Wikipedia:Reference desk/Archives/Mathematics/2009 August 19

= August 19 =

Is this polynomial most extreme?
The polynomial

$$ P(x):=x^{32}+x^{31}+x^{30}+x^{29}+x^{28}+x^{26}+x^{24}+x^{22}+x^{21}+x^{20}+x^{19}+x^{16} +x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^4+1$$

takes on prime or almost-prime (product of two primes) values for x=1*,2*,7*,16,29,30,31,32,33,36,37*,..., where it is prime for the numbers marked (*). Is it likely that there is another polynomial with coefficients in {0,1} such that over half of the values of x for which it is an almost-prime or prime up to some point at least as high as x=33 occur in a string at least five long ending at that point? Julzes (talk) 06:57, 19 August 2009 (UTC)

I took the liberty of formatting your formula with LaTeX --pma (talk) 07:09, 19 August 2009 (UTC)

Note that I suspect that "five" could probably be replaced with "two" and the value 33 reduced also. Julzes (talk) 09:33, 19 August 2009 (UTC)

?Julzes (talk) 04:00, 29 August 2009 (UTC)

Vector perpendicular to plane
So given the equation of a plane. Say... 3x+y-2z=10 How would you find the unit vector orthogonal to it? I ask because I'm not very familiar with vector algebra and this would help me finish a few proofs.--Yanwen (talk) 20:44, 19 August 2009 (UTC)


 * See Surface normal. — JAO • T • C 21:47, 19 August 2009 (UTC)


 * (ec) The vector v = (3, 1, -2) is a normal vector for the plane $$P = \{(x, y, z)|3x+y-2z=10\}$$. This is why:  suppose $$p_1 = (a_1, b_1, c_1)$$ and $$p_2 = (a_2, b_2, c_2)$$ are two points on the plane P;  thus we know $$3a_1 + b_1 - 2c_1 = 3a_2 + b_2 - 2c_2 = 10$$.  Then the vector connecting $$p_1, p_2$$ is $$p_2 - p_1 = (a_2 - a_1, b_2 - b_1, c_2 - c_1)$$, and the dot product of $$p_2 - p_1$$ with v is
 * $$v \cdot (p_2 - p_1) = 3(a_2 - a_1) + 1(b_2 - b_1) - 2(c_2 - c_1) $$
 * $$= (3a_2 + b_2 - 2c_2) - (3a_1 + b_1 - 2c_1) = 10 - 10 = 0 \,$$,
 * so v and $$p_2 - p_1$$ are orthogonal. Therefore v is a normal vector of the plane P.  To get a unit normal vector, just divide v by its length.  Eric.  216.27.191.178 (talk) 21:52, 19 August 2009 (UTC)

Let's say the plane has equation $$ax+by+cz=d$$ for some, not all zero, real numbers a, b, c, and d. A normal vector would be (a,b,c), and so the two unit normal vectors are
 * $${\mathbf N}^{\pm} := \frac{\pm(a,b,c)}{\sqrt{a^2+b^2+c^2}} \ . $$

In fact, if you have some surface S given by an equation f = 0, (where f is a smooth function). So
 * $$ S = \{ (x,y,z) \in \mathbb{R}^3 : f(x,y,z) = 0\} \, $$

then a normal vector to S is given by (fx,fy,fz) where fx, fy and fz are the partial derivatives of f with respect to x, y and z respectively. The surface S will be singular at a point $$(x_0,y_0,z_0) \in S$$ when
 * $$f_x(x_0,y_0,z_0) = f_y(x_0,y_0,z_0) = f_z(x_0,y_0,z_0) = 0 \ . $$

Assuming that not all three partial derivatives are zero we have two unit normals given by
 * $${\mathbf N}^{\pm} := \frac{\pm(f_x,f_y,f_z)}{\sqrt{f_x^2+f_y^2+f_z^2}} \ . $$

In the plane example f(x,y,z) = ax + by + cz - d, and so fx = a, fy = b and fz = c. In your example a = 3, b = 1 and c = -2 so
 * $$ {\mathbf N}^{\pm} = \pm \left(\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}} \right) . $$

Dr Dec ( Talk )    15:56, 20 August 2009 (UTC)


 * So for a surface $$ S = \{ (x,y,z) \in \mathbb{R}^3 : x^2-y^2+z^2 = 18\} \, $$
 * Would the normal vector at (3,4,5) be $${\mathbf N}^{\pm} = \frac{\pm(6,-8,10)}{10\sqrt{2}} \ . $$--Yanwen (talk) 21:57, 20 August 2009 (UTC)
 * That looks right to me. Eric.  98.207.86.2 (talk) 06:31, 21 August 2009 (UTC)

Thanks, All!--Yanwen (talk) 12:46, 21 August 2009 (UTC)

No, that's not right! First of all $${\mathbf N}^{\pm}$$ are two vectors: there's a choice of sign. Both $${\mathbf N}^{-}$$ and $${\mathbf N}^{+}$$ are unit normal vectors. They are both unit length and they are both perpendicular to the plane. Notice that $${\mathbf N}^{-} = -{\mathbf N}^{+}$$. The expression
 * $$ {\mathbf N}^{\pm} = \pm \left(\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}} \right) $$

gives two choices of unit normal vector at each point of the plane. Think about the plane z = 0; both (0,0,1) and (0,0,−1) are unit normal vectors. There is always a choice of two. Notice that $$ {\mathbf N}^{\pm}$$ is independent of x, y and z. So if a vector is a unit normal at one point of the plane then it will be normal vector at every point of the plane. So, the two choices of unit normal vector would be
 * $$ {\mathbf N}^{-} = \left(\frac{-3}{\sqrt{14}},\frac{-1}{\sqrt{14}},\frac{2}{\sqrt{14}} \right), $$
 * $$ {\mathbf N}^{+} = \left(\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}} \right) . $$

I'm not sure why you've re-written the vectors with $$10\sqrt{2}$$ in the denominator. It looks more complicated. The simpler form would, IMHO, be the one already given. I hope this helps... Dr Dec ( Talk )    22:30, 21 August 2009 (UTC)


 * I'm not sure exactly what you think is not right here, but you do realize that Yanwen's $$\scriptstyle \frac{\pm(6,-8,10)}{10\sqrt{2}}$$ vectors were explicitly supposed to be normals of the surface $$\scriptstyle x^2-y^2+z^2 = 18$$, not of the plane in the original question, right? — JAO • T • C 00:02, 22 August 2009 (UTC)


 * What isn't right is that he said the unit normal vector is $$\scriptstyle \frac{\pm(6,-8,10)}{10\sqrt{2}}$$. Once you make a choice of sign you get a unit normal vector. The expression $$\scriptstyle \frac{\pm(6,-8,10)}{10\sqrt{2}}$$ gives two unit normal vectors. He seems to have misunderstood my notation and has taken $${\mathbf N}^{\pm}$$ to be a single vector, which it is not: it is a pair of vectors differing by sign. I thought I made that quite clear in my last post.  Dr Dec  ( Talk )    10:32, 22 August 2009 (UTC)


 * Ah, I see now that Yanwen wrote "vector" in place of "vectors". I think it's a fair bet that he already understood that there are two unit normal vectors, but I see your point. — JAO • T • C 11:41, 22 August 2009 (UTC)