April 4

Moved to science desk. --Bowlhover (talk) 20:44, 10 April 2009 (UTC)[reply]

I get ${\displaystyle N^{2}-N}$ , for this expression: ${\displaystyle \delta _{i}^{i}\delta _{j}^{j}-{\delta _{j}^{i)){\delta _{i}^{j))}$ , but my book says ${\displaystyle 2!}$ . Could anyone help? —Preceding unsigned comment added by Re444 (talk • contribs) 10:50, 4 April 2009 (UTC)[reply]

Perhaps your book is taking N to be 2. Algebraist 10:58, 4 April 2009 (UTC)[reply]

This expression is for its equivalent: ${\displaystyle e^{ij}e_{ij))$ (as in this book). Also it states that, ${\displaystyle e^{ijk}e_{ijk}=3!}$ and ${\displaystyle e^{ijkl}e_{ijkl}=4!}$ where e is the e-permutation symbol. --Re444 (talk) 12:12, 4 April 2009 (UTC)[reply]

In the article on transfinite numbers, it says:

If the axiom of choice holds, the next higher cardinal number is aleph-one, ${\displaystyle {\aleph _{1))}$. If not, there may be other cardinals which are incomparable with aleph-one and larger than aleph-zero. But in any case, there are no cardinals between aleph-zero and aleph-one.

Just to make sure I understand, does this mean that, if we don't assume AC, there may exist a set with cardinality (for example) ${\displaystyle {\daleth ))$, for which it would be insensible to say that ${\displaystyle {\daleth ))$ is larger or smaller than ${\displaystyle {\aleph _{1))}$? --superioridad ^(discusión) 14:10, 4 April 2009 (UTC)[reply]

Indeed. Actually, according to our Axiom of choice article, the statement "If two sets are given, then either they have the same cardinality, or one has a smaller cardinality than the other." is equivalent to AC. Without it, all sorts of things can happen. —JAO • T • C 20:17, 4 April 2009 (UTC)[reply]

Whoa, you need the axiom of choice for that? Ok, I am now a believer in the axiom of choice. Eric. 131.215.159.99 (talk) 22:09, 4 April 2009 (UTC)[reply]

You don't just need choice, that is choice (over ZF). Given any set X, by Hartogs' lemma, there is an ordinal that does not inject into X. If we assume the given property, then X must inject into the ordinal, so X is then wellorderable. But then every set is wellorderable, which is the same thing as choice. Algebraist 00:54, 5 April 2009 (UTC)[reply]

Just to make sure that everybody is talking about the same thing, the equivalence above holds when cardinals are defined as equivalence classes of arbitrary sets wrt equipotence. Often cardinals are defined as ordinal numbers which are not equipotent to any smaller ordinal, and of course you do not need the axiom of choice to show that these cardinals are pairwise comparable. — Emil J. 14:01, 6 April 2009 (UTC)[reply]

Of course; in that case one needs choice to prove that the well ordered cardinalities exhaust all the possible cardinalities. — Carl (CBM · talk) 14:07, 6 April 2009 (UTC)[reply]

I know a dyad is a geometrical object with one magnitude and two directions, but how do you read the magnitude and directions, like there is |r| for a vector r for its magnitude, and there is ${\displaystyle {\frac {r}{|r|))}$ for a vector r's direction?The Successor of Physics 14:53, 4 April 2009 (UTC)[reply]

I've read the article on the Cauchy formula for repeated integration; is there a (relatively simple) formula for the following more complex repeated integration?

