Wikipedia:Reference desk/Archives/Mathematics/2006 October 26

= October 26 =

Terms in polynomial expansion
How can you determine the number of terms in a simplified (like terms have been combined) polynomial expansion? I am particularly interested in $${(a+b+c+d)}^{17}$$, as that was on my math contest today, but any general help would be appreciated. Thanks. — Mets 501 (talk) 01:49, 26 October 2006 (UTC)


 * Well for starters the powers of any of the terms should add up to 17. So the question is how many different ways can you make $$a^x b^y c^z d^w$$ where $$x+y+z+w=17$$? I'd have to think about this for a moment to give you an answer to that question though... - Rainwarrior 02:50, 26 October 2006 (UTC)


 * For just three variables $${(a+b+c)^n}$$ should have $$\sum_{i=1}^{n+1}{i}$$, which is an easily computed arithmetic series. To do a fourth as well, it would compound, I think, as something like $$\sum_{j=0}^{n}{\sum_{i=1}^{j+1}{i}}$$? Since 17 is a fairly small number you could calculate the 18 values (for j=0 to n) and sum them yourself, I suppose, but I don't know of an easier way to compute the sum of this series. - Rainwarrior 03:07, 26 October 2006 (UTC)


 * Ah! It's a Tetrahedral number! (The 3 term polynomial problem was a Triangular number.) So... according to that formula, I'd say, 1140? - Rainwarrior 03:35, 26 October 2006 (UTC)


 * Tetrahedral numbers also turn up as a diagonal of Pascal's triangle. So, $${{n+3}\choose{3}}$$ or $${{17+3}\choose{3}}=1140$$. (I suspected there would be a solution related to the choose operation.) - Rainwarrior 04:08, 26 October 2006 (UTC)


 * The general question asks for the number of terms in $$(x_1 + x_2 + \cdots + x_k)^n$$. A good way to tackle this type of question is to find the answers for specific cases and look for a pattern. For instance, if k = 2 then you're looking at $$(a+b)^n$$ which has $$n+1$$ terms according to the binomial theorem. The trivial case k = 1 is also interesting here; there is only one term if k = 1. Look at some more cases: n = 0, n = 1, perhaps n = 2, etc., and put all the results in a table. I think you will recognize the pattern in the table once you have enough entries. -- Jitse Niesen (talk) 03:06, 26 October 2006 (UTC)
 * OK, I'll try it out tomorrow. Thanks for the help! — Mets 501  (talk) 03:10, 26 October 2006 (UTC)


 * See the multinomial theorem, which deals with this in generality, and may be useful to the problem you are considering. Dysprosia 08:42, 26 October 2006 (UTC)


 * In general for a k-termed polynomial raised to the power of n, you should have $$ {k + n - 1} \choose {k - 1} $$, I think. - Rainwarrior 15:48, 26 October 2006 (UTC)


 * And there is an easy explanation: Each term in the expansion of $$(x_1 + x_2 + \cdots + x_k)^n$$, apart from its multiplicity, has the form $$x_1^{p_1} x_2^{p_k} \cdots x_k^{p_k}$$ in which the exponents $$p_i\,\!$$ sum up to $$n\,$$. There is a one-to-one correspondence between the sequences $$p_1, p_k, \ldots, p_k\,\!$$ of non-negative integers with sum equal to $$n\,$$, and the combinations of $$k{-}1\,\!$$ different choices from the set $$\{1, 2, \ldots, n+k{-}1\}\,\!$$. The correspondence is given by the mapping that sends the sequence $$p_1, p_k, \ldots, p_k\,\!$$ to the set $$\{p_1+1, p_1{+}p_2+2, \ldots, p_1{+}p_2{+}\cdots{+}p_{k-1}+k{-}1\}\,\!$$.  --Lambiam Talk  15:09, 27 October 2006 (UTC)

"measureomorphisms"
Homeomorphisms are bijective & cts both ways. Has anything been done w/relations that're bijective outside of sets of measure zero and measurable both ways, say on [0,1]? Thanks,Rich 10:40, 26 October 2006 (UTC)


 * If you "forget" the topological structure, a homeomorphism is a function between the sets of the two spaces involved. What you write suggests that you are thinking of functions between power sets. Is that correct? For that to be a useful notion, you'd need a lot more in the way of restrictions. Did you have any particular function or application in mind? --Lambiam Talk  12:24, 26 October 2006 (UTC)
 * Measurable functions are perfectly well-defined set functions from one measure space to another. Except for the "sets of measure 0" part, you just have (I believe) an morphism-with-inverse in the category of measurable spaces. I don't know enough category theory to say whether that can be called an "isomorphism" in that category or not. I don't know whether this has been studied specifically in the category of measurable spaces. 128.135.72.214 04:23, 27 October 2006 (UTC)


 * So I'm not too sure what a "measurable space" (as opposed to a "measure space") is. Measurable functions don't naturally form a collection of morphisms, because they're not closed under composition. Maybe the condition you really want is that the preimage of a null set is null?  In that case, I think your "measuremorphism" would be a Rudin–Keisler isomorphism between the Boolean algebras created by modding out by null sets. --Trovatore 04:42, 27 October 2006 (UTC)


