+165(x^{2})^{9}(2)^{3}+330(x^{2})^{8}(2)^{4))$

then I looked up the section and decided it should be this instead:

${\displaystyle (x^{2})^{12}+12(x^{2})^{11}(2)+66(x^{2})^{10}(2)^{2}+220(x^{2})^{9}(2)^{3}+495(x^{2})^{8}(2)^{4))$

Well! The fifth term for proposition 1 is (330x2x2x2x2)x¹⁶, for proposition 2 it is (495x2x2x2x2)x¹⁶.

Uh oh! They are both wrong. What happened? X [Mac Davis] (How's my driving?) 22:51, 23 January 2007 (UTC)[reply]

${\displaystyle {\sqrt {-))4(3-{\sqrt {-))9)}$

I stared at this one and just made it up as I went along. I did

${\displaystyle 3{\sqrt {-))4-{\sqrt {-))4{\sqrt {-))9}$

${\displaystyle 3{\sqrt {4))i-{\sqrt {4))i{\sqrt {9))i}$

${\displaystyle 6i-6i^{2))$

${\displaystyle 6+6i}$

Am I correct? Thank you X [Mac Davis] (DESK|How's my driving?) 16:35, 23 January 2007 (UTC)[reply]

Don't forget that :${\displaystyle {\sqrt {4))}$ equals 2 or -2.87.102.6.197 16:47, 23 January 2007 (UTC)[reply]

True, but ${\displaystyle {\sqrt {))}$ just by its lonesomes usually means principal square root; i.e., the positive one. Mac, your calculations look fine to me. –King Bee ^{(T • C)} 18:28, 23 January 2007 (UTC)[reply]

Actually, the initial formula as written doesn't say what you think. TeX applies the square root to the next item, which here is only a minus sign! The Right Thing is

${\displaystyle {\sqrt {-4))\left(3-{\sqrt {-9))\right)=6+6\mathbf {i} .}$

Now the simplification is correct. :-) --KSmrq^T 18:58, 23 January 2007 (UTC)[reply]

What is ${\displaystyle {\sqrt {-))}$ anyway? Just the dot above the letter i? Maybe just the vertical line part of i? =) –King Bee ^{(T • C)} 20:38, 23 January 2007 (UTC)[reply]

What is that about?? ${\displaystyle {\sqrt {-))}$ looks unlikely. X [Mac Davis] (DESK|How's my driving?) 21:05, 23 January 2007 (UTC)[reply]

It was a joke; sorry. –King Bee ^{(T • C)} 21:23, 23 January 2007 (UTC)[reply]

I mean, I think the programmers should have compensated for that possible error. No body is going to write${\displaystyle {\sqrt {-))}$ X [Mac Davis] (How's my driving?) 21:42, 23 January 2007 (UTC)[reply]

The "programmer" to blame for this behavior is Knuth himself; he decided TeX should act this way when he designed it. LaTeX (and AMS-LaTeX) compensated for some of TeX's peculiar design, but this "feature" may be built in too deeply. Most programming languages have lexical tokens which are words or numbers, but TeX (mostly) works at the character level. For example, the input "x^-1" produces

${\displaystyle x^{-}1,\,\!}$

not

${\displaystyle x^{-1}\,\!}$

as you might expect. And a simple fraction can be produced with "\tfrac58", like so:

${\displaystyle {\tfrac {5}{8))\,\!}$

The obvious interpretation of "(a+b)^3" is that the superscript, which we know denotes an exponent, is on the entire parenthesized expression; but TeX doesn't think that way, and only sees it as attached to the "item" before, which is the closing parenthesis. We see semantic units; TeX sees only characters. Thus we write a hypergeometric series in a semantically bizarre fashion, as "{}_2F_1(a,b;c;z)", producing

${\displaystyle {}_{2}F_{1}(a,b;c;z),\,\!}$

with both a prefixed subscript and a postfixed subscript. In MathML we would use the "multiscripts" element in presentation markup. This unusual TeX worldview makes it quite difficult to automatically convert TeX markup to MathML, or to any notation based more on semantics. --KSmrq^T 22:41, 23 January 2007 (UTC)[reply]

To say that Knuth implementing this behaviour is somewhat "bad" is short-sighted. TeX is a typesetting language. It provides this level of flexibility because you should not be constrained to one type of mathematical notation. Say you want to notate a limit approaching from the left/right by a superscript -/+ respectively. The notation 0^- is not usual but it is easily implementable in TeX because it provides this flexibility. The difficulty in conversion that Ksmrq mentions is precisely because TeX is focused primarily on mathematical typesetting, not conveying mathematical meaning. —The preceding unsigned comment was added by 58.163.151.30 (talk) 10:18, 24 January 2007 (UTC).[reply]

I looked very carefully to see where I might have said Knuth's design choice was bad; to my relief, I did not say that. To the contrary, I stuck to descriptions. TeX's view of its input is not what most people expect. That view lures users into common mistakes. It also makes transformation of TeX input to semantic input difficult. Those are facts. I could similarly explore consequences of design choices in MathML, some of which are horrific. For example, the quadratic formula in TeX is

