Determine compositions of functions

[Cyborg31]

Homework Statement

Let h: Z -> Z be defined as follows:

h(n) = 0 if n is even and 1 if n is odd

Determine h^2, h^3, and h^500

Homework Equations

The Attempt at a Solution

If n is even, h(n) = 0 then h(h(n)) = h(0) = 0
If n is odd, h(n) = 1 then h(h(n)) = h(1) = 1

If n is even h(n) = 0 then h(h(h(n))) = h(h(0)) -> h(0) = 0
If n is odd h(n) = 1 then h(h(h(n))) = h(h(1)) -> h(1) = 1

Then for h^500, if n is even then the result is 0 and n is odd then the result is 1.

Can someone tell if that's correct?

Thanks.

 

[tiny-tim]

Hi Cyborg31!

Yes, that's right!

Why, were you worried that it wasn't?

(Of course, you could also write the result: h^2 = h, h^500 = h. )

 

[sinjan.j]

Seem correct, though unnecesarily long.

Already after

If n is even, h(n) = 0 then h(h(n)) = h(0) = 0
If n is odd, h(n) = 1 then h(h(n)) = h(1) = 1

you can conclude h^2 = h and therefore h^k = h for all k >= 1.

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[Cyborg31]

1. Let f, g: R -> R where f(x) = ax + b and g(x) = x^2 - x + 1. If (g o f)(x) = 9x^2 - 9x + 3, determine the values of a and b.

So I got g(ax + b) = (ax + b)^2 - (ax + b) + 1

(a^2*x^2 + 2(a * b)x + b^2) - (ax + b) + 1 = 9x^2 - 9x + 3
(9x^2 + 2(3 * b)x + b^2) - (3x + b) + 1
(9x^2 + 2(3 * 2)x + 4) - (3x + 2) + 1
9x^2 + 12x + 4 - 3x - 2 + 1
9x^2 + 9x + 3

I know a = 3 and b = 2 but I'm not sure if those steps are correct. I don't think I've shown a way as to how I got the 2 for b. I got the a cause of the coefficient of the square. Anyone know if those steps are correct?

2. Let f, g: R -> R where f(x) = ax + b and g(x) = cx + d for any x element of R where a, b, c, and d are real constants. What relationships must be satisfied by the four constants for the equality: (f o g)(x) = (g o f)(x) for all x element of R?

(f o g)(x) = f(g(x)) = f(cx + d) = a(cx + d) + b
(g o f)(x) = g(f(x)) = g(ax + b) = c(ax + b) + d

a(cx + d) + b = c(ax + b) + d

I got a = c and b = d so that the coefficient and the constant matches so that they're equal. But I think I'm missing some steps of work.

Thanks.

 

[Cyborg31]

Actually nm, I figured it out.