I’ve been trying to prove this inequality:
___________
θ > √ 2(1-cos θ)
for 0 < θ <= π; θ being expressed in radians
for a couple of days. I’ve done it empirically, but that doesn’t count in my mind.
I’ve been trying to prove this inequality:
___________
θ > √ 2(1-cos θ)
for 0 < θ <= π; θ being expressed in radians
for a couple of days. I’ve done it empirically, but that doesn’t count in my mind.
I’m having a hard time reading that; it may just be that my browser isn’t displaying the characters properly. I’m seeing an “a”, an “s”, both with some kind of carat thing over them. And I’m seeing 0 < I (with carats) <= I(and some character I don’t recognize at all). What are these things?
probably. I’ll restate my statement by using standard characters.
_______________
theta > √ 2(1-cos theta)
for 0 < theta <= pi; theta being expressed in radians
That definitely helps, I’m still not sure what “a” and “s” are, though; I’m seeing
theta > a^s 2(1 - cos theta).
AWB, you must define parameters for a & s. If, for example, a = 2 and s = 3, then, for theta = pi/2, the inequality does not maintain validity.
sigh… remind me not to use those special characters anymore:
Using computer formula format:
theta > sqrt(2(1-cos(theta)))
0 < theta <= pi
*
Better?
BTW, “sqrt” is the square root function, if you didn’t know.
This a direct result of the Maclaurin series, which states that:
Cos(A) = 1 - A[sup]2[/sup]/2! + A[sup]4[/sup]/4! - A[sup]6[/sup]/6! ···
A is in radians. It’s not terribly difficult to show that for the range you specified, the sum of all the terms after A[sup]2[/sup]/2! is positive, so you get:
Cos(A) < 1 - A[sup]2[/sup]/2!
And after that, all you need is a little Algebra. Maybe someone else has a better solution, that doesn’t require Calculus.