 * I wouldn't be surprised if the problems with composability went away in most interesting cases. Note that a bounded measurable function on [0,1] (in particular, your measureomorphism) defines an element of the C-star algebra L&infin;([0,1]), which is isomorphic to a continuous [0,1]-valued function on a space X.  I wouldn't be surprised if it turned out that your other requirements then forced that [0,1]-valued function to extend to a topological automorphism of X, and that simple case, at least, could be subsumed in the study of the spectrum of commutative C*-algebras.
 * Put slightly simpler, your auto-measureomorphisms of a measure space S might just be C*-automorphisms of the C*-algebra L&infin;(S), at least in some nice cases.
 * RandomP 20:23, 28 October 2006 (UTC)

Equation of midpoint
A line segment of length 6 moves so that A remain on the x-axis and B remains on the line y=2x. Find the equation of the locus of the midpoint M of segment AB.--Patchouli 10:53, 26 October 2006 (UTC)


 * Do your own homework: if you need help with a specific part or concept of your homework, feel free to ask, but please do not post entire homework questions and expect us to give you the answers. Letting someone else do your homework makes you learn nothing in the process, nor does it allow us Wikipedians to fulfill our mission of ensuring that every person on Earth, such as you, has access to the total sum of human knowledge. -- Meni Rosenfeld (talk) 11:05, 26 October 2006 (UTC)


 * If B has y = 0, where is A? What if y = 6? Find a general equation for A given B. From this it is straightforward to determine the behavior of the midpoint. --Pyroclastic 15:06, 26 October 2006 (UTC)
 * If B(0,0)→A(6,0)→M(3,0).
 * If B(3,6)→A(3,0)→M(3,3).

The pairs of M do not help me to find an equation. I can't use the 2 point to find the equation of a line. I need to be able to impose the constraint that the total distance between A and B always remains 6. That would yield a conic. I don't know how to proceed.--Patchouli 16:25, 26 October 2006 (UTC)


 * It cannot be a rotated parabola with A as the focus and y=2x as the directrix because AB is a (straight) line segment. Furthermore, the conic that I am interested in cannot extend indefintely in any direction forasmuch as AB would then exceed 6.  Therefore, it must be a rotated ellipse. --Patchouli 16:28, 26 October 2006 (UTC)


 * Let's call the coordinates of A (xA, yA), and the coordinates of B (xB, yB). I would start by finding equations for xB and yB in terms of xA. Chuck 17:10, 26 October 2006 (UTC)


 * I did find an equation. It was an interesting exercise. You should try it. :-)
 * As you surmise, the result is a rotated ellipse. It is centered at the origin, and can be given by an implicit algebraic equation whose coefficients are small integers (less than 10). The semimajor axis is three times the golden ratio.
 * This should be enough to help you check your work, and even mount a brute force assault.
 * Better still, an ellipse is determined by five points. You already have four: the two above plus their symmetric mates. It's easy to find another. If A is at the origin, where's B?
 * Or, consider the following line of attack.
 * Since A is on the x axis, it has coordinates (a,0). Let the y coordinate of the midpoint be y; then the y coordinate of B must be 2y, and its x coordinate is immediate. From the coordinates of A and B we can use Euclidean distance to write a quadratic equation in a and y constraining the (squared) length. Implicitly, a is a function of y.
 * See where these hints take you, and ask again with more details if you get stuck. --KSmrqT 18:16, 26 October 2006 (UTC)
 * the y coordinate of the midpoint be y; then the y coordinate of B must be 2y
 * You are assuming that AB is vertical, but it isn't always so.


 * Also, what are the x coordinates of B and M?--Patchouli 18:37, 26 October 2006 (UTC)


 * No, I am not assuming AB is vertical. Nor will I hand-feed you everything; I expect you to find the x coordinates of B and M yourself. And ethically, I expect you to inform your instructor that you received assistance. --KSmrqT 20:34, 26 October 2006 (UTC)

Author of the Kramer Theorem
Who is the the author of the Kramer Theorem? I have heard an opinion that it is the Gaon of Vilna (Rabbi Eliyahu Kramer). He lived at the end of the 17th century and is known to have mastered many areas of knowledge, such as algebra, geometry, etc.

Thank you

Olga Yorish
 * They had no electricity in Vilna at that time. -- DLL .. T 21:06, 26 October 2006 (UTC)


 * If you mean Cramer's rule, that would be Gabriel Cramer. Or perhaps you meant the Cramér-Wold theorem ?  In that case, the authors are Herman Ole Andreas Wold and Harald Cramér.  StuRat 00:55, 27 October 2006 (UTC)


 * I see there is also a Kramers' degeneracy theorem in the physics field. Is that the one you mean ?  It appears to be named after Hendrik Anthony Kramers. StuRat 03:59, 27 October 2006 (UTC)