Jump to content

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

From Wikipedia, the free encyclopedia
Mathematics desk
< August 18 << Jul | August | Sep >> Current desk >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


August 19

[edit]

Is this polynomial most extreme?

[edit]

The polynomial

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)[reply]

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

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)[reply]

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

Vector perpendicular to plane

[edit]

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)[reply]

See Surface normal#Calculating a surface normal. —JAOTC 21:47, 19 August 2009 (UTC)[reply]
(ec) The vector v = (3, 1, -2) is a normal vector for the plane . This is why: suppose and are two points on the plane P; thus we know . Then the vector connecting is , and the dot product of with v is
,
so v and 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)[reply]

Let's say the plane has equation 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

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

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 when

Assuming that not all three partial derivatives are zero we have two unit normals given by

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

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

So for a surface
Would the normal vector at (3,4,5) be --Yanwen (talk) 21:57, 20 August 2009 (UTC)[reply]
That looks right to me. Eric. 98.207.86.2 (talk) 06:31, 21 August 2009 (UTC)[reply]

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

No, that's not right! First of all are two vectors: there's a choice of sign. Both and are unit normal vectors. They are both unit length and they are both perpendicular to the plane. Notice that . The expression

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 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

I'm not sure why you've re-written the vectors with 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)[reply]

I'm not sure exactly what you think is not right here, but you do realize that Yanwen's vectors were explicitly supposed to be normals of the surface , not of the plane in the original question, right? —JAOTC 00:02, 22 August 2009 (UTC)[reply]
What isn't right is that he said the unit normal vector is . Once you make a choice of sign you get a unit normal vector. The expression gives two unit normal vectors. He seems to have misunderstood my notation and has taken 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)[reply]
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. —JAOTC 11:41, 22 August 2009 (UTC)[reply]