Math Is Fun Forum

ω - lowercase omega
at least that's how I've seen it

I couldn't get it to work and my TI-89 Titanium couldn't either and neither could the wolfram Integrator.

See the problem, to be more specific, is that I can't figure out how to integrate ƒ(x)... esp. the x^x

The length (aka. magnitude) of the vector is still it's absolute value.

So for example, if you had a vector <1+i, i-1>, the absolute value of your x component is sqrt(2), the absolute value of your y component is sqrt(2), and so the length of the vector is 2.

So it exists in a complex 4-D coordinate system but only exists at the origin on a 2-D real system where the origin of the vector is (0,0).

That works, yes, but if r is the principle variable, then 2pi r is a line and pi r^2 is a parabola, so you aren't dealing with a circle, are you?

I would say it is not a polynomial. Apart from the fact that you cannot iterate a polynomial infinitely many times, consider this:

ƒ(x) = x^2

ƒa(0) will always be 0, and ƒa(1) and ƒa(-1) will always be 1. For all x not equal to 0 such that |x| < 1, ƒa(x) approaches 0 as a approaches ∞, and for all other numbers, ƒa(x) tends to ∞ as a tends to ∞ - that is, it does not exist. So what you're left with in a limiting case is a discontinuous (and nonexistent) function defined only for the domain [-1, 1].

1. General form of a parabola equation is: (y-k)=a*(x-h)^2 where (h,k) is the vertex and a is a scale factor.

That gives y+1 = a*(x+4)^2

Since (0,-5) works for the equation, substitute and solve for a:

-5 + 1 = a*(0+4)^2
-4 = a*16
a = -1/4

so your equation is:

y + 1 = -1/4 * (x+4)^2

2. Let x be one number and (x-10) be the other number...
    So you want to minimize x*(x-10), which is x^2 - 10*x. The lowest point is at the vertex, so the       
    x coordinate of the vertex is one number, and your other number is x-10.
    Vertex occurs at -b/2a = 10/2 = 5, so the two numbers are 5 and -5

Yeah I tried logs and changes of bases and etc but it just simplifies to the original equation.

The only answers are x = 1 and x = 2, which can be proved once you do "find" them...

I really didn't get what you were trying to do with that recursive equation or else I'd help

Genius!

I like cheese