函數8題之(2)

1) f(x) = ax² + bx + cf(m) - f(n)
= a(m² - n²) + b(m - n)
= (m - n) ( a(m+n) + b )
∴
f(x+3) - 3f(x+2) + 3f(x+1) - f(x)
= f(x+3) - f(x) + 3( f(x+1) - f(x+2) )
= 3 ( a(2x+3) + b) + 3 (-1) ( a(2x+3) + b)
= 0
2) a² f(n+2) = b² f(n)a² f(1+2) = b² f(1)
a² f(3) = 0
因 a > 0 ,
故 f(3) = 0
再以 3 代入 n 得 f(5) = 0 ,
再以 5 代入 n 得 f(7) = 0 ,
........................................故當 n 為奇數時 f(n) = 0。a² f(2+2) = b² f(2)
a² f(4) = b²
f(4) = b² / a²
再以 4 代入 n 得 a² f(6) = b² (b² / a²)
f(6) = b⁴/ a⁴ ,
........................................故當 n 為偶數時 f(n) = bⁿ⁻² / aⁿ⁻²。綜合得 f(n) = (1/2)( (b/a)ⁿ + (- b/a)ⁿ ) (a / b)²
3) f(2 + 1/x) = x² - 3x + 7f(2 + x) = (1/x)² - 3(1/x) + 7f(x) = 1 / (x-2)² - 3 / (x-2) + 7 = (7x² - 31x + 35) / (x - 2)²
4) f(x+1) - f(x) = 8x + 5a(x+1)² + b(x+1) - 3 - (ax² + bx - 3) = 8x + 5ax² + 2ax + a + bx + b - 3 - (ax² + bx - 3) = 8x + 52ax + a + b = 8x + 5比較係數得
2a = 8
{
a + b = 5解得 a = 4 , b = 1。