${\displaystyle \int _{a}^{b}dx_{1}f(x_{1}-x_{0})\int _{a}^{b}dx_{2}f(x_{2}-x_{1})...\int _{a}^{b}dx_{n}f(x_{n}-x_{n-1})}$

Thanks in advance for the answers. --Icek (talk) 17:36, 4 April 2009 (UTC)[reply]

I don't see a simple formula, except in the case ${\displaystyle \scriptstyle a=-\infty ,\,b=+\infty }$, where you get ${\displaystyle \scriptstyle \left(\int _{a}^{b}f(x)dx\right)^{n))$. Also: the same for any reals a<b, but with f periodic of period b-a. --pma (talk) 21:26, 5 April 2009 (UTC)[reply]

Hi there - I was wondering how many relations there are in total on a set ${\displaystyle \mathbb {X} }$ of size n? I figure that binary relations give ${\displaystyle 2^{n))$ options for each element, so multiplying together for all n elements we get ${\displaystyle (2^{n})^{n}=2^{n^{2))}$, and similarly ternary relations give 2^(n^2) options for each element because you're looking at relations on ${\displaystyle \mathbb {X} \times \mathbb {X} }$ (that's the product if it's unclear!) - continuing in this way you'd get the total number of relations, from binary to n-ary, equal to ${\displaystyle \sum _{r=1}^{n}2^{n^{r+1))}$, right? Is this correct?

Thanks, Delaypoems (talk) 17:48, 4 April 2009 (UTC)Delaypoems[reply]

The problem (other than the minor detail that your summation index is going one step too high) is that "the total number of relations, from binary to n-ary" does not equal "how many relations there are in total". Just because a set only has n elements doesn't mean you can't define an (n+1)-ary relation on it. —JAO • T • C 20:30, 4 April 2009 (UTC)[reply]

Ahh damn, so then there are an infinite number of relations on it? Delaypoems (talk) 20:09, 5 April 2009 (UTC)[reply]

Yes. Algebraist 20:10, 5 April 2009 (UTC)[reply]

I am attempting to design an Excel spreadsheet that will help with automobile suspension design. I used to know how do do this, but I am old and tired now. To do this I need help with some general equations.

1. The equations that give the xy coordinates of the third point of a triangle when the lengths of the three sides are known and the xy coordinates of the other two points are known.

2. The equation that gives the x₂ coordinate of a line segment of known length where the x₁y₁ coordinates of one end point are known and the other end of the line segment is displaced by some given delta y amount.

3. Given a simple quadrilateral ABCD, and the x_ay_a and x_by_b coordinates of segment AB, and the angle ABC, what are the equations that give the xy coordinates for the other two quadrilateral points C and D?

Are there some good web sites that have general equations like this?

Thanks for your help dudes (and dudettes) —Preceding unsigned comment added by 99.189.121.250 (talk) 23:14, 4 April 2009 (UTC)[reply]

#1: You need to find the intersection of two circles whose centers and radii you know. There are two solutions.

#2: You need to find one leg of a right triangle whose other two sides you know. There are two solutions.

#3: The information given is insufficient for a general quadrilateral. Do you mean a rhombus?

—Tamfang (talk) 03:29, 5 April 2009 (UTC)[reply]

Is the info Tamfang gave for 1 and 2 enough to go on, or do you need some actual equations ? StuRat (talk) 17:48, 6 April 2009 (UTC)[reply]

Thank you for your reply, Tamfang.

I apologize for being unclear on my second question and giving insufficient information (as you pointed out) on my third question.

Question 2 should have read: Given the x₁y₁ and x₂y₂ coordinates of the endpoints of a line segment, what equation that gives the displaced x₂-coordinate of the line segment if the length of the segment is not changed but the y₂-coordinate is displaced to a new location y₂ + delta-y. That is, the length remains the same, but the y₂ coordinate gets moved up or down. (I think I need the intersection of a circle and a horizontal line.)

Yes, that's what you need. (x3-x1)² + (y3-y1)² = (x2-x1)² + (y2-y1)²; solve for x3. (Do you need help with the algebra?) —Tamfang (talk) 04:07, 9 April 2009 (UTC)[reply]

For question 3, I neglected to state that the four line segment lengths are known. I'm guessing that the solution for the simple quadrilateral is the intersection of two circles applied twice, first for point C and then for point D.

No, point C is on a given line, so you want the intersection of that line with a circle around B. (It will be more convenient, with the info given above, to rotate the segment AB around B – see below – and scale it.) Then for D you need to solve a pair of circles around A and C. —Tamfang (talk) 04:07, 9 April 2009 (UTC)[reply]

I'll also need to know the equation that gives the coordinates of the second end of a line if the line is rotated r° about the first end of the line.

