Pebu

Math is easy. What you need to do is find the deflection in a beam made out of some material with the end fixed in such a position with supports here, here, and here. In general, find out where that beam will fail first... Oh, throw in some cyclic loading while you're at it...

star882

Quote:

But then again, I'm an English major, and I am done with Math for the rest of my life.

Well, English is just another form of math. When you combine letters to make words, combine words to make sentences, and combine sentences to make paragraphs, you just concatenated some numbers. See the Java program in my sig to see an example! In that program, the "+" means concatenate, not add. (When I write concate on paper as a symbol, I use the "+" sign enclosed in a circle to make sure that it doesn't get confused with add.)

__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

TCPA would take your freedom! Say NO!
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

HDTV the way it should be:
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

Originally Posted by A friend of mine who has a Linux kernel named after his girlfriend.

If I was VirtualBox, I could load my virtualization module into Hannah and boot up another kernel in the same address space.

Driftless

Originally Posted by Pebu

Oh, throw in some cyclic loading while you're at it...

you had to do it.......

ericcumbee

Pi the Answer is Pi

jmckeown

If you just want advice, then first try proving that at the maximum possible
value of yx^2 + zy^2 + xz^2, one of x, y, and z can be 0. (This is before
proving what the maximum possible value is). If you want what I think is the
full solution, then read on.

I've used != to mean "is not equal to".

Because of the cyclic symmetry, we can assume that the magnitude of y is between
that of x and z. That is, either x <= y <= z or x >= y >= z. Because of this:

(y - x)(y - z) <= 0 (with equality if and only if y = x or y = z)
y^2 + xz - xy - yz <= 0
y^2 + xz <= xy + yz
y^2 + xz <= 2xy + yz (with equality if and only if (y = x or y = z) and
(x = 0 or y = 0))
zy^2 + xz^2 <= y2xz + yz^2 (with equality if and only if ((y = x or y = z) and
(x = 0 or y = 0)) or z = 0)
yx^2 + zy^2 + xz^2 <= yx^2 + y2xz + yz^2
yx^2 + zy^2 + xz^2 <= y(x + z)^2

But by the AM-GM inequality:
(y((x + z)/2)((x + z)/2))^(1/3) <= (y + (x + z)/2 + (x + z)/2)/3 (with equality
if and only if y = (x + z)/2, which is equivalent to y = 1/3)
((y(x + z)^2)/4)^(1/3) <= (y + x + z)/3 = 1/3
(y(x + z)^2)/4 <= 1/27
y(x + z)^2 <= 4/27
yx^2 + zy^2 + xz^2 <= 4/27

with equality if and only if y = 1/3 and (((y = x or y = z) and x = 0) or z =
0).
But if x = 0 then y != x and because x + y + z = 1, y != z, so the conditions
equivalent to equality simplify to:
y = 1/3 and z = 0,
which implies x = 2/3.
Because we assumed that the magnitude of y was between that of x and z, we need
to reconstruct the other solutions, where x was the middle number, and where z
was the middle number. These are:
x = 1/3; y = 0; z = 2/3
x = 0; y = 2/3; z = 1/3
by the cyclic symmetry.