Lösungen:
= k| k und 0 . 0! + ... + k . k! = (k + 1)! - 1 |
Sei
k . Dann ist:
0 . 0! + ... + k . k! | = | (k + 1)! - 1 | |
0 . 0! + ... + k . k! + (k + 1)(k + 1)! | = | (k + 1)! - 1 + (k + 1)(k + 1)! | |
0 . 0! + ... + k . k! + (k + 1)(k + 1)! | = | (k + 1)!(k + 2) - 1 | |
0 . 0! + ... + k . k! + (k + 1)(k + 1)! | = | (k + 2)! - 1 |
= k| 0 + 1 + ... + k = |
1 + ... + k | = | ||
1 + ... + k + (k + 1) | = | + (k + 1) = |
f (x + 1) = f (x) + (x + 1)2 und f (0) = 0. |
Es ist also
f (x) = a1x + a2x2 + a3x3. Also ist
f (x + 1) | = | a1(x + 1) + a2(x + 1)2 + a3(x + 1)3 | |
= | (a1 + a2 + a3) + (a1 +2a2 +3a3)x + (a2 +3a3)x2 + a3x3 |
f (x + 1) | = | f (x) + (x + 1)2 | |
= | a1x + a2x2 + a3x3 + x2 + 2x + 1 | ||
= | 1 + (a1 +2)x + (a2 +1)x2 + a3x3 |
|
Dies ergibt a3 = , a2 = und a1 = . Also ist f (x) = x3 + x2 + x. |
Wenn es daher ein Polynom mit der gewünschten Eigenschaft gibt, so muss es f
sein.
Tatsächlich hat f die verlangte Eigenschaft.
Denn es ist
f (x + 1) - f (x) | = | (x + 1)3 + (x + 1)2 + - f (x) | |
= | x2 +2x + 1 = (x + 1)2 |
f (x + 1) - f (x) - (x + 1)3 | = | 0 |
a1 + a2 + a3 + a4 | = | 1 | |
2a2 +3a3 +4a4 | = | 3 | |
3a3 +6a4 | = | 3 | |
4a4 | = | 1 |
f (x) = a1x1 + a2x2 + a3x3 + a4x4 + a5x5 |
a1(x + 1) + a2(x + 1)2 + a3(x + 1)3 + a4(x + 1)4 + a5(x + 1)5 | = | ||
ax + a2x2 + a3x3 + a4x4 + a5x5 + x4 +4x3 +6x2 + 4x + 1 |
a1 + a2 + a3 + a4 + a5 | = | 1 | |
2a2 +3a3 +4a4 +5a5 | = | 4 | |
3a3 +6a4 +10a5 | = | 6 | |
4a4 +10a5 | = | 4 | |
5a5 | = | 1 |
f (x) = x + x3 + x4 + x5 |