Do this in three steps. Translate to a temporary coordinate system around the center of rotation (x0,y0):

u = x1-x0, v = y1-y0

Rotate around this new zero:

p = u*cos(rot) - v*sin(rot), q = u*sin(rot) + v*cos(rot)

Translate back:

xnew = p+x0, ynew = q+y0

(Likely someone will tell us how to express all three steps in a single matrix.) —Tamfang (talk) 04:07, 9 April 2009 (UTC)[reply]

To StuRat: Yes, I could use some equations. I've given away all my math books. Thank you. —Preceding unsigned comment added by 99.189.121.250 (talk) 00:46, 9 April 2009 (UTC)[reply]

*ghasp* When I decided I wasn't going to be a programmer after all, I gave away most of my CS books but kept most of the math books. —Tamfang (talk) 04:07, 9 April 2009 (UTC)[reply]

Thanks for your help, TamFang. The algebra is not as much of a problem as will be trying to code it into Excel cell equations, creating an iterative table of values for full suspension travel and graphing the results. I would try to do it all in Visual Basic, but I'm afraid I have forgotten more VB than I learned in the first place. —Preceding unsigned comment added by 99.189.121.250 (talk) 22:38, 9 April 2009 (UTC)[reply]

Hello, I'm studying for a complex analysis test by doing problems from the book. Here's one I think I have sort of figured out but maybe I do not understand the details exactly.

Where does this series converge?

${\displaystyle \sum _{k=0}^{\infty }p(k)\cdot z^{k))$ where p is a polynomial.

My thought is this converges for all |z| < 1 and that's it. But, I'm not exactly sure how the root test would work. I guess with a one term polynomial, ${\displaystyle k^{n))$, that's easy:

${\displaystyle \varlimsup |k^{n}|^{1/k}=\varlimsup |k^{1/k}|^{n}=1^{n}=1}$

And, on D(0, 1), ${\displaystyle z=e^{i\theta ))$ so the absolute value of the terms is

${\displaystyle |k^{n}||e^{i\theta }|^{k}=|k^{n}|\,}$

which goes to infinity, so there is no convergence on the boundary.

But, say I'm dealing with p(x) = x^2 + 7x + 5. Does only the highest order term matter? I know it does in some things, such as the limit of a rational function. But, does the root change things?

StatisticsMan (talk) 23:59, 4 April 2009 (UTC)[reply]

Heh, oops, I guess with this the ratio test is going to work out more easily and then it's just the situation I was describing. For p(x) = x^2 + 7x + 5, we have

${\displaystyle \lim {\frac {x^{2}+9x+13}{x^{2}+7x+5))=1}$

so that the radius of convergence is 1. And, this generalizes. Don't worry, I have another question for you. See below. (never mind, I figured that one out as well)StatisticsMan (talk) 00:04, 5 April 2009 (UTC)[reply]

If p(k) = k² + 7k + 5, then why not just do this:

${\displaystyle \sum _{k=0}^{\infty }(k^{2}+7k+5)z^{k}=\sum _{k=0}^{\infty }k^{2}z^{k}+7\sum _{k=0}^{\infty }kz^{k}+5\sum _{k=0}^{\infty }z^{k}.}$

Then you can apply your result for k = n to the three cases n = 0, 1, 2. Michael Hardy (talk) 01:37, 6 April 2009 (UTC)[reply]

Note that for |z|<1 the sum is ${\displaystyle \scriptstyle {\frac {z^{2}+2z-5}{(z-1)^{3))))$. In general for a (non-zero) polynomial p of degree r the seris ${\displaystyle \scriptstyle \sum _{k=0}^{\infty }p(k)\cdot z^{k))$ has radius of convergence 1, and the sum is a rational function of the form ${\displaystyle \scriptstyle {\frac {P(z)}{(z-1)^{r+1))))$.--pma (talk) 10:39, 6 April 2009 (UTC)[reply]