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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#Calculating a 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

${𝐍}^{\pm }:=\frac{\pm (a,b,c)}{\sqrt{a^2+b^2+c^2}}.$

In fact, if you have some surface S given by an equation f = 0, (where f is a smooth function). So

$S=\{(x,y,z)\in {ℝ}^3:f(x,y,z)=0\},$

then a normal vector to S is given by (f_x,f_y,f_z) where f_x, f_y and f_z 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

${𝐍}^{\pm }:=\frac{\pm (f_x,f_y,f_z)}{\sqrt{f_x^2+f_y^2+f_z^2}}.$

In the plane example f(x,y,z) = ax + by + cz - d, and so f_x = a, f_y = b and f_z = c. In your example a = 3, b = 1 and c = -2 so

${𝐍}^{\pm }=\pm (\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}}).$

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

So for a surface $S=\{(x,y,z)\in {ℝ}^3:x^2-y^2+z^2=18\},$

Would the normal vector at (3,4,5) be ${𝐍}^{\pm }=\frac{\pm (6,-8,10)}{10\sqrt{2}}.$--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 ${𝐍}^{\pm }$ are two vectors: there's a choice of sign. Both ${𝐍}^{-}$ and ${𝐍}^{+}$ are unit normal vectors. They are both unit length and they are both perpendicular to the plane. Notice that ${𝐍}^{-}=-{𝐍}^{+}$. The expression

${𝐍}^{\pm }=\pm (\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}})$

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 ${𝐍}^{\pm }$ 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

${𝐍}^{-}=(\frac{-3}{\sqrt{14}},\frac{-1}{\sqrt{14}},\frac{2}{\sqrt{14}}),$

${𝐍}^{+}=(\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}}).$

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 $\frac{\pm (6,-8,10)}{10\sqrt{2}}$ vectors were explicitly supposed to be normals of the surface $x^2-y^2+z^2=18$, 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 $\frac{\pm (6,-8,10)}{10\sqrt{2}}$. Once you make a choice of sign you get a unit normal vector. The expression $\frac{\pm (6,-8,10)}{10\sqrt{2}}$ gives two unit normal vectors. He seems to have misunderstood my notation and has taken ${𝐍}^{\pm }$ 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)