"\frac{-b \pm \sqrt{b^2 - 4ac)){2a}"

to produce

${\displaystyle {\frac {-b\pm {\sqrt {b^{2}-4ac))}{2a)).}$

The same thing in (presentation) MathML is

"<mfrac>

<mrow>

<mrow>

<mo>-</mo>

<mi>b</mi>

</mrow>

<mo>&PlusMinus;</mo>

<msqrt>

<msup>

<mi>b</mi>

<mn>2</mn>

</msup>

<mo>-</mo>

<mrow>

<mn>4</mn>

<mo>&InvisibleTimes;</mo>

<mi>a</mi>

<mo>&InvisibleTimes;</mo>

<mi>c</mi>

</mrow>

</msqrt>

</mrow>

<mrow>

<mn>2</mn>

<mo>&InvisibleTimes;</mo>

<mi>a</mi>

</mrow>

</mfrac>"

Can anyone spot a questionable design decision? I knew that you could!

But in both cases the designers have what (to them, at least) seemed like good justifications for their decisions. --KSmrq^T 16:55, 24 January 2007 (UTC)[reply]

Sorry, I didn't mean to address that comment to you, but to Mac Davis above you. I thought I fixed my indenting... —The preceding unsigned comment was added by 58.163.151.30 (talk) 08:05, 25 January 2007 (UTC).[reply]

How do I find the nth partial sum of a geometric sequence? I used S_n = a₁(1-rⁿ)/(1-r), where r is three. If I wanted to find the fifth, I would just replace n with 5 and, a₁ with 5 as the problem states. I get 302.5, and this is not the correct answer! What am I doing wrong? X [Mac Davis] (How's my driving?) 21:42, 23 January 2007 (UTC)[reply]

The sum of the first n terms of a geometric series is given in English as (first term - [n +1]^th term)/(1 - common ratio). Give it a try! –King Bee ^{(T • C)} 22:01, 23 January 2007 (UTC)[reply]

Thanks! I got it. Where did that come from? X [Mac Davis] (How's my driving?) 22:18, 23 January 2007 (UTC)[reply]

The insight is simple, once you see it. Write out the terms

${\displaystyle \mathrm {Sum} (a,r;n)=a+ar+ar^{2}+ar^{3}+\cdots +ar^{(n-1)}\,\!}$

The factor of a can be pulled out using the distributive law, so it's all about powers of r. Now multiply by r.

${\displaystyle r\,\mathrm {Sum} (a,r;n)=0+ar+ar^{2}+ar^{3}+\cdots +ar^{(n-1)}+ar^{n}\,\!}$

Essentially, each term shifts one place to the right. Now subtract.

${\displaystyle r\,\mathrm {Sum} (a,r;n)-\mathrm {Sum} (a,r;n)=ar^{n}-a\,\!}$

Factor the left-hand side and divide to get the closed form.

${\displaystyle \mathrm {Sum} (a,r;n)=a{\frac {r^{n}-1}{r-1))}$

We use the same idea when we convert a recurring decimal like

${\displaystyle 0.{\overline {142857))\,\!}$

to a fraction.

${\displaystyle {\begin{aligned}x&{}=0.{\overline {142857))\\1000000x&{}=142857.{\overline {142857))\\999999x&{}=142857\\x&{}={\tfrac {142857}{999999))\\&{}={\tfrac {1}{7))\end{aligned))}$

Essentially, we shift and subtract to kill the repetition. --KSmrq^T 23:04, 23 January 2007 (UTC)[reply]

Why can the factors of a be taken out? X [Mac Davis] (How's my driving?) 23:50, 23 January 2007 (UTC)[reply]

What factors are you referring to? (Or what line are you unsure of?) If you're asking how the terms can be taken out to get line 3 in KSmrq's explanation, you subtract line 1 from line 2 (subtract the left side from the left side, the right side from the right side). 134.10.40.183 06:58, 24 January 2007 (UTC)[reply]

The formula you gave at the beginning is also correct - you have probably just entered the numbers wrong (your denominator was -4 when it should have been -2). The formula will be less confusing if you rewrite it as ${\displaystyle a_{1}{\frac {r^{n}-1}{r-1))}$. -- Meni Rosenfeld (talk) 14:44, 24 January 2007 (UTC)[reply]

Wait a minute, how is ${\displaystyle a_{1}{\frac {r^{n}-1}{r-1))}$ and ${\displaystyle a{\frac {r^{n+1}-1}{r-1))}$ the same thing? X [Mac Davis] (How's my driving?) 18:36, 24 January 2007 (UTC)[reply]

KSmrq made a small mistake in his derivation - the partial sum consisting of n terms only goes up to the (n-1)th power of r, because the first term is simply a. I've corrected his post so it gets the desired result now. Readro 17:53, 29 January 2007 (UTC)[reply]