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

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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

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

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 (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

${\mathbf {N} }^{\pm }:={\frac {\pm (f_{x},f_{y},f_{z})}{\sqrt {f_{x}^{2}+f_{y}^{2}+f_{z}^{2}}}}\ .$

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