10
[tex] f(x) = \dfrac{x-1}{x} = 1 - \dfrac{1}{x} [/tex]
[tex]f(f(x)) =1 - \dfrac{1}{1 - \frac{1}{x}} = 1 - \dfrac{x}{x-1} = \dfrac{x-1-x}{x-1}=\dfrac{1}{1-x} [/tex]
[tex]f(f(f(x)) = 1 - \dfrac{1}{\frac{1}{1 - x}} = x [/tex]
[tex]f^4(x) = 1-\frac{1}{x} =f(x) [/tex]
[tex]f^{3k}(x)=x [/tex]
[tex]2017 = 3(672)+1 [/tex]
[tex]f^{2017}(x) = f(f^{3(672)}(x))=f(x)=\dfrac{x-1}{x} [/tex]
11
f(x)=f(x+1)+f(x-1)
f(20)=15
f(15) - 5 = 15
f(15) = 20
f(16)=f(15)+f(17)=20 +f(17)
f(17)=f(16)+f(18)= 20+f(17)+f(18)
f(18)=-20
f(18)=f(17)+f(19)=-20
f(19)=f(20)+f(18)=15 + f(18)=15 + -20 = -5
f(19) = -5
-20 = f(17) + -5
f(17) = -15
f(16) = 20 + -15= 5
f(16) = 5
Check:
f(15)=20, f(16)=5, f(17)=-15, f(18)=-20, f(19)=-5, f(20)=15
Each has to be the sum of the neighbors.
5=20+-15, -15=5+-20, -20=-15+-5, -5=-20+15, all good
f(x)=f(x+1)+f(x-1)
f(x+1)=f(x)+f(x+2)
f(x)=f(x)+f(x+2)+f(x-1)
f(x+2)=-f(x-1)
f(x+3)=-f(x)
f(x)=g[x mod 6]
where g is the the table
g(0)=-20, g(1)=-5, g(2)=15, g(3)=20, g(4)=5, g(5)=-15
20152015 = 1 mod 6
f(20152015) = g(1) = -5
12
f(0 - 0) + f(0 + 0) = (1/2) f(2(0)) + (1/2) f(2(0))
2f(0) = f(0)
f(0) = 0
f(1) = 1
f(1 - 0) + f(1 + 0) = (1/2) f(2(1)) + (1/2) f(2(0))
2f(1) = (1/2) f(2)
f(2) = 4 f(1)
f(2) = 4
f(2-0) + f(2+0) = (1/2) f(4) + (1/2) f(0)
8 = (1/2) f(4)
f(4) = 16
f(2-1) + f(2+1) = (1/2) f(4) + (1/2) f(2)
f(1) + f(3) = (1/2) f(4) + (1/2) f(2)
f(3) = (1/2) f(4) + (1/2) f(2) - f(1)
f(3) = (1/2) 16 + (1/2) 4 - 1 = 8 + 2 - 1 = 9
f(3) = 9
13
P(x) = x^2 + 2007 x + 1
Let x=P(x)
x = x^2 + 2007 x + 1
0 = x^2 + 2006 x + 1
[tex]x = -1003 \pm \sqrt{1003^2-1} = -1003 \pm 2 \sqrt{251502}[/tex]
Two real fixed points. P(x)=x
Given any x such that P(x)=x, then P(P(x))=x, and it's still x after composing P n times.
10) The only thing I can provide to help you is:
f°ⁿ(x) = f(f°ⁿ⁻¹(x))
Hopefully, that can help you to figure out the answer
*********************************************************************************
11) Answer: f(15) = 20, f(16) = 5, f(17) = -15, f(18) = -20, f(19) = -5
Explanation:
f(x) = f(x + 1) + f(x - 1) f(20) = 15
f(15) - 5 = 15 ⇒ f(15) = 20
f(16) = f(16 + 1) + f(16 - 1)
f(16) = f(17) + f(15)
f(16) = f(17) + 20
f(17) = f(17 + 1) + f(17 - 1)
f(17) = f(18) + f(16)
f(17) = f(18) + f(17) + 20
0 = f(18) + 20
-20 = f(18)
f(19) = f(19 + 1) + f(19 - 1)
f(19) = f(20) + f(18)
f(19) = 15 + -20
f(19) = -5
f(18) = f(18 + 1) + f(18 - 1)
f(18) = f(19) + f(17)
-20 = -5 + f(17)
-15 = f(17)
f(16) = f(17) + 20
f(16) = -15 + 20
f(16) = 5
*****************************************************************************************
12a) Answer: 0
Explanation:
f(m - n) + f(m + n) = [tex]\frac{1}{2}[/tex]f(2m) + [tex]\frac{1}{2}[/tex]f(2n)
f(0) + f(2m) = [tex]\frac{1}{2}[/tex]f(2(m)) + [tex]\frac{1}{2}[/tex]f(2(m))
f(0) + f(2m) = f(2m)
f(0) = 0
*****************************************************************************************
12b) Answer: f(2) = 4, f(3) = [tex]\frac{1}{2}[/tex]f(4) + 1
Explanation:
f(m - n) + f(m + n) = [tex]\frac{1}{2}[/tex]f(2m) + [tex]\frac{1}{2}[/tex]f(2n)
Let m = 1 and n = 0, then
f(1 - 0) + f(1 + 0) = [tex]\frac{1}{2}[/tex]f(2(1)) + [tex]\frac{1}{2}[/tex]f(2(0))
f(1) + f(1) = [tex]\frac{1}{2}[/tex]f(2) + [tex]\frac{1}{2}[/tex]f(0)
2f(1) = [tex]\frac{1}{2}[/tex]f(2) + [tex]\frac{1}{2}[/tex](0)
2f(1) = [tex]\frac{1}{2}[/tex]f(2)
2[1] = [tex]\frac{1}{2}[/tex]f(2)
(2)2 = (2)[tex]\frac{1}{2}[/tex]f(2)
4 = f(2)
Let m = 2 and n = 1, then
f(2 - 1) + f(2 + 1) = [tex]\frac{1}{2}[/tex]f(2(2)) + [tex]\frac{1}{2}[/tex]f(2(1))
f(1) + f(3) = [tex]\frac{1}{2}[/tex]f(4) + [tex]\frac{1}{2}[/tex]f(2)
1 + f(3) = [tex]\frac{1}{2}[/tex]f(4) + [tex]\frac{1}{2}[/tex](4)
1 + f(3) = [tex]\frac{1}{2}[/tex]f(4) + 2
f(3) = [tex]\frac{1}{2}[/tex]f(4) + 1
